Proof. \(\eqref{el:he1} \Rightarrow \eqref{el:he2} \Rightarrow \eqref{el:he3} \Rightarrow \eqref{el:he1}\). The \(\eqref{el:he2} \Rightarrow \eqref{el:he3}\) step follows from Lemma 20. ◻

Proof. Let \(e_1, e_2, \ldots , e_k, f_1, f_2, \ldots , f_k\) be a basis of \(M\) given by Lemma 20, and let \(d_i = \lambda(f_i,f_i)\) (so \(d_1, d_2, \ldots , d_k\) are the diagonal entries of \(D\)). Let \(f'_i = d_ie_i - f_i\), and define \(J : M \rightarrow M\) by setting \(J(e_i)=e_i\) and \(J(f_i)=f'_i\). ◻

Proof. Let \(e_1, e_2, \ldots , e_k, f_1, f_2, \ldots , f_k\) be a basis of \(M\) given by Lemma 20, and let \(d_i = \lambda(f_i,f_i)\). We will also use the notation \(e_i\), \(f_i\) for the basis of the first component in \(M \oplus M\) and \(M \oplus \mathbb{Z}^{2k}\), and the basis of the second component of \(M \oplus M\) will be denoted by \(\bar{e}_i\) and \(\bar{f}_i\). Let \(a_1, a_2, \ldots , a_k, b_1, b_2, \ldots , b_k\) denote the standard basis of \(\mathbb{Z}^{2k}\). With this notation let \(I(e_i)=e_i+b_i\), \(I(f_i)=f_i+d_ib_i\), \(I(\bar{e}_i)=-b_i\) and \(I(\bar{f}_i)=f_i-a_i\). ◻

Proof. If an isomorphism \(I\) exists, it induces a commutative diagram \[\xymatrix@=10pt{ L^* & & M^* \ar[ll] & & M \ar[ll]_-{\mathop{\mathrm{ad}}\nolimits\lambda} \ar[rr]^-{\mu} \ar[dl] \ar[dd]^(.6){I} |!{[dl];[rr]}\hole & & Q \ar@{=}[dd] \\ & & & M/L \ar[ulll]^(.65){\cong} \ar[urrr] \ar[dd] & & & \\ (L')^* \ar[uu]^(.4){I_0^*} & & (M')^* \ar[ll] \ar[uu]^(.4){I^*} |!{[ur];[uull]}\hole & & M' \ar[ll]_(.25){\mathop{\mathrm{ad}}\nolimits\lambda'} |!{[ul];[dl]}\hole \ar[rr]^-{\mu'} \ar[dl] & & Q \\ & & & M'/L' \ar[ulll]^-{\cong} \ar[urrr] & & & }\] hence \(\hat{\mu}'_{L'} = \hat{\mu}_L \circ I_0^*\).

For the other direction suppose that \(\hat{\mu}'_{L'} = \hat{\mu}_L \circ I_0^* : (L')^* \rightarrow Q\). Since \(L \leq M\) and \(L' \leq M'\) are direct summands, they have direct complements \(N \leq M\) and \(N' \leq M'\). By Lemma 19 \(N \cong L^*\), and similarly \(N' \cong (L')^*\). Hence \(I_0^*\) can be regarded as an isomorphism \(I_0^* : N' \rightarrow N\), and the condition means that \(\mu' \big| _{N'} = \mu \big| _N \circ I_0^* : N' \rightarrow Q\).

Let \(r = \mathop{\mathrm{rk}}\nolimits N (= \mathop{\mathrm{rk}}\nolimits N' = \mathop{\mathrm{rk}}\nolimits L = \mathop{\mathrm{rk}}\nolimits L')\). Let \(f'_1, f'_2, \ldots , f'_r\) be a basis of \(N'\), and let \(f_i = I_0^*(f'_i)\). The basis \(f_1, f_2, \ldots , f_r\) of \(N \cong L^*\) determines a dual basis \(e_1, e_2, \ldots , e_r\) of \(L\) (satisfying \(\lambda(e_i,f_j)=0\) if \(i \neq j\) and \(\lambda(e_i,f_i)=1\)). Similarly, \(L'\) has a basis \(e'_1, e'_2, \ldots , e'_r\) which is dual to the basis \(f'_1, f'_2, \ldots , f'_r\) of \(N'\).

The isomorphism \(N \cong L^*\) is given by the composition \(N \rightarrow M \rightarrow M^* \rightarrow L^*\), hence \(f_j \in N\) corresponds to \(\mathop{\mathrm{ad}}\nolimits\lambda(f_j) \big| _L \in L^*\). So we have \(\lambda'(f'_j,I_0(e_i)) = (\mathop{\mathrm{ad}}\nolimits\lambda'(f'_j) \big| _{L'})(I_0(e_i)) = (I_0^*(\mathop{\mathrm{ad}}\nolimits\lambda'(f'_j) \big| _{L'}))(e_i) = (\mathop{\mathrm{ad}}\nolimits\lambda(f_j) \big| _L)(e_i) = \lambda(f_j,e_i) = \lambda'(f'_j,e'_i)\) for every \(i,j\) (using the definition of \(f_j\) and the dual bases). Since \(N' \cong (L')^*\), this implies that \(I_0(e_i)=e'_i\) for every \(i\).

Next we define the elements \(\bar{f}_1, \bar{f}_2, \ldots , \bar{f}_r \in M\) recursively. Let \(\bar{f}_1 = f_1\), and let \(\bar{f}_{i+1} = f_{i+1} - \sum_{j=1}^i \lambda(\bar{f}_j,f_{i+1})e_j - \left\lfloor \frac{\lambda(f_{i+1},f_{i+1})}{2} \right\rfloor e_{i+1}\). By induction we see that \(e_1, e_2, \ldots , e_r, \bar{f}_1, \bar{f}_2, \ldots , \bar{f}_i\) is a basis of \(L \oplus \left< f_1, f_2, \ldots , f_i \right>\), therefore \(e_1, e_2, \ldots , e_r, \bar{f}_1, \bar{f}_2, \ldots , \bar{f}_r\) is a basis of \(M\). By construction \(\lambda(e_i,\bar{f}_j)=\lambda(e_i,f_j)\) for every \(i,j\), \(\lambda(\bar{f}_i,\bar{f}_j)=0\) if \(i \neq j\), and \(\lambda(\bar{f}_i,\bar{f}_i)=0\) if \(\lambda(f_i,f_i)\) is even and \(\lambda(\bar{f}_i,\bar{f}_i)=1\) if \(\lambda(f_i,f_i)\) is odd. Furthermore \(\mu(\bar{f}_i)=\mu(f_i)\). We define the elements \(\bar{f}'_1, \bar{f}'_2, \ldots , \bar{f}'_r \in M'\) analogously.

Let \(I : M \rightarrow M'\) be the homomorphism given by \(I(e_i) = e'_i\) and \(I(\bar{f}_i) = \bar{f}'_i\). This is an isomorphism of abelian groups, and \(I \big| _L = I_0\). We need to check that \(I\) is a morphism of extended quadratic forms. For every \(i\) we have \(\mu(e_i) = 0 = \mu'(e'_i)\) and \(\mu(\bar{f}_i) = \mu(f_i) = \mu \circ I_0^*(f'_i) = \mu'(f'_i) = \mu'(\bar{f}'_i)\), therefore \(\mu = \mu' \circ I\). For every \(i,j\) \(\lambda(e_i,e_j) = 0 = \lambda'(e'_i,e'_j)\). If \(i \neq j\), then \(\lambda(e_i,\bar{f}_j) = 0 = \lambda'(e'_i,\bar{f}'_j)\) and \(\lambda(\bar{f}_i,\bar{f}_j) = 0 = \lambda'(\bar{f}'_i,\bar{f}'_j)\). For every \(i\) we have \(\lambda(e_i,\bar{f}_i) = 1 = \lambda'(e'_i,\bar{f}'_i)\). Finally, both \(\lambda(\bar{f}_i,\bar{f}_i)\) and \(\lambda'(\bar{f}'_i,\bar{f}'_i)\) are either \(0\) or \(1\). And since \(\underline{\tilde{M}}\) and \(\underline{\tilde{M}}{}'\) are geometric, \(\varrho_2 \circ \lambda(\bar{f}_i,\bar{f}_i) = v \circ \mu(\bar{f}_i) = v \circ \mu' \circ I(\bar{f}_i) = v \circ \mu'(\bar{f}'_i) = \varrho_2 \circ \lambda'(\bar{f}'_i,\bar{f}'_i)\), so this means that \(\lambda(\bar{f}_i,\bar{f}_i) = \lambda'(\bar{f}'_i,\bar{f}'_i)\). Therefore \(\lambda(x,y) = \lambda'(I(x),I(y))\) for every \(x,y \in M\), so \(I\) is an isomorphism of extended quadratic forms. ◻

Proof. Since \(\mathop{\mathrm{rk}}\nolimits L = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits H = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits H' = \mathop{\mathrm{rk}}\nolimits L'\), there is an isomorphism \(I_0 : L \rightarrow L'\). Since \(\mu = \mu' = 0\), we also have \(\hat{\mu}_L = \hat{\mu}'_{L'} = 0\). So by Lemma 34 there is an isomorphism \(I : \underline{H}\rightarrow \underline{H}{}'\) such that \(I \big| _L = I_0\), in particular \(I(L)=L'\). ◻

Proof. a) Since \(F\) is free and \(g\) is surjective, the homomorphism \(f\) can be lifted to a homomorphism \(\bar{f} : F \rightarrow G\) such that \(f = g \circ \bar{f}\). Let \(\bar{f}_0 : F_0 \rightarrow \mathop{\mathrm{Ker}}\nolimits g \subseteq G\) be a surjective homomorphism from some free abelian group \(F_0\) to \(\mathop{\mathrm{Ker}}\nolimits g\). Let \(F_1 = F \oplus F_0\) and \(\bar{f}_1 = \bar{f} + \bar{f}_0 : F_1 = F \oplus F_0 \rightarrow G\).

We prove that \(\bar{f}_1\) is surjective. Let \(y \in G\) be an arbitrary element, then there is an \(x \in F\) such that \(f(x) = g(y)\) (because \(f\) is surjective). Let \(y_0 = \bar{f}(x) \in \mathop{\mathrm{Im}}\nolimits\bar{f} \subseteq \mathop{\mathrm{Im}}\nolimits\bar{f}_1\), then \(y-y_0 \in \mathop{\mathrm{Ker}}\nolimits g = \mathop{\mathrm{Im}}\nolimits\bar{f}_0 \subseteq \mathop{\mathrm{Im}}\nolimits\bar{f}_1\), therefore \(y = y_0 + (y-y_0) \in \mathop{\mathrm{Im}}\nolimits\bar{f}_1\).

Since \(G\) is free, \(\bar{f}_1\) has a right inverse, ie.there is a homomorphism \(c : G \rightarrow F_1\) such that \(\bar{f}_1 \circ c = \mathop{\mathrm{id}}\nolimits_G\). This also implies that \(F_1 = \mathop{\mathrm{Ker}}\nolimits\bar{f}_1 \oplus \mathop{\mathrm{Im}}\nolimits c \cong \mathop{\mathrm{Ker}}\nolimits\bar{f}_1 \oplus G\).

Let \(F' = F_0 \oplus G\) and \(G' = F_1\). Then \(F \oplus F' = F \oplus (F_0 \oplus G) \cong F_1 \oplus G\) and \(G \oplus G' = G \oplus F_1\), and we can define the homomorphism \(h : F_1 \oplus G \rightarrow G \oplus F_1\) by the formula \(h(x,y)= (\bar{f}_1(x),x+c(y))\).

To show that \(h\) is an isomorphism, we use the decomposition \(F_1 \cong \mathop{\mathrm{Ker}}\nolimits\bar{f}_1 \oplus G\). For \((x_1,y_2) \in \mathop{\mathrm{Ker}}\nolimits\bar{f}_1 \oplus G \cong F_1\) we have \(\bar{f}_1(x_1,y_2)=y_2\), and for \(y \in G\) we have \(c(y) = (0,y) \in \mathop{\mathrm{Ker}}\nolimits\bar{f}_1 \oplus G \cong F_1\). Hence under the decompositions \(F \oplus F' \cong F_1 \oplus G \cong (\mathop{\mathrm{Ker}}\nolimits\bar{f}_1 \oplus G) \oplus G\) and \(G \oplus G' = G \oplus F_1 \cong G \oplus (\mathop{\mathrm{Ker}}\nolimits\bar{f}_1 \oplus G)\) the map \(h\) is given by the formula \(h((x_1,y_2),y) = (y_2, (x_1,y_2+y))\), therefore it is an isomorphism.

Finally, for any \(x \in F\), \(x_0 \in F_0\) and \(y \in G\) we have \((g+0) \circ h((x,x_0),y) = g \circ \bar{f}_1(x,x_0) = g \circ (\bar{f} + \bar{f}_0)(x,x_0) = g \circ \bar{f}(x) = f(x) = (f+0)(x,(x_0,y))\), where we used that \(g \circ \bar{f}_0 = 0\), because \(\mathop{\mathrm{Im}}\nolimits\bar{f}_0 = \mathop{\mathrm{Ker}}\nolimits g\). Therefore \((f+0) = (g+0) \circ h\).

b) Since \(A\) is free, there are maps \(c : A \rightarrow F\) and \(d : A \rightarrow G\) such that \(f \circ c = g \circ d = \mathop{\mathrm{id}}\nolimits_A\). We get the decompositions \(F = \mathop{\mathrm{Ker}}\nolimits f \oplus \mathop{\mathrm{Im}}\nolimits c\) and \(G = \mathop{\mathrm{Ker}}\nolimits g \oplus \mathop{\mathrm{Im}}\nolimits d\), where \(\mathop{\mathrm{Im}}\nolimits c \cong \mathop{\mathrm{Im}}\nolimits d \cong A\). The subgroups \(\mathop{\mathrm{Ker}}\nolimits f\) and \(\mathop{\mathrm{Ker}}\nolimits g\) are free, and \(\mathop{\mathrm{rk}}\nolimits\mathop{\mathrm{Ker}}\nolimits f = \mathop{\mathrm{rk}}\nolimits F - \mathop{\mathrm{rk}}\nolimits A = \mathop{\mathrm{rk}}\nolimits G - \mathop{\mathrm{rk}}\nolimits A = \mathop{\mathrm{rk}}\nolimits\mathop{\mathrm{Ker}}\nolimits g\), therefore there is an isomorphism \(h_0 : \mathop{\mathrm{Ker}}\nolimits f \rightarrow \mathop{\mathrm{Ker}}\nolimits g\). Then we can take \(h = h_0 \oplus (d \circ f) \big| _{\mathop{\mathrm{Im}}\nolimits c} : F = \mathop{\mathrm{Ker}}\nolimits f \oplus \mathop{\mathrm{Im}}\nolimits c \rightarrow G = \mathop{\mathrm{Ker}}\nolimits g \oplus \mathop{\mathrm{Im}}\nolimits d\). ◻

Proof. a) Since the homomorphisms \(\mu : M \rightarrow Q\) and \(\mu' : M' \rightarrow Q\) are surjective, the same is true for the induced maps \(M/L \rightarrow Q\) and \(M'/L' \rightarrow Q\), and for \(\hat{\mu}_L : L^* \rightarrow Q\) and \(\hat{\mu}'_{L'} : (L')^* \rightarrow Q\). By Lemma 37 a) there are free abelian groups \(A, A'\) and an isomorphism \(h : (L')^* \oplus A' \rightarrow L^* \oplus A\) such that \((\hat{\mu}'_{L'} +0) = (\hat{\mu}_L +0) \circ h : (L')^* \oplus A' \rightarrow Q\). Let \(k = \mathop{\mathrm{rk}}\nolimits A\) and \(l = \mathop{\mathrm{rk}}\nolimits A'\).

Let \(\underline{\tilde{M}}= (\tilde{M}, \tilde{\lambda}, \tilde{\mu})\) denote the direct sum \(\underline{M}\oplus \underline{H}_{2k}\), that is \(\tilde{M} = M \oplus \mathbb{Z}^{2k}\), \(\tilde{\lambda} = \lambda \oplus \bigl[ \begin{smallmatrix} 0 & I_k \\ I_k & 0 \end{smallmatrix} \bigr]\) and \(\tilde{\mu} = \mu \oplus 0\). Similarly let \(\underline{\tilde{M}}{}'= (\tilde{M}', \tilde{\lambda}', \tilde{\mu}') = \underline{M}{}'\oplus \underline{H}_{2l}\). Let \(\tilde{L} = L \oplus (\{ 0 \} \times \mathbb{Z}^k)\) and \(\tilde{L}' = L' \oplus (\{ 0 \} \times \mathbb{Z}^l)\), these are lagrangians in \(\underline{\tilde{M}}\) and \(\underline{\tilde{M}}{}'\) respectively. Hence the new extended quadratic forms \(\underline{\tilde{M}}\) and \(\underline{\tilde{M}}{}'\) are again metabolic, free, (full) and geometric. Moreover, we have \(\hat{\tilde{\mu}}_{\tilde{L}} = \hat{\mu}_L + 0 : \tilde{L}^* \cong L^* \oplus (\{ 0 \} \times \mathbb{Z}^k)^* \rightarrow Q\) and \(\hat{\tilde{\mu}}'_{\tilde{L}'} = \hat{\mu}'_{L'} + 0 : (\tilde{L}')^* \cong (L')^* \oplus (\{ 0 \} \times \mathbb{Z}^l)^* \rightarrow Q\).

If we fix some isomorphisms \(A \cong (\{ 0 \} \times \mathbb{Z}^k)^*\) and \(A' \cong (\{ 0 \} \times \mathbb{Z}^l)^*\), then \(h\) can be regarded as an isomorphism \(h : (\tilde{L}')^* \rightarrow \tilde{L}^*\) such that \(\hat{\tilde{\mu}}'_{\tilde{L}'} = \hat{\tilde{\mu}}_{\tilde{L}} \circ h\). Let \(I_0 = h^* : \tilde{L} \rightarrow \tilde{L}'\). Then by Lemma 34 there is an isomorphism \(I : \underline{\tilde{M}}\rightarrow \underline{\tilde{M}}{}'\) such that \(I \big| _{\tilde{L}} = I_0\), in particular \(I(\tilde{L}) = \tilde{L}'\).

b) In this case we can use part b) of Lemma 37 instead of part a). Hence we can take \(A = A' = 0\), so the statement of part a) holds with \(k = l = 0\). ◻

Proof. Suppose that \(f \in \mathop{\mathrm{Aut}}\nolimits(\underline{N})\). Then \(f\) is a generator of \(\mathop{\mathrm{RU}}\nolimits(\underline{N},K)\) if and only if \(F^{-1} \circ f \circ F\) is a generator of \(\mathop{\mathrm{RU}}\nolimits(\underline{M},L)\). ◻

Proof. We have \(\mathop{\mathrm{RU}}\nolimits(\underline{M},K) = f \circ \mathop{\mathrm{RU}}\nolimits(\underline{M},L) \circ f^{-1} = \mathop{\mathrm{RU}}\nolimits(\underline{M},L)\) by Lemma 41 and because \(f \in \mathop{\mathrm{RU}}\nolimits(\underline{M},L)\). ◻

Proof. Since \(\Delta_{\mathop{\mathrm{id}}\nolimits}\) is the diagonal, \((\Phi \oplus \Phi)(\Delta_{\mathop{\mathrm{id}}\nolimits}) = \Delta_{\mathop{\mathrm{id}}\nolimits}\). ◻

Proof. Denote the rank of \(\underline{M}\) by \(2k\). Let \(e_1, e_2, \ldots , e_k, f_1, f_2, \ldots , f_k\) be a basis of \(M\) given by Lemma 20, and let \(d_i = \lambda(f_i,f_i)\). We will use the notation \(e_i\), \(f_i\) when \(M\) is regarded as the underlying abelian group of \(\underline{M}\), and the same elements will be denoted by \(\bar{e}_i\) and \(\bar{f}_i\) when \(M\) is regarded as the underlying abelian group of \(-\underline{M}\). Let \(a_1, a_2, \ldots , a_k, b_1, b_2, \ldots , b_k\) denote the standard basis of \(\mathbb{Z}^{2k}\). Let \(I : \underline{M}\oplus \underline{M}\rightarrow \underline{M}\oplus \underline{H}_{2k}\) and \(J : \underline{M}\rightarrow -\underline{M}\) denote the isomorphisms from Lemmas 32 and 31 respectively.

The composition of the flip map \(\underline{M}\oplus (-\underline{M}) \rightarrow (-\underline{M}) \oplus \underline{M}\), \(J^{-1} \oplus \mathop{\mathrm{id}}\nolimits_M\) and \(I\) is an isomorphism \(F : \underline{M}\oplus (-\underline{M}) \rightarrow \underline{M}\oplus \underline{H}_{2k}\) given by \(F(e_i)=-b_i\), \(F(f_i)=f_i-a_i\), \(F(\bar{e}_i)=e_i+b_i\) and \(F(\bar{f}_i)=d_ie_i-f_i\). Therefore \[\begin{gather} F(L \oplus L) = F(\left< e_1, \ldots , e_k, \bar{e}_1, \ldots , \bar{e}_k \right>) = \\ = \left< -b_1, \ldots , -b_k, e_1+b_1, \ldots , e_k+b_k \right> = \left< b_1, \ldots , b_k, e_1, \ldots , e_k \right> = L \oplus (\{0\} \times \mathbb{Z}^k) \end{gather}\] and \[\begin{gather} F(\Delta_{\mathop{\mathrm{id}}\nolimits}) = F(\left< e_1+\bar{e}_1, \ldots , e_k+\bar{e}_k, f_1+\bar{f}_1, \ldots , f_k+\bar{f}_k \right>) = \\ = \left< e_1, \ldots , e_k, d_1e_1-a_1, \ldots , d_ke_k-a_k \right> = \left< e_1, \ldots , e_k, a_1, \ldots , a_k \right> = L \oplus (\mathbb{Z}^k \times \{0\}) \text{\,.} \end{gather}\] If \(\Sigma : \mathbb{Z}^{2k} \rightarrow \mathbb{Z}^{2k}\) denotes the flip map (ie.\(\Sigma(a,b)=(b,a)\) for \(a,b \in \mathbb{Z}^k\)), then \(\mathop{\mathrm{id}}\nolimits_M \oplus \Sigma \in \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{H}_{2k}, L \oplus (\{0\} \times \mathbb{Z}^k))\). Since \((\mathop{\mathrm{id}}\nolimits_M \oplus \Sigma)(L \oplus (\{0\} \times \mathbb{Z}^k)) = L \oplus (\mathbb{Z}^k \times \{0\})\), we have \(\mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{H}_{2k}, L \oplus (\{0\} \times \mathbb{Z}^k)) = \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{H}_{2k}, L \oplus (\mathbb{Z}^k \times \{0\}))\) by Lemma 42. Therefore \[\begin{align} \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus (-\underline{M}), \Delta_{\mathop{\mathrm{id}}\nolimits}) &= F^{-1} \circ \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{H}_{2k}, L \oplus (\mathbb{Z}^k \times \{0\})) \circ F = \\ &= F^{-1} \circ \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{H}_{2k}, L \oplus (\{0\} \times \mathbb{Z}^k)) \circ F = \\ &= \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus (-\underline{M}), L \oplus L) \end{align}\] by Lemma 41. ◻

Proof. By Lemma 43 and Proposition 44 we have \(\Phi \oplus \Phi \in \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus (-\underline{M}), L \oplus L)\). This implies that \[\mathop{\mathrm{id}}\nolimits_M \oplus \Phi \oplus \Phi \in \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{M}\oplus (-\underline{M}), L \oplus L \oplus L)\] and that \[\Phi \oplus \mathop{\mathrm{id}}\nolimits_M \oplus \Phi \in \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{M}\oplus (-\underline{M}), L \oplus L \oplus L)\] By composing the latter with the inverse of the former we get that \[\Phi \oplus \Phi^{-1} \oplus \mathop{\mathrm{id}}\nolimits_M \in \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{M}\oplus (-\underline{M}), L \oplus L \oplus L)\] hence \(\Phi \oplus \Phi^{-1} \in \mathop{\mathrm{RU}}\nolimits_{\mathop{\mathrm{st}}\nolimits}(\underline{M}\oplus \underline{M}, L \oplus L)\). ◻

Proof. First we define a function \(F : \ell_{2q+1}(Q,v) \rightarrow \ell^T_{2q+1}(Q,v,R)\). Let \(\underline{R}= (R, 0, 0)\), it is a metabolic extended quadratic form of rank \(0\) over any group, geometric with respect to any homomorphism. If \((\underline{M}; L, V) \in \mathfrak{g}_{2q+1}(Q,v)\), then \((\underline{M}; L, V) \oplus (\underline{R}; R, 0) \in \mathfrak{g}^T_{2q+1}(Q,v,R)\), let \(F([\underline{M}; L, V]) = [\underline{M}\oplus \underline{R}; L \oplus R, V \oplus 0]\).

Next we define a function \(G : \ell^T_{2q+1}(Q,v,R) \rightarrow \ell_{2q+1}(Q,v)\). Suppose that \((\underline{N}; K, W) \in \mathfrak{g}^T_{2q+1}(Q,v,R)\), where \(\underline{N}= (N, \lambda, \mu)\). Since \(\mu \big| _{\mathop{\mathrm{Tor}}\nolimits N} = 0\) (because \(\underline{N}\) has a T-lagrangian), we can define \(\underline{\bar{N}}\). Moreover, \(\bar{K} = K / \mathop{\mathrm{Tor}}\nolimits N\) is a lagrangian in \(\underline{\bar{N}}\), and \(\bar{W} = (W \oplus \mathop{\mathrm{Tor}}\nolimits N) / \mathop{\mathrm{Tor}}\nolimits N\) is a half-rank direct summand, hence \((\underline{\bar{N}}; \bar{K}, \bar{W}) \in \mathfrak{g}_{2q+1}(Q,v)\). Let \(G([\underline{N}; K, W])=[\underline{\bar{N}}; \bar{K}, \bar{W}]\).

It is immediate from the definitions that \(G \circ F = \mathop{\mathrm{id}}\nolimits_{\ell_{2q+1}(Q,v)}\). For the other composition, let \([\underline{N}; K, W] \in \ell^T_{2q+1}(Q,v,R)\), then \(F \circ G([\underline{N}; K, W]) = [\underline{\bar{N}}\oplus \underline{R}; \bar{K} \oplus R, \bar{W} \oplus 0]\). Since \(W\) is a free direct summand, \(\pi_N\) has a right inverse \(I_0 : \bar{N} \rightarrow N\) such that \(W \leq \mathop{\mathrm{Im}}\nolimits I_0\). By choosing an isomorphism \(\mathop{\mathrm{Tor}}\nolimits N \cong R\) we can extend \(I_0\) to an isomorphism \(I : \bar{N} \oplus R \rightarrow N\). This is also an isomorphism \(\underline{\bar{N}}\oplus \underline{R}\cong \underline{N}\) of extended quadratic forms, and by construction \(I(\bar{W} \oplus 0) = I_0(\bar{W})=W\). Since \(K\) is a T-lagrangian, we have \(K = (K \cap \mathop{\mathrm{Im}}\nolimits I_0) \oplus \mathop{\mathrm{Tor}}\nolimits N\), so \(I(\bar{K} \oplus R) = I_0(\bar{K}) \oplus \mathop{\mathrm{Tor}}\nolimits N = (K \cap \mathop{\mathrm{Im}}\nolimits I_0) \oplus \mathop{\mathrm{Tor}}\nolimits N = K\). This shows that \([\underline{\bar{N}}\oplus \underline{R}; \bar{K} \oplus R, \bar{W} \oplus 0] = [\underline{N}; K, W]\), hence \(F \circ G = \mathop{\mathrm{id}}\nolimits_{\ell^T_{2q+1}(Q,v,R)}\). ◻

Proof. It is enough to prove the statement when \(\Psi\) is a generator of \(\mathop{\mathrm{RU}}\nolimits(\underline{M},L)\).

If \(\Psi(L) = L\), then \(\Psi\) is an isomorphism between \((\underline{M}; L, V)\) and \((\underline{M}; L, \Psi(V))\).

If there is an isomorphism \(I : \underline{M}\rightarrow \underline{H}_2 \oplus \underline{M}{}'\) such that \(I(L) = ( \{0 \} \times \mathbb{Z}) \oplus L'\) for some lagrangian \(L'\) in \(\underline{M}{}'\) and \(\Psi = I^{-1} \circ (\sigma \oplus \mathop{\mathrm{id}}\nolimits_{M'}) \circ I\), then \(I \circ \Psi(L) = (\sigma \oplus \mathop{\mathrm{id}}\nolimits_{M'}) \circ I(L) = (\sigma \oplus \mathop{\mathrm{id}}\nolimits_{M'})(( \{0 \} \times \mathbb{Z}) \oplus L') = (\mathbb{Z}\times \{0 \} ) \oplus L'\). So \[\begin{align} (\underline{M}; L, V) &\cong (\underline{H}_2 \oplus \underline{M}{}'; I \circ \Psi(L), I \circ \Psi(V)) = \\ &= (\underline{H}_2 \oplus \underline{M}{}'; (\mathbb{Z}\times \{0 \} ) \oplus L', I \circ \Psi(V)) \sim \\ &\sim (\underline{H}_2 \oplus \underline{M}{}'; ( \{0 \} \times \mathbb{Z}) \oplus L', I \circ \Psi(V)) = \\ &= (\underline{H}_2 \oplus \underline{M}{}'; I(L), I \circ \Psi(V)) \cong \\ &\cong (\underline{M}; L, \Psi(V)) \end{align}\] where we used the isomorphisms \(I \circ \Psi\) and \(I\). Therefore the statement holds for every generator of \(\mathop{\mathrm{RU}}\nolimits(\underline{M},L)\). ◻

Proof. As in Proposition 44, if \(\Sigma\) denotes the flip map, then \(\mathop{\mathrm{id}}\nolimits_M \oplus \Sigma \in \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{H}_{2k}, L \oplus (\{0\} \times \mathbb{Z}^k))\) and \((\mathop{\mathrm{id}}\nolimits_M \oplus \Sigma)(V \oplus (\mathbb{Z}^k \times \{0\})) = V \oplus (\{0\} \times \mathbb{Z}^k)\). Hence by Lemma 64, we have \((\underline{M}; L, V) \sim (\underline{M}; L, V) \oplus (\underline{H}_{2k}; \{ 0 \} \times \mathbb{Z}^k, \mathbb{Z}^k \times \{ 0 \}) \sim (\underline{M}; L, V) \oplus (\underline{H}_{2k}; \{ 0 \} \times \mathbb{Z}^k, \{ 0 \} \times \mathbb{Z}^k)\). ◻

Proof. It is clear that there is a well-defined induced operation \(\oplus\) on \(\ell_{2q+1}(Q,v)\), which is associative and commutative (cf.[23]). We need to show that it has a zero element. (Note that \(\underline{H}_{2k}\) is not full if \(Q \neq 0\), so \(\mathcal{H}_{2k} \not \in \mathfrak{g}_{2q+1}(Q,v)\).) We will do this in 3 steps.

First, we show that there is a quasi-formation of the form \((\underline{M}; L, L)\) in the class \(\mathfrak{g}_{2q+1}(Q,v)\) (in particular \(\mathfrak{g}_{2q+1}(Q,v)\) and \(\ell_{2q+1}(Q,v)\) are non-empty). Let \(\mu_0 : A \rightarrow Q\) be a surjective homomorphism from some free abelian group \(A\) to \(Q\). Let \(k = \mathop{\mathrm{rk}}\nolimits A\), and let \(f_1, f_2, \ldots , f_k\) be a basis of \(A\). Let \(L\) be another rank-\(k\) free abelian group with basis \(e_1, e_2, \ldots , e_k\). Let \(M = L \oplus A\). Let \(\lambda : M \times M \rightarrow \mathbb{Z}\) be the bilinear function given by the block matrix \[\begin{bmatrix} 0 & I_k \\ I_k & D \end{bmatrix}\] (in the basis \(e_1, e_2, \ldots , e_k, f_1, f_2, \ldots , f_k\)), where \(D\) is a diagonal matrix such that the diagonal entry corresponding to \(f_i\) is \(0\) if \(v \circ \mu_0(f_i)=0\) and \(1\) if \(v \circ \mu_0(f_i)=1\) (hence \(\varrho_2 \circ \lambda(f_i,f_i) = v \circ \mu_0(f_i)\)). Let \(\mu : M \rightarrow Q\) be the homomorphism such that \(\mu \big| _L = 0\) and \(\mu \big| _A = \mu_0\). Then \(\underline{M}= (M, \lambda, \mu)\) is an extended quadratic form over \(Q\), and it is free and full. It is also geometric, because for any \(x = \sum_{i=1}^k r_ie_i + \sum_{i=1}^k s_if_i \in M\) we have \[\begin{gather} \varrho_2 \circ \lambda(x,x) = \varrho_2 \left( 2\sum_{i=1}^k r_is_i + \sum_{i=1}^k s_i^2 \lambda(f_i,f_i) \right) = \sum_{i=1}^k \varrho_2(s_i^2) \varrho_2 \circ \lambda(f_i,f_i) = \\ = \sum_{i=1}^k \varrho_2(s_i) \varrho_2 \circ \lambda(f_i,f_i) = \sum_{i=1}^k \varrho_2(s_i) v \circ \mu_0(f_i) = \sum_{i=1}^k v(s_i\mu_0(f_i)) = v \circ \mu(x) \end{gather}\] Furthermore, \(L\) is a lagrangian in \(M\), therefore \((\underline{M}; L, L) \in \mathfrak{g}_{2q+1}(Q,v)\).

Second, we show that if \((\underline{M}; L, L), (\underline{M}{}'; L', L') \in \mathfrak{g}_{2q+1}(Q,v)\), then \((\underline{M}; L, L) \sim (\underline{M}{}'; L', L')\). By Theorem 38 a) there is an isomorphism \(I : \underline{M}\oplus \underline{H}_{2k} \rightarrow \underline{M}{}'\oplus \underline{H}_{2l}\) such that \(I(L \oplus (\{ 0 \} \times \mathbb{Z}^k)) = L' \oplus (\{ 0 \} \times \mathbb{Z}^l)\) for some \(k,l \geq 0\). So we have \[\begin{align} (\underline{M}; L, L) &\sim (\underline{M}; L, L) \oplus (\underline{H}_{2k}; \{ 0 \} \times \mathbb{Z}^k, \{ 0 \} \times \mathbb{Z}^k) = \\ &= (\underline{M}\oplus \underline{H}_{2k}; L \oplus (\{ 0 \} \times \mathbb{Z}^k), L \oplus (\{ 0 \} \times \mathbb{Z}^k)) \cong \\ &\cong (\underline{M}{}'\oplus \underline{H}_{2l}; L' \oplus (\{ 0 \} \times \mathbb{Z}^l), L' \oplus (\{ 0 \} \times \mathbb{Z}^l)) = \\ &= (\underline{M}{}'; L', L') \oplus (\underline{H}_{2l}; \{ 0 \} \times \mathbb{Z}^l, \{ 0 \} \times \mathbb{Z}^l) \sim \\ &\sim (\underline{M}{}'; L', L') \end{align}\] using Lemma 65 and the isomorphism \(I\). Therefore all quasi-formations of the form \((\underline{M}; L, L)\) are in the same equivalence class, which we will denote by \(\alpha \in \ell_{2q+1}(Q,v)\).

Third, we show that \(\alpha\) is a zero element in \(\ell_{2q+1}(Q,v)\). Suppose that \((\underline{M}; L, V) \in \mathfrak{g}_{2q+1}(Q,v)\), then \((\underline{M}; L, L) \in \mathfrak{g}_{2q+1}(Q,v)\) and \([\underline{M}; L, L] = \alpha\). Let \(I : \underline{M}\oplus \underline{M}\rightarrow \underline{M}\oplus \underline{H}_{2k}\) be the isomorphism constructed in the proof of Lemma 32, satisfying \(I(L \oplus L) = L \oplus (\{ 0 \} \times \mathbb{Z}^k)\). We also define the basis elements \(e_i,f_i,\bar{e}_i,\bar{f}_i,a_i,b_i\) and the integers \(d_i\) as in Lemma 32. Let \(v_1, v_2, \ldots , v_k\) be a basis of \(V\), then \(v_i = \sum_{j=1}^k (p_{ij}e_j + q_{ij}f_j)\) for some \(p_{ij}, q_{ij} \in \mathbb{Z}\). We have \(I(v_i) = \sum_{j=1}^k (p_{ij}I(e_j) + q_{ij}I(f_j)) = \sum_{j=1}^k (p_{ij}(e_j+b_j) + q_{ij}(f_j+d_jb_j)) = \sum_{j=1}^k (p_{ij}e_j + q_{ij}f_j) + \sum_{j=1}^k (p_{ij} + q_{ij}d_j)b_j = v_i + \sum_{j=1}^k (p_{ij} + q_{ij}d_j)b_j\) and \(I(\bar{e}_i)=-b_i\). Thus \[\begin{gather} I(V \oplus L) = I(\left< v_1, \ldots , v_k, \bar{e}_1, \ldots , \bar{e}_k \right>) = \\ = \left< v_1 + \sum_{j=1}^k (p_{1j} + q_{1j}d_j)b_j, \ldots , v_k + \sum_{j=1}^k (p_{kj} + q_{kj}d_j)b_j, -b_1, \ldots , -b_k \right> = \\ = \left< v_1, \ldots , v_k, b_1, \ldots , b_k \right> = V \oplus (\{ 0 \} \times \mathbb{Z}^k) \text{\,.} \end{gather}\] Using the isomorphism \(I\) and Lemma 65 we get the following: \[\begin{align} (\underline{M}; L, V) \oplus (\underline{M}; L, L) &= (\underline{M}\oplus \underline{M}; L \oplus L, V \oplus L) \cong \\ &\cong (\underline{M}\oplus \underline{H}_{2k}; L \oplus (\{ 0 \} \times \mathbb{Z}^k), V \oplus (\{ 0 \} \times \mathbb{Z}^k)) = \\ &= (\underline{M}; L, V) \oplus (\underline{H}_{2k}; \{ 0 \} \times \mathbb{Z}^k, \{ 0 \} \times \mathbb{Z}^k) \sim \\ &\sim (\underline{M}; L, V) \end{align}\] This means that \([\underline{M}; L, V] \oplus \alpha = [\underline{M}; L, V]\). Since this is true for any \([\underline{M}; L, V] \in \ell_{2q+1}(Q,v)\), \(\alpha\) is a zero element (hence the zero element) in \(\ell_{2q+1}(Q,v)\). ◻

Proof. We have \((\underline{M}; L, L) \in \mathfrak{g}_{2q+1}(Q,v)\) and by Proposition 66 b) it represents the zero element in \(\ell_{2q+1}(Q,v)\). The same is true for \((\underline{M}\oplus \underline{N}; L \oplus K, L \oplus K) = (\underline{M}; L, L) \oplus (\underline{N}; K, K)\). Therefore we have \((\underline{M}; L, V) \sim (\underline{M}; L, V) \oplus ((\underline{M}; L, L) \oplus (\underline{N}; K, K)) \cong ((\underline{M}; L, V) \oplus (\underline{N}; K, K)) \oplus (\underline{M}; L, L) \sim (\underline{M}; L, V) \oplus (\underline{N}; K, K)\). ◻

Proof. By Definition 40 there is a metabolic form \(\underline{N}\) with a lagrangian \(K\) such that \(\Psi \oplus \mathop{\mathrm{id}}\nolimits_N \in \mathop{\mathrm{RU}}\nolimits(\underline{M}\oplus \underline{N}, L \oplus K)\). By Lemma 69 it is enough to show that \((\underline{M}\oplus \underline{N}; L \oplus K, V \oplus K) \sim (\underline{M}\oplus \underline{N}; L \oplus K, \Psi(V) \oplus K)\), which holds by Lemma 64. ◻

Proof. By Theorem 38 a) there is an isomorphism \(\Phi : \underline{M}\oplus \underline{H}_{2k} \rightarrow \underline{M}\oplus \underline{H}_{2k}\) such that \(\Phi(L \oplus (\{ 0 \} \times \mathbb{Z}^k)) = K \oplus (\{ 0 \} \times \mathbb{Z}^k)\) for some \(k \geq 0\). Let \(\underline{\tilde{M}}= \underline{M}\oplus \underline{H}_{2k}\), \(\tilde{L} = L \oplus (\{ 0 \} \times \mathbb{Z}^k)\), \(\tilde{K} = K \oplus (\{ 0 \} \times \mathbb{Z}^k)\) and \(\tilde{V} = V \oplus (\mathbb{Z}^k \times \{ 0 \})\), then \([\underline{M}; L, V] = [\underline{\tilde{M}}; \tilde{L}, \tilde{V}]\), \([\underline{M}; K, V] = [\underline{\tilde{M}}; \tilde{K}, \tilde{V}]\) and (by Lemma 65) \([\underline{M}; K, L] = [\underline{\tilde{M}}; \tilde{K}, \tilde{L}]\).

By Theorem 45 \(\Phi \oplus \Phi^{-1} \in \mathop{\mathrm{RU}}\nolimits_{\mathop{\mathrm{st}}\nolimits}(\underline{\tilde{M}}\oplus \underline{\tilde{M}}, \tilde{K} \oplus \tilde{K})\). Therefore we have \[\begin{align} (\underline{\tilde{M}}; \tilde{K}, \tilde{L}) \oplus (\underline{\tilde{M}}; \tilde{L}, \tilde{V}) &\cong (\underline{\tilde{M}}; \tilde{K}, \tilde{L}) \oplus (\underline{\tilde{M}}; \tilde{K}, \Phi(\tilde{V})) = \\ &= (\underline{\tilde{M}}\oplus \underline{\tilde{M}}; \tilde{K} \oplus \tilde{K}, \tilde{L} \oplus \Phi(\tilde{V})) \sim \\ &\sim (\underline{\tilde{M}}\oplus \underline{\tilde{M}}; \tilde{K} \oplus \tilde{K}, \tilde{K} \oplus \tilde{V}) = \\ &= (\underline{\tilde{M}}; \tilde{K}, \tilde{K}) \oplus (\underline{\tilde{M}}; \tilde{K}, \tilde{V}) \sim \\ &\sim (\underline{\tilde{M}}; \tilde{K}, \tilde{V}) \end{align}\] using the isomorphism \(\Phi\) and Lemmas 70 and 69. ◻

Proof. First we show that every element of \(L_{2q+1}(Q,v)\) has an inverse. If \([\underline{M}; L, K] \in L_{2q+1}(Q,v)\), then \(K\) is a lagrangian in \(\underline{M}\), so \((\underline{M}; K, L) \in \mathfrak{g}_{2q+1}(Q,v)\). By Theorem 71 \([\underline{M}; L, K] \oplus [\underline{M}; K, L] = [\underline{M}; L, L]\), which is the zero element, therefore \([\underline{M}; K, L]\) is the inverse of \([\underline{M}; L, K]\).

Next we show that every invertible element is contained in \(L_{2q+1}(Q,v)\). For an \((\underline{M}; L, V) \in \mathfrak{g}_{2q+1}(Q,v)\) let \(\alpha(\underline{M}; L, V)\) denote the subgroup \(\mu(V) \leq Q\) (where \(\underline{M}= (M, \lambda, \mu)\)). Then \(\alpha(\underline{M}; L, V)\) is invariant under the elementary equivalences that define \(\sim\), so it induces a well-defined function on \(\ell_{2q+1}(Q,v)\). Moreover, we have \(\alpha([\underline{M}; L, V] \oplus [\underline{M}{}'; L', V']) = \alpha([\underline{M}; L, V]) + \alpha([\underline{M}{}'; L', V'])\) and \(\alpha([\underline{M}; L, L]) = 0 \leq Q\). This implies that if \([\underline{M}; L, V]\) is invertible, then \(\alpha([\underline{M}; L, V]) = 0\), that is, \(\mu \big| _V = 0\). Similarly, let \(\beta(\underline{M}; L, V) \leq \mathbb{Z}\) denote the image of the homomorphism \(V \otimes V \rightarrow \mathbb{Z}\) corresponding to the bilinear function \(\lambda \big| _{V \times V} : V \times V \rightarrow \mathbb{Z}\). Then \(\beta\) also induces a well-defined function on \(\ell_{2q+1}(Q,v)\), we have \(\beta([\underline{M}; L, V] \oplus [\underline{M}{}'; L', V']) = \beta([\underline{M}; L, V]) + \beta([\underline{M}{}'; L', V'])\) and \(\beta([\underline{M}; L, L]) = 0 \leq \mathbb{Z}\). So if \([\underline{M}; L, V]\) is invertible, then \(\beta([\underline{M}; L, V]) = 0\), that is, \(\lambda \big| _{V \times V} = 0\). This means that if \([\underline{M}; L, V]\) is invertible, then \(V\) is a lagrangian, therefore \([\underline{M}; L, V] \in L_{2q+1}(Q,v)\). ◻

Proof. Kreck [1] defines \(l_{2q+1}(e)\) as follows: an element of \(l_{2q+1}(e)\) is represented by a pair \((\underline{H}_{2k}, V)\) for some \(k \geq 0\), where \(V\) is a half-rank summand in \(\underline{H}_{2k}\). The equivalence relation on such pairs is generated by the elementary equivalences \((\underline{H}_{2k}, V) \sim (\underline{H}_{2k} \oplus \underline{H}_2, V \oplus (\mathbb{Z}\times \{0 \} ))\) and \((\underline{H}_{2k}, V) \sim (\underline{H}_{2k}, \Phi(V))\) for \(\Phi \in RU(\underline{H}_{2k}, \{ 0 \} \times \mathbb{Z}^k)\).

We define \(F : l_{2q+1}(e) \rightarrow \ell_{2q+1}(0,0)\) by \(F([\underline{H}_{2k}, V]) = [\underline{H}_{2k}; \{0 \} \times \mathbb{Z}^k, V]\), this map is well-defined by Lemma 64.

To define \(G : \ell_{2q+1}(0,0) \rightarrow l_{2q+1}(e)\), note that if \([\underline{H}; L, V] \in \ell_{2q+1}(0,0)\), then \(\underline{H}\) is hyperbolic (see Lemma 24), so by Corollary 36 there is a \(k \geq 0\) and an isomorphism \(I : \underline{H}\rightarrow \underline{H}_{2k}\) such that \(I(L) = \{0 \} \times \mathbb{Z}^k\). Let \(G([\underline{H}; L, V]) = [\underline{H}_{2k}, I(V)]\), we will check that this is well-defined. If \(I : \underline{H}\rightarrow \underline{H}_{2k}\) and \(I' : \underline{H}\rightarrow \underline{H}_{2k}\) are two isomorphisms as above, then \(I' \circ I^{-1}(\{0 \} \times \mathbb{Z}^k) = \{0 \} \times \mathbb{Z}^k\), so \(I' \circ I^{-1} \in RU(\underline{H}_{2k}, \{ 0 \} \times \mathbb{Z}^k)\) and \([\underline{H}_{2k}, I(V)] = [\underline{H}_{2k}, I'(V)]\). If \((\underline{H}; L, V) \cong (\underline{H}{}'; L', V')\), then \(I' : \underline{H}{}'\rightarrow \underline{H}_{2k}\) can be chosen to be the isomorphism \(\underline{H}{}'\rightarrow \underline{H}\) composed with \(I : \underline{H}\rightarrow \underline{H}_{2k}\), so that \((\underline{H}_{2k}, I(V)) = (\underline{H}_{2k}, I'(V'))\). If \((\underline{H}{}'; L', V') = (\underline{H}; L, V) \oplus \mathcal{H}_2\), then we can choose \(I' = I \oplus \mathop{\mathrm{id}}\nolimits_{\mathbb{Z}^2}\), so that \((\underline{H}_{2k+2}, I'(V')) = (\underline{H}_{2k} \oplus \underline{H}_2, I(V) \oplus (\mathbb{Z}\times \{0 \} )) \sim (\underline{H}_{2k}, I(V))\). Finally consider the equivalence \((\underline{H}{}'\oplus \underline{H}_2; L \oplus (\{ 0 \} \times \mathbb{Z}), V) \sim (\underline{H}{}'\oplus \underline{H}_2; L \oplus (\mathbb{Z}\times \{ 0 \}), V)\). If \(I : \underline{H}{}'\oplus \underline{H}_2 \rightarrow \underline{H}_{2k}\) is an isomorphism such that \(I(L \oplus (\{ 0 \} \times \mathbb{Z})) = \{0 \} \times \mathbb{Z}^k\), then \(I \circ (\mathop{\mathrm{id}}\nolimits_{H'} \oplus \sigma)(L \oplus (\mathbb{Z}\times \{ 0 \})) = \{0 \} \times \mathbb{Z}^k\) and \(I \circ (\mathop{\mathrm{id}}\nolimits_{H'} \oplus \sigma) \circ I^{-1} \in RU(\underline{H}_{2k}, \{ 0 \} \times \mathbb{Z}^k)\), therefore \([\underline{H}_{2k}, I(V)] = [\underline{H}_{2k}, I \circ (\mathop{\mathrm{id}}\nolimits_{H'} \oplus \sigma)(V)]\).

\(F\) and \(G\) are obviously inverses of each other, and they preserve direct sums, therefore they are isomorphisms.

The pair \((\underline{H}_{2k}, V)\) represents an element in \(L_{2q+1}\) if and only if \(V\) is a lagrangian in \(\underline{H}_{2k}\), which holds if and only if \([\underline{H}_{2k}; \{0 \} \times \mathbb{Z}^k, V] \in L_{2q+1}(0,0)\), showing that \(F\) restricts to an isomorphism \(L_{2q+1} \rightarrow L_{2q+1}(0,0)\). (Alternatively, note that \(L_{2q+1}(0,0)\) and \(L_{2q+1}\) are both equal to the group of invertible elements in the corresponding monoid.) ◻

Proof. The quasi-formation \((\underline{H}; K, J)\) represents an element of \(L_{2q+1}(0,0)\). By Proposition 74 and Remark 75 \(L_{2q+1}(0,0) \cong 0\), so \((\underline{H}; K, J) \sim (\underline{H}; K, K)\) in \(\mathfrak{g}_{2q+1}(0,0)\). These elements are also equivalent in \(\bar{\mathfrak{g}}_{2q+1}(0,0)\), and hence in \(\bar{\mathfrak{g}}_{2q+1}(Q,v)\), by our observations in Remark 56 (in particular, the naturality of \(\sim\) under the unique morphism \((0,0) \rightarrow (Q,v)\) in the slice category). Since \(\sim\) respects direct sums, \((\underline{M}; L, V) \oplus (\underline{H}; K, J) \sim (\underline{M}; L, V) \oplus (\underline{H}; K, K)\) in \(\bar{\mathfrak{g}}_{2q+1}(Q,v)\), and hence in \(\mathfrak{g}_{2q+1}(Q,v)\). By applying Lemma 69 we get that \((\underline{M}; L, V) \oplus (\underline{H}; K, J) \sim (\underline{M}; L, V)\). ◻

Proof. Let \([\underline{M}; L, K]\) be an arbitrary element in \(L_{2q+1}(Q,v)\).

Let \(J = K \cap L\). Since \(K\) and \(L\) are direct summands in \(M\), the same holds for \(J\) (see Lemma 15). Let \(X\) be a direct complement to \(J\) in \(M\), \(K' = K \cap X\) and \(L' = L \cap X\), then \(K = J \oplus K'\) and \(L = J \oplus L'\).

Let \(M' = J \oplus X^{\perp}\). It follows from Lemma 18 that this is a subgroup of \(M\) and \(\lambda \big| _{M' \times M'}\) is nonsingular. Let \(H = (M')^{\perp}\), by Lemma 17 \(M = M' \oplus H\). Let \(\underline{M}{}'= (M', \lambda \big| _{M' \times M'}, \mu \big| _{M'})\) and \(\underline{H}= (H, \lambda \big| _{H \times H}, \mu \big| _H)\), then \(\underline{M}= \underline{M}{}'\oplus \underline{H}\), and since \(\underline{M}\) is nonsingular, \(\underline{H}\) is nonsingular too.

Next we show that \(J\) is a lagrangian in \(\underline{M}{}'\). It is a direct summand in \(M'\) by the latter’s definition. We have \(\lambda \big| _{J \times J} = 0\) and \(\mu \big| _J = 0\), because \(J \leq L\). Finally \(\mathop{\mathrm{rk}}\nolimits M' = \mathop{\mathrm{rk}}\nolimits J + \mathop{\mathrm{rk}}\nolimits X^{\perp} = \mathop{\mathrm{rk}}\nolimits J + (\mathop{\mathrm{rk}}\nolimits M - \mathop{\mathrm{rk}}\nolimits X) = \mathop{\mathrm{rk}}\nolimits J + \mathop{\mathrm{rk}}\nolimits J = 2 \mathop{\mathrm{rk}}\nolimits J\). So \(J\) is a lagrangian in \(\underline{M}{}'\), and this means that \(\underline{M}{}'\) is metabolic.

Next we show that \(L'\) is a lagrangian in \(\underline{H}\). We have \(L' = L \cap X = L^{\perp} \cap X \leq J^{\perp} \cap X = (J \oplus X^{\perp})^{\perp} = H\). By construction \(L'\) is a direct summand in \(L\), hence in \(M\), hence in \(H\). We have \(\lambda \big| _{L' \times L'} = 0\) and \(\mu \big| _{L'} = 0\), because \(L' \leq L\). Finally \(\mathop{\mathrm{rk}}\nolimits L' = \mathop{\mathrm{rk}}\nolimits L - \mathop{\mathrm{rk}}\nolimits J = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits M - \frac{1}{2} \mathop{\mathrm{rk}}\nolimits M' = \frac{1}{2}(\mathop{\mathrm{rk}}\nolimits M - \mathop{\mathrm{rk}}\nolimits M') = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits H\). Therefore \(L'\) is a lagrangian in \(\underline{H}\), and since \(\underline{H}\) is nonsingular, it is metabolic. We can show similarly that \(K'\) is also a lagrangian in \(\underline{H}\).

Consider the subgroup \(K' \oplus L' \leq H\) (note that \(K' \cap L' = \{ 0 \}\)). Since \(K'\) and \(L'\) are lagrangians in \(H\), we have \(\mathop{\mathrm{rk}}\nolimits(K' \oplus L') = \mathop{\mathrm{rk}}\nolimits H\), therefore \(H / (K' \oplus L')\) is a torsion group. The homomorphism \(\mu \big| _H : H \rightarrow Q\) induces a homomorphism \(H / (K' \oplus L') \rightarrow Q\), because \(\mu \big| _{K' \oplus L'} = 0\). Since \(Q\) is assumed to be free, this induced homomorphism is trivial, therefore \(\mu \big| _H = 0\). This implies that \(\underline{M}{}'\) is full (because \(\underline{M}\) is full). Moreover, since \(\underline{M}\) is geometric (with respect to \(v\)), the same holds for \(\underline{H}\), so \(\underline{H}\) is even. By applying Lemma 24, we get that \(\underline{H}\) is hyperbolic.

Then we have \((\underline{M}; L, K) = (\underline{M}{}'\oplus \underline{H}; J \oplus L', J \oplus K') = (\underline{M}{}'; J, J) \oplus (\underline{H}; L', K') \sim (\underline{M}{}'; J, J)\) by Lemma 77. By Proposition 66 b) \((\underline{M}{}'; J, J)\) represents the zero element in \(\ell_{2q+1}(Q,v)\), and hence in \(L_{2q+1}(Q,v)\). Therefore \([\underline{M}; L, K]\) is the zero element in \(L_{2q+1}(Q,v)\), showing that \(L_{2q+1}(Q,v) \cong 0\). ◻

Proof. Changing the orientation of \(M\) changes the sign of the intersection form \(\lambda\), therefore \(E_q(-M,f) = (H_q(M), -\lambda, \mu)= -E_q(M,f)^*\). ◻

Proof. By the assumption \(f^*(v_q(\xi)) = v_q(f^*(\xi)) = v_q(\nu_M) = v_q(M) \in H^q(M; \mathbb{Z}_2)\).

Let \(x \in H_q(M)\) be an arbitrary element, let \(\alpha \in H^q(M)\) be its Poincaré-dual, and let \(\alpha_2 \in H^q(M; \mathbb{Z}_2)\) be the mod \(2\) reduction of \(\alpha\). Since \(\lambda(x,x) = \left< [M], \alpha \cup \alpha \right> = \left< [M], \alpha^2 \right>\) and \(\alpha_2^2 = Sq^q(\alpha_2) = \alpha_2 \cup v_q(M)\), we have \[\begin{gather} \varrho_2 \circ \lambda(x,x) = \varrho_2(\left< [M], \alpha^2 \right>) = \left< [M], \alpha_2^2 \right> = \left< [M], \alpha_2 \cup v_q(M) \right> = \left< x, v_q(M) \right> = \\ = \left< f_*(x), v_q(\xi) \right> = \left< \mu(x), v_q(\xi) \right> = \hat{v}_q(\xi) \circ \mu(x) \text{\,.} \end{gather}\] Therefore \(E_q(M,f)\) is geometric with respect to \(\hat{v}_q(\xi)\). ◻

Proof. a) and b) are well-known, see eg.[33].

c) Regarding \(F\) as a map of pairs \((W, M) \rightarrow (B, B)\), we get a morphism between the homological long exact sequences, and a commutative diagram: \[\xymatrix{ H_{q+1}(W, M) \ar[r] \ar[d]_-{H_q(F)} & H_q(M) \ar[d]^{H_q(f)} \\ H_{q+1}(B, B) \ar[r] & H_q(B) }\] Since \(H_{q+1}(B, B) \cong 0\), the composition \(H_{q+1}(W, M) \rightarrow H_q(M) \stackrel{\mu}{\rightarrow} H_q(B)\) is trivial. ◻

Proof. Let \(X = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(W, M) \rightarrow H_q(M))\), then \(\mathop{\mathrm{rk}}\nolimits X = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits H_q(M)\), \(\lambda \big| _{X \times X} = 0\) and \(\mu \big| _X = 0\) by Lemma 91.

Let \(Y = \{ y \in H_q(M) \mid ky \in X \text{ for some k \in \mathbb{Z}\setminus \{ 0 \}} \}\). Then \(Y\) is a direct summand in \(H_q(M)\) and \(\mathop{\mathrm{Tor}}\nolimits H_q(M) \leq Y\) by Lemma 15. Since \(X \leq Y\) and \(Y/X\) is a torsion group, \(\mathop{\mathrm{rk}}\nolimits Y = \mathop{\mathrm{rk}}\nolimits X\). We have \(\lambda \big| _{Y \times Y} = 0\), because for any \(y_1, y_2 \in Y\) there is \(k_1, k_2 \in \mathbb{Z}\setminus \{ 0 \}\) such that \(k_1y_1, k_2y_2\in X\), so \(\lambda(k_1y_1, k_2y_2) = 0\).

The homomorphism \(\mu \big| _Y : Y \rightarrow H_q(B)\) induces a homomorphism \(Y/X \rightarrow H_q(B)\), because \(\mu \big| _X = 0\). Since \(Y/X\) is a torsion group and \(H_q(B)\) is free, this induced homomorphism is trivial, therefore \(\mu \big| _Y = 0\).

Therefore \(Y\) is a T-lagrangian in \(E_q(M,f)\). Since \(E_q(M,f)\) is nonsingular, this means that it is metabolic. ◻

Proof. Use the fact that the maps \(F\), \(f_0\) and \(f_1\), as well as the inclusion \(\mathop{\mathrm{sk}}\nolimits_q W \simeq U \rightarrow W\), are \(q\)-connected, \(B\) is simply-connected, and \(q \geq 2\). Moreover, since \(\mathop{\mathrm{sk}}\nolimits_q W\) is a codimension-\((q+1)\) subcomplex in \(U\) and \(W\), the inclusions \(\partial U \simeq U \setminus \mathop{\mathrm{sk}}\nolimits_q W \rightarrow U\) and \(W \setminus \mathop{\mathrm{int}}\nolimits U \simeq W \setminus \mathop{\mathrm{sk}}\nolimits_q W \rightarrow W\) are \(q\)-connected too. (For more details see [23].) ◻

Proof. Since \(U \simeq \mathop{\mathrm{sk}}\nolimits_q W\) and \(\mathop{\mathrm{sk}}\nolimits_q W\) is \(q\)-dimensional, \(H_q(U)\) is free and \(H_{q+1}(U) \cong 0\). Using the exact sequence of the pair \((U, \partial U)\), we get \(\mathop{\mathrm{Tor}}\nolimits H_q(\partial U) \cong \mathop{\mathrm{Tor}}\nolimits H_{q+1}(U, \partial U)\). By the universal coefficient formula we have \(\mathop{\mathrm{Tor}}\nolimits H_{q+1}(U, \partial U) \cong \mathop{\mathrm{Tor}}\nolimits H^{q+2}(U, \partial U)\). By Poincaré-duality and Proposition 93 b) \(H^{q+2}(U, \partial U) \cong H_{q-1}(U) \cong H_{q-1}(B)\). ◻

Proof. Again, we have that \(H_q(U)\) is free, so \(\mathop{\mathrm{Coker}}\nolimits(H_{q+1}(U, \partial U) \rightarrow H_q(\partial U)) \cong \mathop{\mathrm{Im}}\nolimits(H_q(\partial U) \rightarrow H_q(U)) \leq H_q(U)\) is free too. By Lemma 15 \(L = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(U, \partial U) \rightarrow H_q(\partial U))\) is a direct summand in \(H_q(\partial U)\), and it contains \(\mathop{\mathrm{Tor}}\nolimits H_q(\partial U)\). By applying Lemma 91 we also get \(\mathop{\mathrm{rk}}\nolimits L = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits H_q(\partial U)\), \(\lambda \big| _{L \times L} = 0\) and \(\mu \big| _L = 0\). ◻

Proof. The homomorphism \(H_q(\partial U) \rightarrow H_q(U)\) is surjective, because \(H_q(U, \partial U) \cong 0\) (see Proposition 93 d)). Hence \(H_q(\mathop{\mathrm{sk}}\nolimits_q W) \cong H_q(U) \cong \mathop{\mathrm{Im}}\nolimits(H_q(\partial U) \rightarrow H_q(U)) \cong H_q(\partial U) / \mathop{\mathrm{Ker}}\nolimits(H_q(\partial U) \rightarrow H_q(U)) \cong H_q(\partial U) / \mathop{\mathrm{Im}}\nolimits(H_{q+1}(U, \partial U) \rightarrow H_q(\partial U))\). ◻

Proof. It follows from Proposition 93 b) that the boundary map \(H_q(W, M_0) \rightarrow H_{q-1}(M_0)\) is trivial.

By Proposition 93 c) \(H_i(W, U) \cong 0\) for \(i \leq q\), so the boundary homomorphism \(H_q(W, U \sqcup M_0) \rightarrow H_{q-1}(U \sqcup M_0, U)\) in the exact sequence of the triple \((W, U \sqcup M_0, U)\) is an isomorphism. By excision we get the isomorphism in the commutative diagram \[\xymatrix{ H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \ar[r] \ar[d] & H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) \ar[d]^-{\cong} \\ H_q(W, M_0) \ar[r]^-{0} & H_{q-1}(M_0) }\] showing that the map \(H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \rightarrow H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0)\) is trivial. ◻

Proof. By Poincaré-duality it is enough to prove that the homomorphism \(H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) \rightarrow H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1)\) is trivial.

Since \(F\) and \(f_1\) are \(q\)-connected, the inclusion \(M_1 \rightarrow W\) induces an isomorphism on \(H^i\) for \(i \leq q-1\), so the boundary map \(H^{q-1}(M_1) \rightarrow H^q(W, M_1)\) is trivial.

Since \(U \simeq \mathop{\mathrm{sk}}\nolimits_q W\), we have \(H^i(W, U) \cong 0\) for \(i \leq q\), therefore the boundary homomorphism \(H^{q-1}(U \sqcup M_1, U) \rightarrow H^q(W, U \sqcup M_1)\) in the exact sequence of the triple \((W, U \sqcup M_1, U)\) is an isomorphism. By excision we get the isomorphism in the commutative diagram \[\xymatrix{ H^{q-1}(M_1) \ar[r]^-{0} \ar[d]_-{\cong} & H^q(W, M_1) \ar[d] \\ H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) \ar[r] & H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1) }\] showing that the map \(H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) \rightarrow H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1)\) is trivial. ◻

Proof. The universal coefficient formula, applied dually to compute the integral homology of a space \(X\) (with finitely generated cohomology groups) from its cohomology, gives a natural isomorphism \(\mathop{\mathrm{Ext}}\nolimits(H^{i+1}(X),\mathbb{Z}) \stackrel{\cong}{\rightarrow} \mathop{\mathrm{Tor}}\nolimits H_i(X)\) for every \(i\). Therefore it is enough to prove that the homomorphism \(H^{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \rightarrow H^{q+1}(\partial U)\) is an isomorphism.

Consider the diagram \[\xymatrix{ H^{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \ar[r] \ar[d] & H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) \ar[r] \ar[d] & H_q(W, U \sqcup M_1) \ar[d] \\ H^{q+1}(\partial U) \ar[r] & H_{q-1}(\partial U) \ar[r] & H_{q-1}(U) }\] By Poincaré-duality the first square commutates and the horizontal arrows in it are isomorphisms. The second square also commutates, because it is induced by a map \((W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1, M_1) \rightarrow (W, U \sqcup M_1, M_1)\) of triples, and its horizontal arrows are isomorphisms by excision and Proposition 93 b).

It is now enough to prove that the map \(H_q(W, U \sqcup M_1) \rightarrow H_{q-1}(U)\) is an isomorphism. We use the exact sequence of the triple \((W, U \sqcup M_1, M_1)\). We have \(H_{q-1}(W, M_1) \cong 0\) by Proposition 93 e). The map \(H_q(W, M_1) \rightarrow H_q(W, U \sqcup M_1)\) is trivial, because \(H_q(W, U \sqcup M_1) \cong H_{q-1}(M_1)\) (similarly to the isomorphism \(H_q(W, U \sqcup M_0) \cong H_{q-1}(M_0)\) in the proof of Lemma 100) and the composition \(H_q(W, M_1) \rightarrow H_{q-1}(M_1)\) vanishes by Proposition 93 b).

It is now enough to prove that the map \(H_q(W, U \sqcup M_1) \rightarrow H_{q-1}(U)\) is an isomorphism. This holds, because in the exact sequence of the triple \((W, U \sqcup M_1, M_1)\) we have \(H_{q-1}(W, M_1) \cong 0\) and the map \(H_q(W, M_1) \rightarrow H_q(W, U \sqcup M_1) \cong H_{q-1}(M_1)\) is trivial by Proposition 93 e) and b) respectively (where we used the isomorphism from the proof of Lemma 100). ◻

Proof. It follows from Poincaré-duality and the universal coefficient formula that for every \(i\) we have \(\mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) = \mathop{\mathrm{rk}}\nolimits H^{2q+1-i}(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1) = \mathop{\mathrm{rk}}\nolimits H_{2q+1-i}(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1)\), in particular \(\mathop{\mathrm{rk}}\nolimits H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) = \mathop{\mathrm{rk}}\nolimits H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1)\).

Since \(\chi(M_0) = \chi(M_1)\), we have \(\chi(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) = \chi(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1)\), that is, \[\sum_{i=0}^{2q+1} (-1)^i \mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) = \sum_{i=0}^{2q+1} (-1)^i \mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1)\] Hence \[\begin{gather} \mathop{\mathrm{rk}}\nolimits H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) - \mathop{\mathrm{rk}}\nolimits H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1) = \\ = \sum_{i \neq q} (-1)^{i+q+1} (\mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) - \mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1)) \end{gather}\]

If \(i \leq q-1\), then \(H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \cong H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1) \cong 0\) by Proposition 93 e).

If \(i \geq q+1\), then \(\mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1) = \mathop{\mathrm{rk}}\nolimits H_{2q+1-i}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) = \mathop{\mathrm{rk}}\nolimits H_{2q-i}(M_0) = \mathop{\mathrm{rk}}\nolimits H_{2q-i}(B)\) (the second equality follows from the proof of Lemma 100 and the third from Proposition 93 b)). Similarly \(\mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) = \mathop{\mathrm{rk}}\nolimits H_{2q-i}(B)\), therefore \(\mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) - \mathop{\mathrm{rk}}\nolimits H_i(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1) = 0\).

It follows that \(\mathop{\mathrm{rk}}\nolimits H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) - \mathop{\mathrm{rk}}\nolimits H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_1) = 0\). ◻

Proof. \(\underline{M}= E_q(\partial U, F \big| _{\partial U})\) is an extended quadratic form over \(H_q(B)\). It is metabolic, because the intersection form \(\lambda\) is nonsingular, and \(L\) is a T-lagrangian (Proposition 97). It follows from Theorem 102 and Lemmas 103 and 104 that \(V\) is a free half-rank direct summand. Therefore \(\theta_{U,F}\) is a quasi-formation over \(H_q(B)\).

Since \(F\) is a normal \((q{-}1)\)-smoothing, \(F^*(\xi) \cong \nu_W\) and hence \((F \big| _{\partial U})^*(\xi) \cong \nu_{\partial U}\). By Lemma 90 this implies that \(\underline{M}\) is geometric with respect to \(\hat{v}_q(\xi)\). It is also full, because \(F \big| _{\partial U}\) is the composition \(\partial U \rightarrow U \rightarrow W \stackrel{F}{\rightarrow} B\) of three \(q\)-connected maps (using Proposition 93 d), the homotopy equivalence \(U \simeq \mathop{\mathrm{sk}}\nolimits_q W\) and the assumption on \(F\)). By Lemma 95 \(\mathop{\mathrm{Tor}}\nolimits H_q(\partial U) \cong \mathop{\mathrm{Tor}}\nolimits H_{q-1}(B)\). Therefore \(\theta_{U,F}\) is in the class \(\mathfrak{g}^T_{2q+1}(B,\xi) = \mathfrak{g}^T_{2q+1}(H_q(B),\hat{v}_q(\xi),\mathop{\mathrm{Tor}}\nolimits H_{q-1}(B))\). ◻

Proof. By Theorem 110 a) (applied to the composition \(K_1 \stackrel{j}{\rightarrow} K_2 \rightarrow U_2\)), there is a compact codimension-\(0\) submanifold \(U_2' \subset U_2\) with simply-connected boundary and a homotopy equivalence \(K_1 \rightarrow U_2'\) such that the diagram \[\xymatrix{ K_1 \ar[r] \ar[d]_-{j} & U_2' \ar[d] \\ K_2 \ar[r] & U_2 }\] commutes up to homotopy. Since \(j\) is a homotopy equivalence, the inclusion \(U_2' \rightarrow U_2\) is a homotopy equivalence too. The compositions \(K_1 \rightarrow U_1 \rightarrow W\) and \(K_1 \rightarrow U_2' \rightarrow U_2 \rightarrow W\) are homotopic, so Theorem 110 b) can be applied to the homotopy equivalences \(K_1 \rightarrow U_1\) and \(K_1 \rightarrow U_2'\). We get that there is a diffeomorphism \(g : U_1 \rightarrow U_2'\) such that the inclusion \(U_1 \rightarrow W\) and the composition \(U_1 \stackrel{g}{\rightarrow} U_2' \rightarrow W\) are isotopic, and the diagram \[\xymatrix{ K_1 \ar[r] \ar@{=}[d] & U_1 \ar[d]^-{g} \\ K_1 \ar[r] & U_2' }\] commutes up to homotopy. In particular, \(U_2'\) is a regular neighbourhood of an embedding \(K_1 \rightarrow \mathop{\mathrm{int}}\nolimits W\) isotopic to the inclusion \(K_1 \rightarrow W\), so \(\theta_{U_2',F}\) is defined.

We will show that \(\theta_{U_1,F} \cong \theta_{U_2',F}\) and \(\theta_{U_2',F} \cong \theta_{U_2,F}\). We introduce the notation \((\underline{M}; L, V) = \theta_{U_1,F}\), \((\underline{M}{}'; L', V') = \theta_{U_2',F}\) and \((\underline{M}{}''; L'', V'') = \theta_{U_2,F}\).

First we consider \(\theta_{U_1,F}\) and \(\theta_{U_2',F}\). It follows from the isotopy extension theorem that there is a diffeomorphism \(G : W \rightarrow W\) (isotopic to the identity) such that \(G \big| _{U_1} = g\) (hence \(G(U_1)=U_2'\)). Then \(H_q(G \big| _{\partial U_1}) = H_q(g \big| _{\partial U_1}) : H_q(\partial U_1) \rightarrow H_q(\partial U_2')\) is an isomorphism between \(\underline{M}\) and \(\underline{M}{}'\). Moreover, \(G\) induces commutative diagrams \[\xymatrix{ H_{q+1}(U_1, \partial U_1) \ar[r] \ar[d]_{\cong} & H_q(\partial U_1) \ar[d]^{\cong} \\ H_{q+1}(U_2', \partial U_2') \ar[r] & H_q(\partial U_2') } \quad \quad \xymatrix{ H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U_1, \partial U_1 \sqcup M_0) \ar[r] \ar[d]_{\cong} & H_q(\partial U_1) \ar[d]^{\cong} \\ H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U_2', \partial U_2' \sqcup M_0) \ar[r] & H_q(\partial U_2') }\] Hence \(H_q(G \big| _{\partial U_1})\) sends \(L\) to \(L'\) and \(V\) to \(V'\). Therefore \((\underline{M}; L, V) \cong (\underline{M}{}'; L', V')\), via the isomorphism \(H_q(g \big| _{\partial U_1})\).

Next we consider \(\theta_{U_2',F}\) and \(\theta_{U_2,F}\). Let \(C = U_2 \setminus \mathop{\mathrm{int}}\nolimits U_2'\). It follows from the assumptions and the Van Kampen theorem that \(\partial U_2'\), \(\partial U_2\) and \(C\) are all simply-connected. Since the inclusion \(U_2' \rightarrow U_2\) is a homotopy equivalence, \(H_*(C, \partial U_2') \cong H_*(U_2, U_2') \cong 0\), so the inclusion \(\partial U_2' \rightarrow C\) is also a homotopy equivalence. It follows from Poincaré-duality that the same is true for the inclusion \(\partial U_2 \rightarrow C\), so \(C\) is an h-cobordism. This (and excision) implies that the vertical arrows are isomorphisms in the following diagrams, induced by the inclusions: \[\xymatrix{ H_{q+1}(U_2', \partial U_2') \ar[r] \ar[d]_{\cong} & H_q(\partial U_2') \ar[d]^{\cong} \\ H_{q+1}(U_2, C) \ar[r] & H_q(C) \\ H_{q+1}(U_2, \partial U_2) \ar[r] \ar[u]^{\cong} & H_q(\partial U_2) \ar[u]_{\cong} } \quad \quad \xymatrix{ H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U_2', \partial U_2' \sqcup M_0) \ar[r] \ar[d]_{\cong} & H_q(\partial U_2') \ar[d]^{\cong} \\ H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U_2', C \sqcup M_0) \ar[r] & H_q(C) \\ H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U_2, \partial U_2 \sqcup M_0) \ar[r] \ar[u]^{\cong} & H_q(\partial U_2) \ar[u]_{\cong} }\] Thus the composition of the isomorphism \(H_q(\partial U_2') \rightarrow H_q(C)\) and the inverse of \(H_q(\partial U_2) \rightarrow H_q(C)\) is an isomorphism \(H_q(\partial U_2') \rightarrow H_q(\partial U_2)\) that sends \(L'\) to \(L''\) and \(V'\) to \(V''\). Moreover, it is an isomorphism between the extended quadratic forms \(\underline{M}{}'\) and \(\underline{M}{}''\), because the inclusions induce isomorphisms of cohomology rings \(H^*(\partial U_2') \cong H^*(C) \cong H^*(\partial U_2)\), and they commute with the homomorphisms \(H_q(F \big| _{\partial U_2'})\), \(H_q(F \big| _C)\) and \(H_q(F \big| _{\partial U_2})\). Therefore \((\underline{M}{}'; L', V') \cong (\underline{M}{}''; L'', V'')\).

We take \(I\) to be the composition of the above isomorphisms \(H_q(\partial U_1) \rightarrow H_q(\partial U_2') \rightarrow H_q(\partial U_2)\). To complete the proof, we verify the commutativity of the following diagram: \[\xymatrix{ H_q(K_1) \ar[r]^-{\cong} \ar@{=}[d] & H_q(U_1) \ar[d]^-{H_q(g)} & H_q(\partial U_1) \ar[l] \ar[d]^-{H_q(g | _{\partial U_1})} \\ H_q(K_1) \ar[r]^-{\cong} \ar[d]_-{H_q(j)} & H_q(U_2') \ar[d] & H_q(\partial U_2') \ar[l] \ar[d] \\ H_q(K_2) \ar[r]^-{\cong} & H_q(U_2) & H_q(\partial U_2) \ar[l] }\] The squares on the left are obtained by applying \(H_q\) to the earlier homotopy commutative diagrams. The top right square is induced by \(g\) and the inclusions. Finally, the bottom right square commutes, because both compositions are equal to the composition \(H_q(\partial U_2') \rightarrow H_q(C) \rightarrow H_q(U_2)\) induced by the inclusions. ◻

Proof. Part a) is a special case of Wall’s embedding theorem [34]. Part b) follows from the uniqueness part of the embedding theorem combined with results of Hudson [35]. ◻

Proof. Let \(c_1\) and \(c_2\) be arbitrary CW-decompositions of \(W\). Then they have stabilisations (see Definition 113) \(c_1'\) and \(c_2'\) that satisfy the conditions of Lemma 115. By Lemmas 114 and 108 we have \(\theta_{c_1,F} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}\theta_{c_1',F} \cong \theta_{c_2',F} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}\theta_{c_2,F}\). Therefore the stable isomorphism class of \(\theta_{c,F}\) is independent of the choice of \(c\). ◻

Proof. It is enough to prove this when \(c'\) is obtained from \(c\) by a single stabilisation step.

Let \(K\) and \(K'\) denote the \(q\)-skeletons of \(W\) with respect to \(c\) and \(c'\), then \(K' = K \vee S^q\). Let \(U\) and \(U'\) be regular neighbourhoods of \(K\) and \(K'\), we can choose these such that \(U' \approx U \natural (S^q \times D^{q+1})\). Let \((\underline{M}; L, V) = \theta_{U,F}\) and \((\underline{M}{}'; L', V') = \theta_{U',F}\).

Since \(\partial U' \approx U \# (S^q \times S^q)\), we have \(H_q(\partial U') \cong H_q(\partial U) \oplus \mathbb{Z}^2\) and \(\underline{M}{}'\cong \underline{M}\oplus \underline{H}_2\), we will use these canonical isomorphisms to identify these groups and extended quadratic forms. The inclusion \(H_q(\partial U) \rightarrow H_q(\partial U')\) under this identification is in fact the composition \(H_q(\partial U) \cong H_q(\partial U \setminus \mathop{\mathrm{int}}\nolimits D^{2q}) \rightarrow H_q(\partial U')\), where the homomorphism \(H_q(\partial U \setminus \mathop{\mathrm{int}}\nolimits D^{2q}) \rightarrow H_q(\partial U')\) is induced by the inclusion of spaces. Thus the inclusion \(U \rightarrow U'\) induces commutative diagrams \[\xymatrix{ H_{q+1}(U, \partial U) \ar[r] \ar[d] & H_q(\partial U) \ar[d] \\ H_{q+1}(U', \partial U') \ar[r] & H_q(\partial U') }\] and \[\xymatrix{ H_{q+1}(W, U \sqcup M_0) \cong H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) \ar[r] \ar[d] & H_q(\partial U) \ar[d] \\ H_{q+1}(W, U' \sqcup M_0) \cong H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U', \partial U' \sqcup M_0) \ar[r] & H_q(\partial U') }\] hence \(L \leq L'\) and \(V \leq V'\).

If we fix some basepoint \(x \in S^q\), the submanifold \(\{ x \} \times D^{q+1} \subset S^q \times D^{q+1} \subseteq U'\) represents an element of \(H_{q+1}(U', \partial U')\), and its boundary is represented by the submanifold \(\{ x \} \times S^q \subset S^q \times S^q \subseteq \partial U'\). This shows that \((0,(0,1)) \in L' \leq H_q(\partial U') \cong H_q(\partial U) \oplus \mathbb{Z}^2\), therefore \(L \oplus (\{ 0 \} \times \mathbb{Z}) \leq L'\). Since \(L'\) and \(L \oplus (\{ 0 \} \times \mathbb{Z})\) are both T-lagrangians in \(\underline{M}{}'\), we get that \(L \oplus (\{ 0 \} \times \mathbb{Z})\) is a direct summand in \(L'\), and they have the same torsion subgroup and the same rank, and this implies that \(L' = L \oplus (\{ 0 \} \times \mathbb{Z})\).

Similarly, the embedded \(D^{q+1}\) that we used to construct the stabilisation \(c'\) from \(c\) represents an element of \(H_{q+1}(W, U' \sqcup M_0) \cong H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U', \partial U' \sqcup M_0)\), and its boundary is represented by \(S^q \times \{ x \} \subset S^q \times S^q \subseteq \partial U'\), therefore \(V \oplus (\mathbb{Z}\times \{ 0 \}) \leq V'\). Since \(V'\) and \(V \oplus (\mathbb{Z}\times \{ 0 \})\) are both free half-rank direct summands in \(\underline{M}{}'\), this implies that \(V' = V \oplus (\mathbb{Z}\times \{ 0 \})\).

Therefore \((\underline{M}{}'; L', V') \cong (\underline{M}\oplus \underline{H}_2; L \oplus (\{ 0 \} \times \mathbb{Z}), V \oplus (\mathbb{Z}\times \{ 0 \})) = (\underline{M}; L, V) \oplus \mathcal{H}_2\). Since \(\theta_{c,F}\) and \(\theta_{c',F}\) are the isomorphism classes of \(\theta_{U,F}\) and \(\theta_{U',F}\) respectively, this means that \(\theta_{c',F} \cong \theta_{c,F} \oplus \mathcal{H}_2\). ◻

Proof. Let \(K_i\) denote the \(q\)-skeleton of \(W\) with respect to \(c_i\) and let \(g_i : K_i \rightarrow W\) denote the inclusion. Then the homomorphisms \(H_q(g_i) : H_q(K_i) \rightarrow H_q(W)\) are surjective, so by Lemma 37 a) there are free abelian groups \(A_i\) and an isomorphism \(J : H_q(K_1) \oplus A_1 \rightarrow H_q(K_2) \oplus A_2\) such that \((H_q(g_1)+0) = (H_q(g_2)+0) \circ J : H_q(K_1) \oplus A_1 \rightarrow H_q(W)\).

Let \(k_i = \mathop{\mathrm{rk}}\nolimits A_i\) and let \(c_i'\) be a \(k_i\)-fold stabilisation of \(c_i\). Then \(K_i' \approx K_i \vee \bigvee^{k_i} S^q\), so \(H_q(K_i') \cong H_q(K_i) \oplus A_i\), and \(J\) can be regarded as an isomorphism \(J : H_q(K_1') \rightarrow H_q(K_2')\) such that \(H_q(g_1') = H_q(g_2') \circ J : H_q(K_1') \rightarrow H_q(W)\), where \(g_i' : K_i' \rightarrow W\) is the inclusion.

By Lemma 116 there is a map \(j : K_1' \rightarrow K_2'\) such that \(g_1' \simeq g_2' \circ j : K_1' \rightarrow W\) and \(H_q(j) = J\). Since \(H_h(g_1') = H_h(g_2') \circ H_h(j)\) and \(H_h(g_1')\) and \(H_h(g_2')\) are isomorphisms if \(h \leq q-1\), the same is true for \(H_h(j)\). As \(H_q(j) = J\) is also an isomorphism, and \(K_1'\) and \(K_2'\) are simply-connected and \(q\)-dimensional, \(j\) is a homotopy equivalence. ◻

Proof. The map \(f : R \rightarrow X\) has a lift \(\tilde{f} : R \rightarrow S\) such that \(g \circ \tilde{f} \simeq f\), because the obstructions to the existence of such a lift are in the groups \(H^i(R;\pi_{i-1}(\mathop{\mathrm{hofib}}\nolimits(g)))\), which are trivial, because \(g\) is \(q\)-connected and \(R\) is \(q\)-dimensional. Let \(J_0 = J - H_q(\tilde{f}) : H_q(R) \rightarrow H_q(S)\).

The pair \((X, S)\) is \(q\)-connected (where \(S\) is regarded as a subspace of \(X\) via \(g\)), so \(\pi_{q+1}(X, S) \cong H_{q+1}(X, S)\). The Hurewicz-homomorphisms induce the following commutative diagram: \[\xymatrix{ & H_q(R) \ar[d]^-{J_0} \ar@{-->}[dl] & \\ H_{q+1}(X, S) \ar[r] & H_q(S) \ar[r]^-{H_q(g)} & H_q(X) \\ \pi_{q+1}(X, S) \ar[u]^-{\cong} \ar[r] & \pi_q(S) \ar[u]_-{h} \ar[r]^-{\pi_q(g)} & \pi_q(X) \ar[u] }\] We have \(H_q(g) \circ J_0 = H_q(g) \circ (J - H_q(\tilde{f})) = H_q(g) \circ J - H_q(g) \circ H_q(\tilde{f}) = H_q(f) - H_q(f) = 0\) and \(H_q(R)\) is free, so \(J_0\) can be lifted to a homomorphism \(H_q(R) \rightarrow H_{q+1}(X, S)\). By composing this homomorphism with the inverse of the isomorphism \(\pi_{q+1}(X, S) \rightarrow H_{q+1}(X, S)\) and the boundary map, we get a homomorphism \(\hat{J}_0 : H_q(R) \rightarrow \pi_q(S)\) such that \(h \circ \hat{J}_0 = J_0\). Moreover, it follows from the construction of \(\hat{J}_0\) that \(\pi_q(g) \circ \hat{J}_0 = 0\), that is \(\mathop{\mathrm{Im}}\nolimits(\hat{J}_0) \leq \mathop{\mathrm{Ker}}\nolimits(\pi_q(g))\).

Let \((C_*(R), \partial_*)\) be the cellular chain complex of \(R\). Since \(\mathop{\mathrm{Im}}\nolimits(\partial_q) \cong C_q(R) / \mathop{\mathrm{Ker}}\nolimits(\partial_q)\) is a subgroup of \(C_{q-1}(R)\), it is free, hence \(H_q(R) = \mathop{\mathrm{Ker}}\nolimits(\partial_q)\) is a direct summand in \(C_q(R)\). Therefore \(\hat{J}_0 : H_q(R) \rightarrow \mathop{\mathrm{Ker}}\nolimits(\pi_q(g))\) can be extended (arbitrarily) to a homomorphism \(\bar{J}_0 : C_q(R) \rightarrow \mathop{\mathrm{Ker}}\nolimits(\pi_q(g)) \leq \pi_q(S)\).

Given a \(q\)-cell \(e^q\) of \(R\) and an element \([a] \in \pi_q(S)\), we can modify the map \(\tilde{f}\) by adding \([a]\) to \(\tilde{f} \big| _{e^q}\), ie.by replacing \(\tilde{f} \big| _{e^q}\) with the composition \(e^q \rightarrow e^q \vee S^q \stackrel{\tilde{f} \vee a}{\longrightarrow} S\). This operation changes the homomorphism \(H_q(\tilde{f}) : H_q(R, \mathop{\mathrm{sk}}\nolimits_{q-1} R) \rightarrow H_q(S, \mathop{\mathrm{sk}}\nolimits_{q-1} S)\) by adding the image of \(h([a]) \in H_q(S)\) in \(H_q(S, \mathop{\mathrm{sk}}\nolimits_{q-1} S)\) to the image of the generator \([e^q] \in H_q(R, \mathop{\mathrm{sk}}\nolimits_{q-1} R) \cong C_q(R)\). Moreover, if \([a] \in \mathop{\mathrm{Ker}}\nolimits(\pi_q(g))\), then the homotopy class of \(g \circ \tilde{f}\) does not change, because \(g \circ a\) is nullhomotopic, so the composition \(e^q \rightarrow e^q \vee S^q \stackrel{\tilde{f} \vee a}{\longrightarrow} S \stackrel{g}{\rightarrow} X\) is homotopic to \(g \circ \tilde{f} \big| _{e^q}\) (rel boundary).

Let \(j : R \rightarrow S\) be the map obtained from \(\tilde{f}\) by adding \(\bar{J}_0([e^q_i]) \in \pi_q(S)\) to \(\tilde{f} \big| _{e^q_i}\) for every \(q\)-cell \(e^q_i\) of \(R\). This changes \(H_q(\tilde{f}) : H_q(R, \mathop{\mathrm{sk}}\nolimits_{q-1} R) \cong C_q(R) \rightarrow H_q(S, \mathop{\mathrm{sk}}\nolimits_{q-1} S)\) by adding the composition \(C_q(R) \stackrel{\bar{J}_0}{\rightarrow} \pi_q(S) \stackrel{h}{\rightarrow} H_q(S) \rightarrow H_q(S, \mathop{\mathrm{sk}}\nolimits_{q-1} S)\). Therefore \(H_q(j) = H_q(\tilde{f}) + h \circ \bar{J}_0 \big| _{H_q(R)} : H_q(R) \rightarrow H_q(S)\). Hence \[H_q(j) = H_q(\tilde{f}) + h \circ \bar{J}_0 \big| _{H_q(R)} = H_q(\tilde{f}) + h \circ \hat{J}_0 = H_q(\tilde{f}) + J_0 = H_q(\tilde{f}) + J - H_q(\tilde{f}) = J \text{\,.}\] Moreover, since \(\mathop{\mathrm{Im}}\nolimits(\bar{J}_0) \leq \mathop{\mathrm{Ker}}\nolimits(\pi_q(g))\), we have \(g \circ j \simeq g \circ \tilde{f} \simeq f\), therefore \(j\) satisfies both conditions. ◻

Proof. Let \(c_0\) be an arbitrary CW-decomposition of \(W\). Then \(\theta_{W,F}\) is the stable isomorphism class of \(\theta_{c_0,F}\), so \(\theta_{W,F} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}x\) means that there are integers \(k,l \geq 0\) such that \(x \oplus \mathcal{H}_{2k} \cong \theta_{c_0,F} \oplus \mathcal{H}_{2l}\). Let \(c\) be an \(l\)-fold stabilisation of \(c_0\). Then by Lemma 114 \(\theta_{c,F} \cong \theta_{c_0,F} \oplus \mathcal{H}_{2l} \cong x \oplus \mathcal{H}_{2k}\). ◻

Proof. Let \(G : X \rightarrow B\) be a normal bordism between \(F\) and \(F'\). By applying surgery below the middle dimension (see eg.[1]), we can assume that \(G\) is \((q+1)\)-connected. Since the compositions \(W \rightarrow X \rightarrow B\) and \(W' \rightarrow X \rightarrow B\) are \(q\)-connected, this implies that the inclusions \(W \rightarrow X\) and \(W' \rightarrow X\) are \(q\)-connected too. Then there is a Morse-function on the triad \((X, W, W')\) having only critical points of index \(q+1\) (see [36]). This means that \(W'\) can be obtained from \(W\) by a sequence of \(q\)-surgeries over \((B,\xi)\) (with trace \(X\)), so it is enough to prove the statement when \(W'\) is obtained by a single \(q\)-surgery.

Fix an embedding \(S^q \times D^{q+1} \rightarrow W\) where the surgery is done, and let \(W_0 = W \setminus \mathop{\mathrm{int}}\nolimits(S^q \times D^{q+1})\). Then \(W' = W_0 \cup_{S^q \times S^q} D^{q+1} \times S^q\). Take the standard CW-decomposition of \(S^q \times S^q\) into \(4\) cells (coming from the product structure) and extend it to a CW-decomposition \(c_0\) of \(W_0\). Define the equivalence relation \(\underset{W}{\sim}\) on \(S^q \times S^q\) by \((x,y_1) \underset{W}{\sim} (x,y_2)\) for any \(x,y_1,y_2 \in S^q\), then \({W_0} / {\underset{W}{\sim}}\) is diffeomorphic to \(W\) (because a neighbourhood of \(S^q \times D^{q+1}\) in \(W\) can be identified with \(S^q \times \mathbb{R}^{q+1}\), and \((\mathbb{R}^{q+1} \setminus \mathop{\mathrm{int}}\nolimits D^{q+1}) / S^q \approx \mathbb{R}^{q+1}\)). Moreover, \(c_0\) induces a CW-decomposition \(c\) on \({W_0} / {\underset{W}{\sim}} \approx W\), namely we have the standard decomposition of \({S^q \times S^q} / {\underset{W}{\sim}} \approx S^q\) into \(2\) cells, and all other cells remain the same, with adjusted gluing maps. If we fix these CW-decompositions, then \(\mathop{\mathrm{sk}}\nolimits_q W\) is of the form \(K_0 \vee S^q\), where \(K_0\) is the \(q\)-skeleton of \(W_0\) without the two \(q\)-cells of \(S^q \times S^q\). Then a regular neighbourhood \(U\) of \(\mathop{\mathrm{sk}}\nolimits_q W\) is of the form \(U_0 \natural (S^q \times D^{q+1})\), where \(U_0\) is a regular neighbourhood of \(K_0\) and we can assume that the embedding \(S^q \times D^{q+1} \rightarrow U \rightarrow W\) coincides with the embedding previously fixed for surgery (and hence \(F \big| _{S^q \times D^{q+1}}\) is nullhomotopic). This means that \(\partial U \approx \partial U_0 \# (S^q \times S^q)\) and \(F \big| _{S^q \times S^q}\) is nullhomotopic. Let \((\underline{M}; L, V) = \theta_{U,F}\). By Lemmas 86 and 87 we have \(\underline{M}\cong \underline{M}_0 \oplus \underline{H}_2\), where \(\underline{M}_0 = (H_q(\partial U_0), \lambda_0, \mu_0)\), \(\lambda_0\) is the intersection form of \(\partial U_0\) and \(\mu_0 = H_q(F \big| _{\partial U_0})\). Moreover \(L \cong L_0 \oplus (\{ 0 \} \times \mathbb{Z})\), where \(L_0 = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(U_0, \partial U_0) \rightarrow H_q(\partial U_0))\).

Similarly we define the equivalence relation \(\underset{W'}{\sim}\) on \(S^q \times S^q\) by \((x_1,y) \underset{W'}{\sim} (x_2,y)\) for any \(x_1,x_2,y \in S^q\), then \({W_0} / {\underset{W'}{\sim}} \approx W'\). We get a CW-decomposition \(c'\) of \(W'\) such that \(\mathop{\mathrm{sk}}\nolimits_q W' \approx K_0 \vee S^q\), and \(\mathop{\mathrm{sk}}\nolimits_q W'\) has a regular neighbourhood \(U' \approx U_0 \natural (D^{q+1} \times S^q)\), where \(U_0\) is the same regular neighbourhood of \(K_0\) that we used earlier. We can assume that the image of the embedding \(D^{q+1} \times S^q \rightarrow U' \rightarrow W'\) is \(\overline{W' \setminus W_0}\), and the restriction of this embedding to \(S^q \times S^q\) coincides with the previously fixed embedding \(S^q \times S^q \rightarrow W_0\). Let \((\underline{M}{}'; L', V') = \theta_{U',F'}\). We have \(\partial U' = \partial U \approx \partial U_0 \# (S^q \times S^q)\), therefore \(\underline{M}{}'= \underline{M}\cong \underline{M}_0 \oplus \underline{H}_2\) and \(L' \cong L_0 \oplus (\mathbb{Z}\times \{ 0 \})\). Moreover, \(W \setminus \mathop{\mathrm{int}}\nolimits U = W_0 \setminus \mathop{\mathrm{int}}\nolimits U_0 = W' \setminus \mathop{\mathrm{int}}\nolimits U'\), so \(V = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) \rightarrow H_q(\partial U)) = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(W' \setminus \mathop{\mathrm{int}}\nolimits U', \partial U' \sqcup M_0) \rightarrow H_q(\partial U')) = V'\). By the definition of the equivalence relation \(\sim\) (see Definition 55), we have \(\theta_{U,F} \cong (\underline{M}_0 \oplus \underline{H}_2; L_0 \oplus (\{ 0 \} \times \mathbb{Z}), V) \sim (\underline{M}_0 \oplus \underline{H}_2; L_0 \oplus (\mathbb{Z}\times \{ 0 \}), V) \cong \theta_{U',F'}\). Therefore \(\theta_{W,F} \sim \theta_{W',F'}\). ◻

Proof. If \(\theta_{W,F} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{M}\oplus \underline{H}_2; L \oplus ( \{0 \} \times \mathbb{Z}), V)\) and \(x \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{M}\oplus \underline{H}_2; L \oplus (\mathbb{Z}\times \{ 0 \}), V)\) for some \(\underline{M}\), \(L\) and \(V\), then we can apply Lemma 120 below to get a suitable \(F'\).

The case \(\theta_{W,F} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{M}\oplus \underline{H}_2; L \oplus (\mathbb{Z}\times \{ 0 \}), V)\), \(x \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{M}\oplus \underline{H}_2; L \oplus ( \{0 \} \times \mathbb{Z}), V)\) can be reduced to the previous one by applying the automorphism \(\mathop{\mathrm{id}}\nolimits_M \oplus \sigma\) of \(\underline{M}\oplus \underline{H}_2\) (where \(\sigma : \mathbb{Z}^2 \rightarrow \mathbb{Z}^2\) is the flip map).

In general, \(\theta_{W,F}\) is related to \(x\) by a sequence of the above elementary equivalences, so we can get an \(F' : W' \rightarrow B\) by a repeated application of Lemma 120. ◻

Proof. First we construct a normal bordism \(F''\) by applying surgery to \(F\).

By Proposition 117 there is a CW-decomposition \(c\) of \(W\) and an integer \(k \geq 0\) such that \(\theta_{c,F} \cong \mathcal{H}_{2k} \oplus (\underline{M}_0 \oplus \underline{H}_2; L_0 \oplus ( \{0 \} \times \mathbb{Z}), V)\). To simplify notation let \(\underline{\bar{M}}_0 = \underline{H}_{2k} \oplus \underline{M}_0\), \(\bar{L}_0 = ( \{0 \} \times \mathbb{Z}^k) \oplus L_0\) and \(\bar{V} = (\mathbb{Z}^k \times \{0 \} ) \oplus V\). Let \(\bar{M}_0\) denote the underlying group of \(\underline{\bar{M}}_0\). Let \(K\) denote the \(q\)-skeleton of \(W\) with respect to \(c\) and let \(U\) be a regular neighbourhood of \(K\), then by definition \(\theta_{c,F}\) is the isomorphism class of \(\theta_{U,F}\). We fix an isomorphism between \(\theta_{U,F}\) and \((\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus ( \{0 \} \times \mathbb{Z}), \bar{V})\).

Let \(\alpha \in H_q(W)\) be the image of \((0,(1,0)) \in \bar{M}_0 \oplus \mathbb{Z}^2\) under the homomorphism \(H_q(\partial U) \rightarrow H_q(W)\) induced by the inclusion. Since \((0,(1,0))\) is in the hyperbolic component of \(H_q(\partial U)\), we have \(H_q(F \big| _{\partial U})(0,(1,0)) = 0\) and hence \(H_q(F)(\alpha) = 0\).

Consider the commutative diagram \[\xymatrix{ H_{q+1}(B, W) \ar[r] & H_q(W) \ar[r]^-{H_q(F)} & H_q(B) \\ \pi_{q+1}(B, W) \ar[u]^-{\cong} \ar[r] & \pi_q(W) \ar[u] \ar[r]^-{\pi_q(F)} & \pi_q(B) \ar[u] }\] induced by the Hurewicz-homomorphisms (where we regard \(W\) as a subspace of \(B\) via \(F\)). The pair \((B,W)\) is \(q\)-connected, hence \(\pi_{q+1}(B, W) \cong H_{q+1}(B, W)\).

Since \(H_q(F)(\alpha) = 0\), \(\alpha\) is the image of some element in \(H_{q+1}(B, W)\). The image of the corresponding relative homotopy class in \(\pi_{q+1}(B, W)\) under the boundary homomorphism is an element \(\hat{\alpha} \in \pi_q(W)\), which is mapped to \(\alpha\) by the Hurewicz-homomorphism, and it follows from the exactness of the sequence that \(\pi_q(F)(\hat{\alpha}) = 0\).

Since \(q < \frac{1}{2} \dim W\), \(\hat{\alpha}\) can be represented by an embedding \(S^q \times D^{q+1} \rightarrow W\) along which we can perform a surgery over \((B, \xi)\) (see [1]). Let \(F'' : W'' \rightarrow B\) be the result of this surgery, it is a normal bordism between \(f_0\) and \(f_1\) which is normally bordant to \(F\) by construction. Since \(F\) is \(q\)-connected and we did surgery along a \(q\)-sphere, \(F''\) is \(q\)-connected too ([1]).

In the rest of the proof we will show that the obstruction \(\theta_{W'',F''}\) is stably isomorphic to \((\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus (\mathbb{Z}\times \{0 \} ), \bar{V})\), and hence to \((\underline{M}_0 \oplus \underline{H}_2; L_0 \oplus (\mathbb{Z}\times \{0 \} ), V)\). We start by applying Theorem 118, which describes the relationship between \(\theta_{W'',F''}\) and \(\theta_{W,F}\) (more precisely, between certain representatives thereof, which are obtained using specific CW-decompositions).

We construct CW-decompositions \(c'\) and \(c''\) of \(W\) and \(W''\) respectively from a CW-decomposition of \(W \setminus \mathop{\mathrm{int}}\nolimits(S^q \times D^{q+1})\) as in the proof of Theorem 118. Let \(K'\) denote the \(q\)-skeleton of \(W\) with respect to \(c'\), then \(K' \approx K'_0 \vee S^q\) for some \(K'_0\), so \(H_q(K') \cong H_q(K'_0) \oplus \mathbb{Z}\). Since \(S^q \subseteq K'\) represents the sphere along which we do surgery, the inclusion \(K' \rightarrow W\) induces a homomorphism that sends \((0,1) \in H_q(K') \cong H_q(K'_0) \oplus \mathbb{Z}\) to \(\alpha \in H_q(W)\). Let \(U'\) be a regular neighbourhood of \(K'\), then \(U'\) is of the form \(U'_0 \natural (S^q \times D^{q+1})\), and we can assume that the composition \(S^q \times D^{q+1} \rightarrow U' \rightarrow W\) coincides with the embedding used for the surgery. Therefore \(\theta_{U',F} \cong (\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus (\{ 0 \} \times \mathbb{Z}), V')\), where \(\underline{M}{}'_0 = (H_q(\partial U'_0), \lambda'_0, \mu'_0)\), \(\lambda'_0\) is the intersection form of \(\partial U'_0\), \(\mu'_0 = H_q(F \big| _{\partial U'_0})\) and \(L'_0 = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(U'_0, \partial U'_0) \rightarrow H_q(\partial U'_0))\). We also have \(\theta_{c'',F''} \cong (\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus ( \mathbb{Z}\times \{0 \}), V')\), as we saw in the proof of Theorem 118.

Now consider the following diagram: \[\xymatrix{ \theta_{W,F} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus (\{ 0 \} \times \mathbb{Z}), V') \ar@{<->}[r]^-{\sim} \ar@{<->}[d]_-{\mathrel{\underset{^{\scriptstyle{s}}}{\cong}}\;} & (\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus ( \mathbb{Z}\times \{0 \}), V') \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}\theta_{W'',F''} \\ \theta_{W,F} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus ( \{0 \} \times \mathbb{Z}), \bar{V}) \ar@{<->}[r]^-{\sim} & (\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus (\mathbb{Z}\times \{0 \} ), \bar{V}) }\] The top row summarises the information we just got from Theorem 118. The bottom row contains the data appearing in the statement of the lemma (stabilised by adding \(\mathcal{H}_{2k}\)). The stable isomorphism in the left column follows from the well-definedness of \(\theta_{W,F}\) (Theorem 112). Our goal is to prove that \((\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus ( \mathbb{Z}\times \{0 \}), V') \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus (\mathbb{Z}\times \{0 \} ), \bar{V})\).

The stable isomorphism between \((\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus (\{ 0 \} \times \mathbb{Z}), V')\) and \((\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus ( \{0 \} \times \mathbb{Z}), \bar{V})\) is given by an isomorphism \(I : \underline{H}_{2l} \oplus \underline{M}{}'_0 \oplus \underline{H}_2 \rightarrow \underline{H}_{2m} \oplus \underline{\bar{M}}_0 \oplus \underline{H}_2\) for some \(l,m\) that is compatible with the (stabilised) quasi-formations. We will prove that there exists an \(I\) that also determines a stable isomorphism between \((\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus ( \mathbb{Z}\times \{0 \}), V')\) and \((\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus (\mathbb{Z}\times \{0 \} ), \bar{V})\). The key to this is the observation that we have some control over the stable isomorphism produced by Theorem 112. This is due to the freedom in the choice of the isomorphism \(J\) in Lemma 115, which is used in Lemma 116 to construct a \(j\) with \(H_q(j)=J\), and related to \(I\) via the diagram in Lemma 108. So we will construct a suitable \(I\) by going through the proof of Theorem 112 again, this time choosing the isomorphism \(J\) more carefully.

The homomorphism \(H_q(K) \cong (\bar{M}_0 \oplus \mathbb{Z}^2) / (\bar{L}_0 \oplus ( \{0 \} \times \mathbb{Z})) \rightarrow H_q(W)\) is surjective, because \(K\) is a \(q\)-skeleton of \(W\) (and we used Lemma 98 for the isomorphism), and the image of (the coset of) \((0,(1,0))\) is by definition \(\alpha\), so there is an induced homomorphism \(H_q(K) / \left< [0,(1,0)] \right> \cong \bar{M}_0/\bar{L}_0 \rightarrow H_q(W) / \left< \alpha \right>\), which is also surjective.

The homomorphism \(H_q(K') \cong H_q(K'_0) \oplus \mathbb{Z}\rightarrow H_q(W)\) is surjective too, and the image of \((0,1)\) is \(\alpha\), so there is an induced homomorphism \(H_q(K'_0) \rightarrow H_q(W) / \left< \alpha \right>\), which is also surjective.

By Lemma 37 a) there are integers \(l, m \geq 0\) and an isomorphism \(\mathbb{Z}^l \oplus H_q(K'_0) \rightarrow \mathbb{Z}^m \oplus \bar{M}_0/\bar{L}_0\) which commutes with the maps to \(H_q(W) / \left< \alpha \right>\) (where the \(\mathbb{Z}^l\) and \(\mathbb{Z}^m\) components are mapped to \(0\)). We fix such an isomorphism.

Consider the homomorphisms \(\mathbb{Z}^l \oplus H_q(K'_0) \rightarrow H_q(W)\) and \(\mathbb{Z}^l \oplus H_q(K'_0) \cong \mathbb{Z}^m \oplus \bar{M}_0/\bar{L}_0 \rightarrow H_q(W)\) (where the \(\mathbb{Z}^l\) and \(\mathbb{Z}^m\) components are again mapped to \(0\)). After composing with the projection \(H_q(W) \rightarrow H_q(W) / \left< \alpha \right>\), these two homomorphisms become equal, therefore their difference is a homomorphism \(\mathbb{Z}^l \oplus H_q(K'_0) \rightarrow \left< \alpha \right>\). Since \(\left< \alpha \right>\) is cyclic and \(\mathbb{Z}^l \oplus H_q(K'_0)\) is free, this homomorphism factors as a composition \(\mathbb{Z}^l \oplus H_q(K'_0) \rightarrow \mathbb{Z}\rightarrow \left< \alpha \right>\), where the second homomorphism maps \(1\) to \(\alpha\). By adding the first homomorphism of this composition to the previously fixed isomorphism, we get a homomorphism \(\mathbb{Z}^l \oplus H_q(K'_0) \rightarrow \mathbb{Z}^m \oplus \bar{M}_0/\bar{L}_0 \oplus \mathbb{Z}\cong \mathbb{Z}^m \oplus H_q(K)\) which commutes with the maps to \(H_q(W)\). We can extend it to an isomorphism \(J : \mathbb{Z}^l \oplus H_q(K') \cong \mathbb{Z}^l \oplus H_q(K'_0) \oplus \mathbb{Z}\rightarrow \mathbb{Z}^m \oplus H_q(K) \cong \mathbb{Z}^m \oplus \bar{M}_0/\bar{L}_0 \oplus \mathbb{Z}\) by sending \((0,0,1)\) to \((0,0,1)\). This isomorphism \(J\) again commutes with the maps to \(H_q(W)\).

Let \(\tilde{c}\) be an \(m\)-fold stabilisation of \(c\) and \(\tilde{c}'\) be an \(l\)-fold stabilisation of \(c'\). Let \(\tilde{K}\) and \(\tilde{K}'\) denote the \(q\)-skeletons of \(W\) with respect to \(\tilde{c}\) and \(\tilde{c}'\), then \(H_q(\tilde{K}) \cong \mathbb{Z}^m \oplus H_q(K)\) and \(H_q(\tilde{K}') \cong \mathbb{Z}^l \oplus H_q(K')\). Moreover, \(J : H_q(\tilde{K}') \cong \mathbb{Z}^l \oplus H_q(K') \rightarrow H_q(\tilde{K}) \cong \mathbb{Z}^m \oplus H_q(K)\) is an isomorphism which commutes with the homomorphisms to \(H_q(W)\). By applying Lemma 116 as in the proof of Lemma 115, we get a homotopy equivalence \(j : \tilde{K}' \rightarrow \tilde{K}\) that commutes with the inclusions in \(W\) (up to homotopy) such that \(H_q(j)=J\).

Let \(\tilde{U}\) and \(\tilde{U}'\) be regular neighbourhoods of \(\tilde{K}\) and \(\tilde{K}'\) respectively. Since \(\tilde{c}\) and \(\tilde{c}'\) are stabilisations of \(c\) and \(c'\), we have \(\theta_{\tilde{U},F} \cong \mathcal{H}_{2m} \oplus \theta_{U,F}\) and \(\theta_{\tilde{U}',F} \cong \mathcal{H}_{2l} \oplus \theta_{U',F}\) by Lemma 114. Let \(\underline{\tilde{M}}_0 = \underline{H}_{2m} \oplus \underline{\bar{M}}_0\), \(\tilde{L}_0 = ( \{0 \} \times \mathbb{Z}^m) \oplus \bar{L}_0\) and \(\tilde{V} = (\mathbb{Z}^m \times \{0 \} ) \oplus \bar{V}\), then \(\theta_{\tilde{U},F} \cong (\underline{\tilde{M}}_0 \oplus \underline{H}_2; \tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z}), \tilde{V})\). Similarly, let \(\underline{\tilde{M}}{}'_0 = \underline{H}_{2l} \oplus \underline{M}{}'_0\), \(\tilde{L}'_0 = ( \{0 \} \times \mathbb{Z}^l) \oplus L'_0\) and \(\tilde{V}' = (\mathbb{Z}^l \times \{0 \} ) \oplus V'\), then \(\theta_{\tilde{U}',F} \cong (\underline{\tilde{M}}{}'_0 \oplus \underline{H}_2; \tilde{L}'_0 \oplus ( \{0 \} \times \mathbb{Z}), \tilde{V}')\).

We introduce the notation \((\tilde{M}_0, \tilde{\lambda}_0, \tilde{\mu}_0) = \underline{\tilde{M}}_0\) and \((\tilde{M}'_0, \tilde{\lambda}'_0, \tilde{\mu}'_0) = \underline{\tilde{M}}{}'_0\), and let \(\tilde{\lambda} = \tilde{\lambda}_0 \oplus \bigl[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr]\) and \(\tilde{\lambda}' = \tilde{\lambda}'_0 \oplus \bigl[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr]\) be the bilinear functions in \(\underline{\tilde{M}}_0 \oplus \underline{H}_2\) and \(\underline{\tilde{M}}{}'_0 \oplus \underline{H}_2\) respectively. By Lemma 98 we have \(H_q(\tilde{U}) \cong (\tilde{M}_0 \oplus \mathbb{Z}^2) / (\tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z}))\) and \(H_q(\tilde{U}') \cong (\tilde{M}'_0 \oplus \mathbb{Z}^2) / (\tilde{L}'_0 \oplus ( \{0 \} \times \mathbb{Z}))\), and the quotient homomorphisms \(Q : \tilde{M}_0 \oplus \mathbb{Z}^2 \rightarrow (\tilde{M}_0 \oplus \mathbb{Z}^2) / (\tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z}))\) and \(Q' : \tilde{M}'_0 \oplus \mathbb{Z}^2 \rightarrow (\tilde{M}'_0 \oplus \mathbb{Z}^2) / (\tilde{L}'_0 \oplus ( \{0 \} \times \mathbb{Z}))\) are induced by the inclusions \(\partial \tilde{U} \rightarrow \tilde{U}\) and \(\partial \tilde{U}' \rightarrow \tilde{U}'\).

By applying Lemma 108 to the homotopy equivalence \(j : \tilde{K}' \rightarrow \tilde{K}\), we get that \(\theta_{\tilde{U}',F} \cong \theta_{\tilde{U},F}\), via an isomorphism \(I\) that fits into the commutative diagram \[\!\!\!\! \xymatrix{ H_q(\tilde{K}') \cong \mathbb{Z}^l \oplus H_q(K'_0) \oplus \mathbb{Z}\ar[r]^-{\cong} \ar[d]_-{H_q(j) = J} & H_q(\tilde{U}') \cong (\tilde{M}'_0 \oplus \mathbb{Z}^2) / (\tilde{L}'_0 \oplus ( \{0 \} \times \mathbb{Z})) \ar[d] & H_q(\partial \tilde{U}') \cong \tilde{M}'_0 \oplus \mathbb{Z}^2 \ar[d]^-{I} \ar[l]_-{Q'} \\ H_q(\tilde{K}) \cong \mathbb{Z}^m \oplus \bar{M}_0/\bar{L}_0 \oplus \mathbb{Z}\ar[r]^-{\cong} & H_q(\tilde{U}) \cong (\tilde{M}_0 \oplus \mathbb{Z}^2) / (\tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z})) & H_q(\partial \tilde{U}) \cong \tilde{M}_0 \oplus \mathbb{Z}^2 \ar[l]_-{Q} }\] In particular, \(I\) is an isomorphism \(I : \underline{\tilde{M}}{}'_0 \oplus \underline{H}_2 \rightarrow \underline{\tilde{M}}_0 \oplus \underline{H}_2\) such that \(I(\tilde{L}'_0 \oplus ( \{0 \} \times \mathbb{Z})) = \tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z})\) and \(I(\tilde{V}') = I(\tilde{V})\).

It remains to prove that \(I(\tilde{L}'_0 \oplus ( \mathbb{Z}\times \{0 \})) = \tilde{L}_0 \oplus (\mathbb{Z}\times \{0 \} )\), implying that \(I\) determines an isomorphism between \((\underline{\tilde{M}}{}'_0 \oplus \underline{H}_2; \tilde{L}'_0 \oplus ( \mathbb{Z}\times \{0 \}), \tilde{V}')\) and \((\underline{\tilde{M}}_0 \oplus \underline{H}_2; \tilde{L}_0 \oplus (\mathbb{Z}\times \{0 \} ), \tilde{V})\), and hence a stable isomorphism between \((\underline{M}{}'_0 \oplus \underline{H}_2; L'_0 \oplus ( \mathbb{Z}\times \{0 \}), V')\) and \((\underline{\bar{M}}_0 \oplus \underline{H}_2; \bar{L}_0 \oplus (\mathbb{Z}\times \{0 \} ), \bar{V})\).

Under the identifications \(H_q(\tilde{U}') \cong H_q(\tilde{K}')\) and \(H_q(\tilde{U}) \cong H_q(\tilde{K})\), the isomorphism \(J\) maps \([0,(1,0)]\), the coset of \((0,(1,0)) \in \tilde{M}'_0 \oplus \mathbb{Z}^2\) (which corresponds to \((0,0,1) \in \mathbb{Z}^l \oplus H_q(K'_0) \oplus \mathbb{Z}\)) to \([0,(1,0)]\), the coset of \((0,(1,0)) \in \tilde{M}_0 \oplus \mathbb{Z}^2\) (that is \((0,0,1) \in \mathbb{Z}^m \oplus \bar{M}_0/\bar{L}_0 \oplus \mathbb{Z}\)) by construction. Hence \(Q \circ I(0,(1,0)) = J \circ Q'(0,(1,0)) = J([0,(1,0)]) = [0,(1,0)]\), so \(I(0,(1,0)) \in Q^{-1}([0,(1,0)])\), therefore \(I(0,(1,0)) = (a,(1,b))\) for some \(a \in \tilde{L}_0\) and \(b \in \mathbb{Z}\).

Since \(I\) is an isomorphism of extended quadratic forms, we have \(0 = \tilde{\lambda}'((0,(1,0)),(0,(1,0))) = \tilde{\lambda}((a,(1,b))(a,(1,b))) = \tilde{\lambda}_0(a,a)+2b = 2b\) (because \(a \in \tilde{L}_0\) and \(\tilde{L}_0\) is a T-lagrangian in \(\underline{\tilde{M}}_0\)), hence \(b=0\). Therefore \(I(0,(1,0)) = (a,(1,0))\) for some \(a \in \tilde{L}_0\).

In \(\underline{\tilde{M}}{}'_0 \oplus \underline{H}_2\) we have \((\mathbb{Z}\times \{0 \})^{\perp} = \tilde{M}'_0 \oplus (\mathbb{Z}\times \{0 \})\), so \(\tilde{L}'_0 = (\tilde{L}'_0 \oplus ( \{0 \} \times \mathbb{Z})) \cap (\mathbb{Z}\times \{0 \})^{\perp}\). Hence its image under the isomorphism \(I\) is \(I(\tilde{L}'_0) = I(\tilde{L}'_0 \oplus ( \{0 \} \times \mathbb{Z})) \cap I((\mathbb{Z}\times \{0 \})^{\perp}) = (\tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z})) \cap (I(\mathbb{Z}\times \{0 \}))^{\perp} = (\tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z})) \cap \left< (a,(1,0)) \right> ^{\perp}\). For \((x,(0,y)) \in \tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z})\) we have \(\tilde{\lambda}((a,(1,0))(x,(0,y))) = \tilde{\lambda}_0(a,x)+y = y\), so \((\tilde{L}_0 \oplus ( \{0 \} \times \mathbb{Z})) \cap \left< (a,(1,0)) \right> ^{\perp} = \tilde{L}_0\). Therefore \(I(\tilde{L}'_0) = \tilde{L}_0\).

This implies that \(I(\tilde{L}'_0 \oplus ( \mathbb{Z}\times \{0 \})) = \tilde{L}_0 \oplus (\mathbb{Z}\times \{0 \} )\), as required. ◻

Proof. Choose an arbitrary \(c\) and \(U\), we will prove that \(W\) is an h-cobordism if and only if \(\theta_{U,F}\) is elementary.

By Proposition 93 e) we have \(H_i(W, M_0) \cong 0\) for \(i \leq q-1\), and it also follows that if \(i \geq q+2\), then \(H_i(W, M_0) \cong H^{2q+1-i}(W, M_1) \cong 0\) (by Poincaré-duality and the universal coefficient formula). Since \(W\), \(M_0\) and \(M_1\) are all simply-connected, this means that \(W\) is an h-cobordism if and only if \(H_q(W, M_0) \cong H_{q+1}(W, M_0) \cong 0\).

Consider the homological long exact sequence of the triple \((W, W \setminus \mathop{\mathrm{int}}\nolimits U, M_0)\). By Proposition 93 d) we have \(H_q(W, W \setminus \mathop{\mathrm{int}}\nolimits U) \cong 0\). From the proof of Lemma 101 we get a commutative diagram \[\xymatrix{ H^{q-1}(M_1) \ar[rr]^-{0} \ar[d]_-{\cong} \ar[dr]^-{\cong} & & H^q(W, M_1) \\ H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) & H^q(W, U \sqcup M_1) \ar[l]_-{\cong} \ar[ru] }\] so the homomorphism \(H^q(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) \cong H^q(W, U \sqcup M_1) \rightarrow H^q(W, M_1)\) is trivial. By Poincaré-duality, the homomorphism \(H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \rightarrow H_{q+1}(W, M_0)\) is trivial too. So we have the following exact sequence (note that this sequence is a copy of the horizontal exact sequence in [1], but \(U\) is defined differently): \[0 \rightarrow H_{q+1}(W, M_0) \rightarrow H_{q+1}(W, W \setminus \mathop{\mathrm{int}}\nolimits U) \rightarrow H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \rightarrow H_q(W, M_0) \rightarrow 0\] It follows that \(W\) is an h-cobordism if and only if the map \(H_{q+1}(W, W \setminus \mathop{\mathrm{int}}\nolimits U) \rightarrow H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0)\) is an isomorphism.

This map can be written as a composition \(H_{q+1}(W, W \setminus \mathop{\mathrm{int}}\nolimits U) \cong H_{q+1}(U, \partial U) \rightarrow H_q(\partial U) \rightarrow H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0)\). Since \(H_{q+1}(U) \cong H_{q+1}(\mathop{\mathrm{sk}}\nolimits_q W) \cong 0\), the map \(H_{q+1}(U, \partial U) \rightarrow H_q(\partial U)\) is injective, and its image is by definition \(L\). By the exact sequence \[\xymatrix{ 0 \ar[r] & H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) \ar[r] & H_q(\partial U) \ar[r] & H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0) \ar[r] & 0 }\] of Theorem 102 and the definition of \(V\), the restriction of the map \(H_q(\partial U) \rightarrow H_q(W \setminus \mathop{\mathrm{int}}\nolimits U, M_0)\) to \(L\) is an isomorphism if and only if \(H_q(\partial U) \cong L \oplus V\).

Therefore \(W\) is an h-cobordism if and only if \(\theta_{U,F}\) is elementary. ◻

Proof. First suppose that \(F\) is normally bordant to an \(F' : W' \rightarrow B\) such that \(W'\) is an h-cobordism. In this case \(F'\) is \(q\)-connected (because the inclusion \(M_0 \rightarrow W'\) is a homotopy equivalence, and the composition \(M_0 \rightarrow W' \stackrel{F'}{\rightarrow} B\) is equal to \(f_0\), which is \(q\)-connected), so \(\theta_{W',F'}\) is defined. It follows from Proposition 121 that \(\theta_{W',F'}\) is elementary, hence \([\theta_{W',F'}]\) is elementary, and by Theorem 118 \([\theta_{W,F}] = [\theta_{W',F'}]\).

Now suppose that \([\theta_{W,F}]\) is elementary. This means that there is an elementary quasi-formation \(x \in \mathfrak{g}^T_{2q+1}(B,\xi)\) such that \(\theta_{W,F} \sim x\). By Theorem 119 there is a \(q\)-connected normal bordism \(F' : W' \rightarrow B\) which is normally bordant to \(F\) such that \(\theta_{W',F'} \mathrel{\underset{^{\scriptstyle{s}}}{\cong}}x\). Then \(\theta_{W',F'}\) is elementary, so \(W'\) is an h-cobordism by Proposition 121. ◻

Proof. Let \(d_1 : H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) \rightarrow H_q(\partial U)\) and \(d_2 : H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) \rightarrow H_q(M_0)\) denote the components of the boundary homomorphism \(d : H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0) \rightarrow H_q(\partial U \sqcup M_0) \cong H_q(\partial U) \oplus H_q(M_0)\).

By Theorem 102 \(d_1\) is injective and its image is \(V\). In the homological long exact sequence of the triple \((W, U \sqcup M_0, U)\) we have \(H_q(W, U) \cong 0\) (see Proposition 93 c)), therefore \(d_2\) is surjective. Let \(p = - d_2 \circ d_1^{-1} : V \rightarrow H_q(M_0)\), this is a surjective homomorphism.

Let \(X\) denote the image of \(\partial : H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_0 \sqcup M_1) \rightarrow H_q(\partial U \sqcup M_0 \sqcup M_1)\). Since \(\mathop{\mathrm{Im}}\nolimits d \leq X\), we have \(((-\lambda) \oplus \lambda_0) \big| _{\mathop{\mathrm{Im}}\nolimits d \times \mathop{\mathrm{Im}}\nolimits d} = 0\) and \((\mu \oplus \mu_0) \big| _{\mathop{\mathrm{Im}}\nolimits d} = 0\) by Lemma 91, and this implies that \(p\) is a morphism of extended quadratic forms. (For more details see [23].) ◻

Proof. We have \([\bar{\theta}_{W,F}] = [\underline{\bar{M}}; \bar{L}, \bar{V}]\) by definition. To compute it, first we will define a subgroup \(V' \leq M\) such that in \(\underline{\bar{M}}\) we have \(\bar{V}' = \bar{V}^{\perp}\). Then we will use the morphism \(\bar{p} : \underline{\bar{V}}\rightarrow \underline{\bar{E}}_0\) induced by the morphism \(p : \underline{V}\rightarrow \underline{E}_0\) from Lemma 125 and an analogous morphism \(\bar{p}' : \underline{\bar{V}}{}'\rightarrow -\underline{\bar{E}}{}^*_1\) to relate \(\underline{\bar{M}}\) to \(\underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1)\).

First we consider \(-W\), ie.\(W\) with the opposite orientation, this is an oriented cobordism between \(M_1\) and \(M_0\), ie.\(\partial(-W) = M_1 \sqcup (-M_0)\). After changing the orientations \(F\), \(f_0\) and \(f_1\) are still normal \((q{-}1)\)-smoothings over \((B,\xi)\), and the obstruction \(\theta_{-W,F}\) can be defined as before. The CW-decomposition \(c\) of \(W\) that we fixed earlier is also a CW-decomposition of \(-W\), and we can choose the same regular neighbourhood \(U\) of the \(q\)-skeleton, but the orientation of \(-W\) restricts to the negative orientation of \(U\). We will denote by \(\theta_{-U,F}\) the quasi-formation determined by \(-U\) (\(U\) with the opposite orientation) that represents \(\theta_{-W,F}\).

Let \((\underline{M}{}'; L', V') = \theta_{-U,F}\). Then \(\underline{M}{}'= E_q(-\partial U, F \big| _{\partial U}) = -\underline{M}{}^*= (H_q(\partial U), -\lambda, \mu)\) by Lemma 88. By definition we have \(L' = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(-U, -\partial U) \rightarrow H_q(-\partial U)) = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(U, \partial U) \rightarrow H_q(\partial U)) = L\) and \(V' = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(-W \setminus \mathop{\mathrm{int}}\nolimits(-U), -\partial U \sqcup M_1) \rightarrow H_q(-\partial U)) = \mathop{\mathrm{Im}}\nolimits(H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) \rightarrow H_q(\partial U))\). Note that \(M_1\) takes the place of \(M_0\), because \(V\) was defined using the positive boundary component of \(W\) (ie.\(M_0\)) and the positive boundary component of \(-W\) is \(M_1\).

Let \(\underline{\tilde{V}}{}'= (V', -\lambda \big| _{V' \times V'}, \mu \big| _{V'})\). By Lemma 125 there is a morphism of extended quadratic forms \(p' : \underline{\tilde{V}}{}'\rightarrow \underline{E}_1\) such that the underlying homomorphism \(p' : V' \rightarrow H_q(M_1)\) is surjective. Let \(\underline{V}{}'= (V', \lambda \big| _{V' \times V'}, \mu \big| _V') = -(\underline{\tilde{V}}{}')^*\), then \(p'\) is also a morphism \(p' : \underline{V}{}'= -(\underline{\tilde{V}}{}')^* \rightarrow -\underline{E}{}^*_1\).

We have \(p' = - d'_2 \circ (d'_1)^{-1}\), where \(d'_1\) and \(d'_2\) denote the components of the boundary homomorphism \(d' : H_{q+1}(W \setminus \mathop{\mathrm{int}}\nolimits U, \partial U \sqcup M_1) \rightarrow H_q(\partial U \sqcup M_1)\) as in the proof of Lemma 125. We also get that \(\mathop{\mathrm{Im}}\nolimits d' \leq X\), hence \(((-\lambda) \oplus \lambda_0 \oplus \lambda_1) \big| _{\mathop{\mathrm{Im}}\nolimits d \times \mathop{\mathrm{Im}}\nolimits d'} = 0\). This implies that \(\lambda(x,y) = 0\) for every \(x \in V\), \(y \in V'\).

Now we consider \(\underline{\bar{M}}\). Since \(V\) is free, we have \(\underline{\bar{V}}= \underline{V}\) by definition, and \(\bar{V} = \pi(V) \cong V\), therefore \(\underline{\bar{V}}\cong (\bar{V}, \bar{\lambda} \big| _{\bar{V} \times \bar{V}}, \bar{\mu} \big| _{\bar{V}})\). We will in fact identify these two extended quadratic forms, ie.we will regard \(\bar{V} \leq \bar{M}\) as the underlying group of \(\underline{\bar{V}}\) (instead of \(V\)). Similarly we have \(\underline{\bar{V}}{}'\cong (\bar{V}', \bar{\lambda} \big| _{\bar{V}' \times \bar{V}'}, \bar{\mu} \big| _{\bar{V}'})\).

The surjective morphism \(p : \underline{V}\rightarrow \underline{E}_0\) induces a surjective morphism \(\bar{p} : \underline{\bar{V}}\rightarrow \underline{\bar{E}}_0\). Since \(\bar{E}_0\) is free, there is a homomorphism \(\bar{i} : \bar{E}_0 \rightarrow \bar{V}\) such that \(\bar{p} \circ \bar{i} = \mathop{\mathrm{id}}\nolimits_{\bar{E}_0}\). This is a morphism of extended quadratic forms \(\bar{i} : \underline{\bar{E}}_0 \rightarrow \underline{\bar{V}}\), because \(\bar{p}\) is a morphism of extended quadratic forms. Similarly, \(p' : \underline{V}{}'\rightarrow -\underline{E}{}^*_1\) induces a surjective morphism \(\bar{p}' : \underline{\bar{V}}{}'\rightarrow -\underline{\bar{E}}{}^*_1\), which has a right inverse \(\bar{i}' : -\underline{\bar{E}}{}^*_1 \rightarrow \underline{\bar{V}}{}'\).

Let \(Z = p^{-1}(\mathop{\mathrm{Tor}}\nolimits H_q(M_0)) \leq V\), then \(\bar{Z} = \mathop{\mathrm{Ker}}\nolimits\bar{p} \leq \bar{V}\). Let \(\bar{Y} = \mathop{\mathrm{Im}}\nolimits\bar{i}\), then \(\bar{V} = \bar{Y} \oplus \bar{Z}\). We have \(\bar{\mu}(x) = \bar{\mu}_0(\bar{p}(x)) = \bar{\mu}_0(0) = 0\) and \(\bar{\lambda}(x,y) = \bar{\lambda}_0(\bar{p}(x), \bar{p}(y)) = 0\) for every \(x \in \bar{Z}\), \(y \in \bar{V}\). Therefore \(\bar{\lambda} \big| _{\bar{Z} \times \bar{Z}} = 0\) and \(\bar{\mu} \big| _{\bar{Z}} = 0\), and \(\underline{\bar{V}}= \underline{\bar{Y}}\oplus \underline{\bar{Z}}\), where \(\underline{\bar{Y}}= (\bar{Y}, \bar{\lambda} \big| _{\bar{Y} \times \bar{Y}}, \bar{\mu} \big| _{\bar{Y}})\) and \(\underline{\bar{Z}}= (\bar{Z},0,0)\). Moreover, \(\bar{i} : \underline{\bar{E}}_0 \rightarrow \underline{\bar{Y}}\) is an isomorphism of extended quadratic forms.

Similarly, let \(Z' = (p')^{-1}(\mathop{\mathrm{Tor}}\nolimits H_q(M_1)) \leq V'\) so that \(\bar{Z}' = \mathop{\mathrm{Ker}}\nolimits\bar{p}'\), and let \(\bar{Y}' = \mathop{\mathrm{Im}}\nolimits\bar{i}'\). If \(\underline{\bar{Y}}{}'= (\bar{Y}', \bar{\lambda} \big| _{\bar{Y}' \times \bar{Y}'}, \bar{\mu} \big| _{\bar{Y}'})\) and \(\underline{\bar{Z}}{}'= (\bar{Z}',0,0)\), then \(\underline{\bar{V}}{}'= \underline{\bar{Y}}{}'\oplus \underline{\bar{Z}}{}'\) and \(\bar{i}' : -\underline{\bar{E}}{}^*_1 \rightarrow \underline{\bar{Y}}{}'\) is an isomorphism of extended quadratic forms.

Since \(\lambda(x,y) = 0\) for every \(x \in V\), \(y \in V'\), we get that \(\bar{\lambda}(x,y) = 0\) for every \(x \in \bar{V}\), \(y \in \bar{V}'\), in particular \(\bar{V}' \leq \bar{V}^{\perp}\). Since \(\bar{V}'\) is a direct summand in \(\bar{M}\) and \(\mathop{\mathrm{rk}}\nolimits\bar{V}' = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits\underline{M}= \mathop{\mathrm{rk}}\nolimits\bar{V}^{\perp}\), this implies that \(\bar{V}' = \bar{V}^{\perp}\). Similarly \(\bar{V} = (\bar{V}')^{\perp}\).

Since \(\underline{\bar{V}}= \underline{\bar{Y}}\oplus \underline{\bar{Z}}\) and \(\bar{\lambda}(x,y) = 0\) for every \(x \in \bar{Z}\), \(y \in \bar{V}\), we have \(\bar{Z} \leq \bar{V} \cap \bar{V}^{\perp}\). On the other hand, \(\underline{\bar{Y}}\cong \underline{\bar{E}}_0\) is nonsingular, so \(\bar{Y} \cap \bar{Y}^{\perp} = \{ 0 \}\), hence \(\bar{Y} \cap \bar{V}^{\perp} = \{ 0 \}\). Therefore \(\bar{V} \cap \bar{V}^{\perp} = \bar{Z}\). Similarly, \(\bar{V}' \cap (\bar{V}')^{\perp} = \bar{Z}'\). Together with our earlier observations this means that \(\bar{Z} = \bar{V} \cap \bar{V}' = \bar{Z}'\). So we have \(\bar{V} = \bar{Y} \oplus \bar{Z}\), \(\bar{V}' = \bar{Y}' \oplus \bar{Z}\) and \(\bar{V} \cap \bar{V}' = \bar{Z}\), which implies that \(\bar{Y} \cap \bar{Y}' = \{ 0 \}\).

Let \(\bar{N} = \bar{Y} \oplus \bar{Y}'\) and \(\underline{\bar{N}}= (\bar{N}, \bar{\lambda} \big| _{\bar{N} \times \bar{N}}, \bar{\mu} \big| _{\bar{N}}) = \underline{\bar{Y}}\oplus \underline{\bar{Y}}{}'\). Then \(\bar{i} \oplus \bar{i}' : \underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1) \rightarrow \underline{\bar{N}}\) is an isomorphism, so \(\underline{\bar{N}}\) is nonsingular, hence by Lemma 17 we have \(\bar{M} = \bar{N} \oplus \bar{N}^{\perp}\). Let \(\bar{N}' = \bar{N}^{\perp}\) and \(\underline{\bar{N}}{}'= (\bar{N}', \bar{\lambda} \big| _{\bar{N}' \times \bar{N}'}, \bar{\mu} \big| _{\bar{N}'})\), then \(\underline{\bar{M}}= \underline{\bar{N}}\oplus \underline{\bar{N}}{}'\).

We already saw that \(\bar{\lambda} \big| _{\bar{Z} \times \bar{Z}} = 0\) and \(\bar{\mu} \big| _{\bar{Z}} = 0\). Also, \(\bar{Z}\) is a direct summand in \(\bar{V}\), hence in \(\bar{M}\), hence in \(\bar{N}'\). Finally, \(\mathop{\mathrm{rk}}\nolimits\bar{Z} = \mathop{\mathrm{rk}}\nolimits\bar{V} - \mathop{\mathrm{rk}}\nolimits\bar{Y} = \frac{1}{2}(\mathop{\mathrm{rk}}\nolimits\underline{\bar{M}}- \mathop{\mathrm{rk}}\nolimits\underline{\bar{N}}) = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits\underline{\bar{N}}{}'\). Therefore \(\bar{Z}\) is a lagrangian in \(\underline{\bar{N}}{}'\).

By assumption \(\bar{K}\) is a lagrangian in \(\underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1)\), so \((\bar{i} \oplus \bar{i}')(\bar{K})\) is a lagrangian in \(\underline{\bar{N}}\). Therefore \((\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}\) is a lagrangian in \(\underline{\bar{N}}\oplus \underline{\bar{N}}{}'= \underline{\bar{M}}\).

Now we can compute \([\bar{\theta}_{W,F}]\). \[\begin{align} \bar{\theta}_{W,F} &\mathrel{\underset{^{\scriptstyle{s}}}{\cong}}(\underline{\bar{M}}; \bar{L}, \bar{V}) \sim \\ &\sim (\underline{\bar{M}}; \bar{L}, (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}) \oplus (\underline{\bar{M}}; (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}, \bar{V}) = \\ &= (\underline{\bar{M}}; \bar{L}, (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}) \oplus (\underline{\bar{N}}\oplus \underline{\bar{N}}{}'; (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}, \bar{Y} \oplus \bar{Z}) = \\ &= (\underline{\bar{M}}; \bar{L}, (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}) \oplus (\underline{\bar{N}}; (\bar{i} \oplus \bar{i}')(\bar{K}), \bar{Y}) \oplus (\underline{\bar{N}}{}'; \bar{Z}, \bar{Z}) \sim \\ &\sim (\underline{\bar{M}}; \bar{L}, (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}) \oplus (\underline{\bar{N}}; (\bar{i} \oplus \bar{i}')(\bar{K}), \bar{Y}) \cong \\ &\cong (\underline{\bar{M}}; \bar{L}, (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}) \oplus (\underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1); \bar{K}, \bar{E}_0) \end{align}\] where we used Theorem 71, Lemma 69 and the isomorphism \(\bar{i} \oplus \bar{i}'\). Let \(x = [\underline{\bar{M}}; \bar{L}, (\bar{i} \oplus \bar{i}')(\bar{K}) \oplus \bar{Z}] \in L_{2q+1}(B,\xi)\), then \([\bar{\theta}_{W,F}] = x \oplus [\underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1); \bar{K}, \bar{E}_0]\). ◻

Proof of Theorem 2. First we apply surgery below the middle dimension to get a \(q\)-connected normal bordism \(F : W \rightarrow B\) which is normally bordant to \(F_0\) (see [1]). We have \(H_i(M_0) \cong H_i(M_1)\) for every \(i\) (by the assumptions that \(f_0\) and \(f_1\) are \(q\)-connected and \(E_q(M_0,f_0) \cong E_q(M_1,f_1)\), and Poincaré-duality), hence \(\chi(M_0)=\chi(M_1)\). In this setting the surgery obstruction \(\theta_{W,F} \in \mathscr{s}^T_{2q+1}(B,\xi)\) is defined. Let \([\bar{\theta}_{W,F}] \in \ell_{2q+1}(B,\xi)\) denote the element corresponding to \([\theta_{W,F}] \in \ell^T_{2q+1}(B,\xi)\) under the bijection of Proposition 59.

We define \(\underline{E}_i\), \(\bar{E}_i\) and \(\underline{\bar{E}}_i\) as before (using that \(H_q(B)\) is free). By assumption there is an isomorphism \(I : \underline{E}_0 \rightarrow \underline{E}_1\), it induces an isomorphism \(\bar{I} : \underline{\bar{E}}_0 \rightarrow \underline{\bar{E}}_1\). By Lemma 29 b) \(\Delta^*_{\bar{I}} \leq \bar{E}_0 \oplus \bar{E}_1\) is a lagrangian in \(\underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1)\). By Theorem 128 and Remark 129 we have \([\bar{\theta}_{W,F}] = [\underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1); \Delta^*_{\bar{I}}, \bar{E}_0]\).

We have \(\Delta^*_{\bar{I}} \oplus \bar{E}_0 = \bar{E}_0 \oplus \bar{E}_1\), because \(\Delta^*_{\bar{I}}\) is the anti-diagonal in \(\bar{E}_0 \oplus \bar{E}_1\) (with respect to the isomorphism \(\bar{I}\)). This means that \([\underline{\bar{E}}_0 \oplus (-\underline{\bar{E}}{}^*_1); \Delta^*_{\bar{I}}, \bar{E}_0]\), and hence \([\bar{\theta}_{W,F}]\) is elementary. Therefore \([\theta_{W,F}]\), the corresponding element of \(\ell^T_{2q+1}(B,\xi)\), is elementary too (see Remark 63). Then by Proposition 122 \(F\) is normally bordant to a normal bordism \(F' : W' \rightarrow B\) such that \(W'\) is an h-cobordism. ◻

Proof. For the existence of such a normal map we need to show that \(\nu_M : M \rightarrow BSO\) can be lifted to a map \(f : M \rightarrow BSO\left< q, A\right>\). Since \(M\) is \((q{-}1)\)-connected, it is homotopy equivalent to a CW complex whose \(q\)-skeleton is a wedge of \(q\)-spheres. So the restriction of \(\nu_M\) to the \(q\)-skeleton amounts to a choice of some elements in \(\pi_q(BSO)\), which are in \(\mathop{\mathrm{Im}}\nolimits(\pi_q(\nu_M))\). Since \(\mathop{\mathrm{Im}}\nolimits(\pi_q(\nu_M))\) is contained in \(A = \mathop{\mathrm{Im}}\nolimits(\pi_q(g) : \pi_q(BSO\left< q, A\right>) \rightarrow \pi_q(BSO))\), this restriction can be lifted to \(BSO\left< q, A\right>\). The obstructions to lifting the map on the higher-dimensional cells are in the groups \(H^i(M;\pi_{i-1}(F))\) for \(i > q\), where \(F\) denotes the fibre of \(g : BSO\left< q, A\right> \rightarrow BSO\). By the definition of \(BSO\left< q, A\right>\) we have \(\pi_i(F) \cong 0\) for \(i \geq q\), so all of the obstructions vanish. Therefore there is a map \(f : M \rightarrow BSO\left< q, A\right>\) which lifts \(\nu_M\) (and hence it is covered by a bundle map). Moreover, \(\pi_i(M) \cong \pi_i(BSO\left< q, A\right>) \cong 0\) for \(i < q\), and if \(A = \mathop{\mathrm{Im}}\nolimits(\pi_q(\nu_M))\), then \(\pi_q(f)\) is surjective (because \(\mathop{\mathrm{Im}}\nolimits(\pi_q(g) \circ \pi_q(f)) = \mathop{\mathrm{Im}}\nolimits(\pi_q(\nu_M)) = A\) and \(\pi_q(g) : \pi_q(BSO\left< q, A\right>) \rightarrow A\) is an isomorphism), so \(f\) is \(q\)-connected.

For the uniqueness we need to show that the space of normal maps \(\nu_M \rightarrow \xi_{q,A}\) is path-connected. This space is homeomorphic to the space of bundle maps \(F(\nu_M) \rightarrow F(\xi_{q,A})\), where \(F(\nu_M)\) and \(F(\xi_{q,A})\) denote the oriented frame bundles of \(\nu_M\) and \(\xi_{q,A}\) respectively. Since \(F(\nu_M)\) and \(F(\xi_{q,A})\) are principal \(SO\)-bundles, a continuous map \(F(\nu_M) \rightarrow F(\xi_{q,A})\) is a bundle map if and only if it is \(SO\)-equivariant.

By [37] the space of \(SO\)-equivariant maps has the following equivalent description. There is a well-defined map \(F(\nu_M) \times_{SO} F(\xi_{q,A}) \rightarrow F(\nu_M) / SO\) given by \([x,y] \mapsto [x]\), and it is the projection of a locally trivial bundle over \(F(\nu_M) / SO = M\) with fibre \(F(\xi_{q,A})\) (which is associated to the principal \(SO\)-bundle \(F(\nu_M)\)). A section \(s : M \rightarrow F(\nu_M) \times_{SO} F(\xi_{q,A})\) determines a well-defined \(SO\)-equivariant map \(\bar{f}_s : F(\nu_M) \rightarrow F(\xi_{q,A})\) given by \(\bar{f}_s(x)=y\) whenever \([x,y] \in \mathop{\mathrm{Im}}\nolimits(s)\). This correspondence is a homeomorphism between the space of sections of \(F(\nu_M) \times_{SO} F(\xi_{q,A})\) and the space of \(SO\)-equivariant maps \(F(\nu_M) \rightarrow F(\xi_{q,A})\).

So it is enough to show that the space of sections of \(F(\nu_M) \times_{SO} F(\xi_{q,A})\) is path-connected. Given any two sections, the obstructions to the existence of a homotopy between them are in the groups \(H^i(\Sigma M;\pi_{i-1}(F(\xi_{q,A}))) \cong H^{i-1}(M;\pi_{i-1}(F(\xi_{q,A})))\). Since \(M\) is \((q{-}1)\)-connected, these groups vanish if \(i \leq q\). Hence it is enough to show that \(\pi_i(F(\xi_{q,A})) \cong 0\) for \(i \geq q\).

Since \(\xi_{q,A}\) is the pullback of the universal oriented vector bundle \(\gamma \rightarrow BSO\), there is a commutative diagram between the homotopical long exact sequences of the associated principal \(SO\)-bundles: \[\xymatrix{ \pi_{i+1}(BSO\left< q, A\right>) \ar[r] \ar[d] & \pi_i(SO) \ar[r] \ar[d]_-{\cong} & \pi_i(F(\xi_{q,A})) \ar[r] \ar[d] & \pi_i(BSO\left< q, A\right>) \ar[r] \ar[d] & \pi_{i-1}(SO) \ar[d]_-{\cong} \\ \pi_{i+1}(BSO) \ar[r]^-{\cong} & \pi_i(SO) \ar[r] & \pi_i(F(\gamma)) \ar[r] & \pi_i(BSO) \ar[r]^-{\cong} & \pi_{i-1}(SO) }\] Here \(F(\gamma)=ESO\) is contractible, and by the definition of \(BSO\left< q, A\right>\) the map \(\pi_i(BSO\left< q, A\right>) \rightarrow \pi_i(BSO)\) is an isomorphism for \(i > q\) and injective for \(i=q\). Therefore by the \(4\)-lemma we have \(\pi_i(F(\xi_{q,A})) \cong 0\) for \(i \geq q\). ◻

Proof. The bordism class \(\eta_V(M)\) is represented by a normal map \(f : M \rightarrow BSO\left< q, A\right>\) and \(\eta_{2q}^A(\Sigma)\) is represented by a normal map \(f_0 : \Sigma \rightarrow BSO\left< q, A\right>\). The connected sum \(M' := M \# \Sigma\) is formed using embeddings \(D^{2q} \rightarrow M\) and \(D^{2q} \rightarrow \Sigma\). When composed with \(f\) and \(f_0\) respectively, these embeddings give homotopic maps \(D^{2q} \rightarrow BSO\left< q, A\right>\) (covered by bundle maps). The homotopy \(D^{2q} \times I \rightarrow BSO\left< q, A\right>\), together with the maps \(f \circ \pi_1 : M \times I \rightarrow BSO\left< q, A\right>\) and \(f_0 \circ \pi_1 : \Sigma \times I \rightarrow BSO\left< q, A\right>\) (where \(\pi_1\) denotes the projection to the first component), defines a normal bordism \(W := ((M \sqcup \Sigma) \times I) \cup (D^{2q} \times I) \rightarrow BSO\left< q, A\right>\) between \(f \sqcup f_0 : M \sqcup \Sigma \rightarrow BSO\left< q, A\right>\) and \(f' := f \# f_0 : M' \rightarrow BSO\left< q, A\right>\). This shows that \(f'\) represents \(\eta_V(M) + \eta_{2q}^A(\Sigma)\). By Proposition 131 \(f'\) also represents \(\eta_V(M')\), therefore \(\eta_V(M \# \Sigma) = \eta_V(M) + \eta_{2q}^A(\Sigma)\). ◻

Proof. Let \(V = M \setminus \mathop{\mathrm{int}}\nolimits D^{2q}\) and suppose that \(\Sigma \in I(M)\). Then \(M \# \Sigma \approx M\), so \(\eta_V(M \# \Sigma) = \eta_V(M)\). By Proposition 134, \(\eta_V(M \# \Sigma) = \eta_V(M) + \eta_{2q}^A(\Sigma)\), hence \(\eta_{2q}^A(\Sigma)=0\). ◻

Proof. Suppose that \(M,M' \in C^{2q}_V\) and \(\eta_V(M)=\eta_V(M')\). This means that the unique normal maps \(f : M \rightarrow BSO\left< q, A\right>\) and \(f' : M' \rightarrow BSO\left< q, A\right>\) are normally bordant. By Proposition 131 these maps are normal \((q{-}1)\)-smoothings. Since \(M,M' \in C^{2q}_V\), there are diffeomorphisms \(j : V \rightarrow M \setminus \mathop{\mathrm{int}}\nolimits D^{2q}\) and \(j' : V \rightarrow M' \setminus \mathop{\mathrm{int}}\nolimits D^{2q}\).

a) Since \(M \setminus \mathop{\mathrm{int}}\nolimits D^{2q} \approx V \approx M' \setminus \mathop{\mathrm{int}}\nolimits D^{2q}\), we have \(\chi(M)=\chi(M')\). So all conditions of [1] are satisfied, which implies that \(M \approx M'\).

b) By excision \(j\) and \(j'\), regarded as embeddings \(j : V \rightarrow M\) and \(j' : V \rightarrow M'\), are \((2q-1)\)-connected. In particular, \(H_q(j)\) and \(H_q(j')\) are isomorphisms. Since \(\partial V \approx S^{2q-1}\), the map \(H_q(V) \rightarrow H_q(V, \partial V)\) is an isomorphism, so there is an intersection form \(H_q(V) \times H_q(V) \rightarrow \mathbb{Z}\). The isomorphism \(H_q(j) : H_q(V) \rightarrow H_q(M)\) preserves the intersection form, because every homology class in \(H_q(M)\) has a representative in \(j(V) \subset M\). We get similarly that \(H_q(j') : H_q(V) \rightarrow H_q(M')\) also preserves the intersection form. Moreover, \(f \circ j\) and \(f' \circ j'\) are both normal maps \(V \rightarrow BSO\left< q, A\right>\), so by Proposition 131 they are homotopic. This implies that \(H_q(f \circ j) = H_q(f' \circ j')\), hence \(H_q(f) = H_q(f') \circ H_q(j') \circ H_q(j)^{-1}\). Therefore \(H_q(j') \circ H_q(j)^{-1} : H_q(M) \rightarrow H_q(M')\) is an isomorphism between the Q-forms \(E_q(f)\) and \(E_q(f')\).

We have \(H_q(BSO\left< q, A\right>) \cong \pi_q(BSO\left< q, A\right>) \cong A \leq \pi_q(BSO)\). If \(q \equiv 0\), \(4\) or \(6\) mod \(8\), then \(\pi_q(BSO)\) is free, so \(H_q(BSO\left< q, A\right>)\) is also free. So in these cases all conditions of Theorem 2 are satisfied, therefore \(M \approx M'\). ◻

Proof. Let \(V = M \setminus \mathop{\mathrm{int}}\nolimits D^{2q}\) and suppose that \(\Sigma \in \mathop{\mathrm{Ker}}\nolimits(\eta_{2q}^A)\). Then \(\eta_V(M) = \eta_V(M) + \eta_{2q}^A(\Sigma) = \eta_V(M \# \Sigma)\) by Proposition 134. By Proposition 136 a) or b) this implies that \(M \approx M \# \Sigma\). ◻

Proof of Theorem 8. We have \(I(M) \leq \mathop{\mathrm{Ker}}\nolimits(\eta_{2q}^A)\) by Proposition 135 and \(I(M) \geq \mathop{\mathrm{Ker}}\nolimits(\eta_{2q}^A)\) by Theorem 137 b). ◻

Proof of Theorem 4. Suppose that \(q = 4\) or \(8\), and let \(M\) be a \((q{-}1)\)-connected \(2q\)-manifold. Let \(A = \mathop{\mathrm{Im}}\nolimits(\pi_q(\nu_M): \pi_q(M) \rightarrow \pi_q(BSO)) \leq \pi_q(BSO) \cong \mathbb{Z}\), then \(A = \alpha_M\mathbb{Z}\) for a unique non-negative integer \(\alpha_M\). By Theorems 8 and 7 \(I(M) = \mathop{\mathrm{Ker}}\nolimits(\eta_{2q}^A)\), which is \(0\) if \(4 | \alpha_M\) and \(\Theta_{2q}\) otherwise.

Let \(j = \frac{q}{4}\) and let \(d_M\) denote the divisibility of the Pontryagin class \(p_j(M) \in H^{4j}(M)\). By Bott-Milnor [38] we have (for any \((4j{-}1)\)-connected manifold) that \(d_M = a_j(2j{-}1)!\alpha_M\), where \(a_j=1\) if \(j\) is even and \(a_j=2\) if \(j\) is odd. Therefore \(4 | \alpha_M\) if and only if \(8 | d_M\) (for \(q=4\)) or \(24 | d_M\) (for \(q=8\)). ◻

Proof of Theorem 11. a) First suppose that \(g : N \rightarrow B\) represents an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f)\). Then there is a normal bordism between \(f\) and \(g\), and the proof of [1] shows that it can be turned into an s-cobordism by the compatible subtraction of some tori (and, since \(\chi(N)=\chi(M)\), the same number of tori are subtracted at both ends). The subtraction of one torus changes the Q-form of \(f\) or \(g\) by adding the hyperbolic form \(\underline{H}_2\), so by Remark 3 we have \(E_q(N,g) \oplus \underline{H}_{2k} \cong E_q(M,f) \oplus \underline{H}_{2k}\) for some \(k \geq 0\), ie.\(E_q(N,g)\) represents an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\). It also follows from Remark 3 that the isomorphism class of \(E_q(N,g)\) is independent of the choice of the representative \(g : N \rightarrow B\), so taking the Q-form induces a well-defined map \(E_q : \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f) \rightarrow \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\).

The argument from the proof of [13] shows that this map is surjective. Namely, suppose that \(\underline{E}\) represents an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\), ie.\(\underline{E}\oplus \underline{H}_{2k} \cong E_q(M,f) \oplus \underline{H}_{2k}\) for some \(k \geq 0\). By performing \(k\) trivial \((q{-}1)\)-surgeries on \(f\) we get a normal \((q{-}1)\)-smoothing with Q-form \(E_q(M,f) \oplus \underline{H}_{2k}\), and after performing \(k\) further \(q\)-surgeries along a basis of a lagrangian in the \(\underline{H}_{2k}\) component of \(\underline{E}\oplus \underline{H}_{2k}\) (which has been identified with \(E_q(M,f) \oplus \underline{H}_{2k}\)), we get a \(g : N \rightarrow B\) representing an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f)\), with \(E_q(N,g) \cong \underline{E}\).

Finally suppose that \(H_q(B)\) is free. If \(g : N \rightarrow B\) and \(g' : N' \rightarrow B\) represent elements of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f)\), then they are normally bordant. So by Theorem 2, if \(E_q(N,g) \cong E_q(N',g')\), then \(g\) and \(g'\) represent the same element. This means that the map \(E_q : \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f) \rightarrow \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\) is also injective.

b) Suppose that \(h \in A^{(B,\xi)}_{[M,f]}\) and \((X,\lambda,\mu)\) represents an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\). We define the action of \(h\) on \((X,\lambda,\mu)\) to be given by \((X,\lambda,h \circ \mu)\). The isomorphism class of \((X,\lambda,h \circ \mu)\) is determined by that of \((X,\lambda,\mu)\), so we only need to check that it is in \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\). By the definition of \(A^{(B,\xi)}_{[M,f]}\), the automorphism \(h\) of \(H_q(B)\) is induced by some \(\bar{h} \in \mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]}\). By the surjectivity of \(E_q\), we have \((X,\lambda,\mu) \cong E_q(N,g)\) for some \(g : N \rightarrow B\) representing an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f)\). Recall that \(\mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]}\) acts on \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f)\), in particular \(\bar{h}\) sends \(g\) to \(\bar{h} \circ g : N \rightarrow B\). Therefore \((X,\lambda,h \circ \mu) \cong E_q(N,\bar{h} \circ g)\) represents an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\), as required.

In fact, we can define an action of \(\mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]}\) on \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\), where an element \(\bar{h} \in \mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]}\) acts via the action of its induced automorphism \(h \in A^{(B,\xi)}_{[M,f]}\). Then we have \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f)) / \mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]} = \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f)) / A^{(B,\xi)}_{[M,f]}\). Moreover, \(E_q : \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f) \rightarrow \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\) is \(\mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]}\)-equivariant. Hence it induces a map \[\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,f) / \mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]} \cong \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M) \rightarrow \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f)) / \mathop{\mathrm{Aut}}\nolimits(B,\xi)_{[M,f]} = \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f)) / A^{(B,\xi)}_{[M,f]}\] Since \(E_q\) is surjective in general, and bijective if \(H_q(B)\) is free, the same is true for the induced map. ◻

Proof. a) In this case \(\mathop{\mathrm{Ker}}\nolimits\mu = 0\) and \(0^{\perp}=M\).

b) Any isomorphism \(\underline{M}\rightarrow \underline{M}{}'\) restricts to an isomorphism \(\mathop{\mathrm{Ker}}\nolimits\mu \rightarrow \mathop{\mathrm{Ker}}\nolimits\mu'\), and hence an isomorphism \((\mathop{\mathrm{Ker}}\nolimits\mu)^{\perp} \rightarrow (\mathop{\mathrm{Ker}}\nolimits\mu')^{\perp}\), which preserves the restrictions of \(\lambda\) and \(\mu\).

c) Since \(\mu' = 0\), we have \(\mathop{\mathrm{Ker}}\nolimits(\mu \oplus \mu') = (\mathop{\mathrm{Ker}}\nolimits\mu) \oplus M'\), and since \(\underline{M}{}'\) is nonsingular, \(((\mathop{\mathrm{Ker}}\nolimits\mu) \oplus M')^{\perp} = ((\mathop{\mathrm{Ker}}\nolimits\mu) \oplus 0)^{\perp} \cap (0 \oplus M')^{\perp} = ((\mathop{\mathrm{Ker}}\nolimits\mu) \oplus 0)^{\perp} \cap (M \oplus 0) = (\mathop{\mathrm{Ker}}\nolimits\mu)^{\perp}\oplus 0 \leq M \oplus 0 \leq M \oplus M'\).

d) follows immediately from parts b) and c). ◻

Proof. If \(a=b=0\), then let \(\bar{a}=\bar{b}=1\), otherwise define \(\bar{a} = \frac{a}{\gcd(a,b)}\) and \(\bar{b} = \frac{b}{\gcd(a,b)}\). Then in all cases \(a = \bar{a} \gcd(a,b)\), \(b = \bar{b} \gcd(a,b)\), \(a\bar{b} = \bar{a}b = \mathop{\mathrm{lcm}}\nolimits(a,b)\) and \(\gcd(\bar{a},\bar{b}) = 1\).

If \(\underline{E}\oplus \underline{H}_{2k} \cong \underline{E}_{a,b} \oplus \underline{H}_{2k}\) for some \(k \geq 0\), then \(\underline{E}\) has rank \(2\), and its bilinear function has signature \(0\), hence it is indefinite. Moreover, it is even, so by the classification of indefinite forms (see [27]), it is isomorphic to \(\bigl[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr]\). Therefore \(\underline{E}\cong \underline{E}_{c,d}\) for some \(c,d \in \mathbb{Z}\).

For any pair of extended quadratic forms \((M,\lambda,\mu)\) and \((M',\lambda',\mu')\) over an abelian group \(Q\), if \((M,\lambda,\mu) \oplus \underline{H}_{2k} \cong (M',\lambda',\mu') \oplus \underline{H}_{2k}\) for some \(k \geq 0\), then \(\mathop{\mathrm{Im}}\nolimits(\mu)=\mathop{\mathrm{Im}}\nolimits(\mu') \leq Q\). In the case of \(\underline{E}_{a,b}\) we have \(\mathop{\mathrm{Im}}\nolimits[a,b] = a\mathbb{Z}+ b\mathbb{Z}= \gcd(a,b)\mathbb{Z}\leq \mathbb{Z}\), so we get that \(\gcd(a,b)=\gcd(c,d)\).

By Lemma 139 d) we have \(\kappa(\underline{E}_{a,b}) \cong \kappa(\underline{E}_{c,d})\). By Lemma 142 below, \(a=b=0 \Leftrightarrow \kappa(\underline{E}_{a,b}) \cong (0,0,0) \Leftrightarrow \kappa(\underline{E}_{c,d}) \cong (0,0,0) \Leftrightarrow c=d=0\), and in this case \(ab=0=cd\). Otherwise \(\kappa(\underline{E}_{a,b}) \cong (\mathbb{Z},2\bar{a}\bar{b},2\mathop{\mathrm{lcm}}\nolimits(a,b))\) and \(\kappa(\underline{E}_{c,d}) \cong (\mathbb{Z},2\bar{c}\bar{d},2\mathop{\mathrm{lcm}}\nolimits(c,d))\) (where \(\bar{c}\) and \(\bar{d}\) are defined analogously to \(\bar{a}\) and \(\bar{b}\)). An isomorphism between \((\mathbb{Z},2\bar{a}\bar{b},2\mathop{\mathrm{lcm}}\nolimits(a,b))\) and \((\mathbb{Z},2\bar{c}\bar{d},2\mathop{\mathrm{lcm}}\nolimits(c,d))\) sends \(1\) to either \(1\) or \(-1\), and since it preserves the bilinear function, we get that in both cases \(2\bar{a}\bar{b} = 2\bar{c}\bar{d}\). This implies that \(2ab = 2\bar{a}\bar{b}\gcd(a,b)^2 = 2\bar{c}\bar{d}\gcd(c,d)^2 = 2cd\), hence \(ab=cd\).

In the other direction, if \(c\) and \(d\) are integers such that \(\gcd(a,b)=\gcd(c,d)\) and \(ab=cd\), then also \(\mathop{\mathrm{lcm}}\nolimits(a,b) = \mathop{\mathrm{lcm}}\nolimits(c,d)\). Therefore it is enough to show that \(\underline{E}_{a,b} \oplus \underline{H}_2 \cong \underline{E}_{\mathop{\mathrm{lcm}}\nolimits(a,b),\gcd(a,b)} \oplus \underline{H}_2\).

Since \(\gcd(\bar{a},\bar{b}) = 1\), there are \(\alpha, \beta \in \mathbb{Z}\) such that \(\alpha\bar{a} + \beta\bar{b} = 1\). Note that \[\begin{bmatrix} \beta\bar{b}^2 & \alpha\bar{a}^2 & -\alpha\beta\bar{a}\bar{b} & \bar{a}\bar{b} \\ \alpha & \beta & \alpha\beta & -1 \\ \beta\bar{b} & -\beta\bar{a} & \beta^2\bar{b} & \bar{a} \\ -\alpha\bar{b} & \alpha\bar{a} & \alpha^2\bar{a} & \bar{b} \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} \beta\bar{b}^2 & \alpha & \beta\bar{b} & -\alpha\bar{b} \\ \alpha\bar{a}^2 & \beta & -\beta\bar{a} & \alpha\bar{a} \\ -\alpha\beta\bar{a}\bar{b} & \alpha\beta & \beta^2\bar{b} & \alpha^2\bar{a} \\ \bar{a}\bar{b} & -1 & \bar{a} & \bar{b} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}\] and \[[a,b,0,0] \begin{bmatrix} \beta\bar{b}^2 & \alpha & \beta\bar{b} & -\alpha\bar{b} \\ \alpha\bar{a}^2 & \beta & -\beta\bar{a} & \alpha\bar{a} \\ -\alpha\beta\bar{a}\bar{b} & \alpha\beta & \beta^2\bar{b} & \alpha^2\bar{a} \\ \bar{a}\bar{b} & -1 & \bar{a} & \bar{b} \end{bmatrix} = [\mathop{\mathrm{lcm}}\nolimits(a,b),\gcd(a,b),0,0]\] So if \(I : \mathbb{Z}^4 \rightarrow \mathbb{Z}^4\) is the homomorphism determined by the square matrix in the second equality, then \(I\) is a morphism \(\underline{E}_{\mathop{\mathrm{lcm}}\nolimits(a,b),\gcd(a,b)} \oplus \underline{H}_2 \rightarrow \underline{E}_{a,b} \oplus \underline{H}_2\). Moreover, since \(I^*(\underline{E}_{a,b} \oplus \underline{H}_2) = \underline{E}_{\mathop{\mathrm{lcm}}\nolimits(a,b),\gcd(a,b)} \oplus \underline{H}_2\) is nonsingular (and \(\mathop{\mathrm{rk}}\nolimits\underline{E}_{\mathop{\mathrm{lcm}}\nolimits(a,b),\gcd(a,b)} \oplus \underline{H}_2 = \mathop{\mathrm{rk}}\nolimits\underline{E}_{a,b} \oplus \underline{H}_2\)), \(I\) is an isomorphism. Therefore \(\underline{E}_{a,b} \oplus \underline{H}_2 \cong \underline{E}_{\mathop{\mathrm{lcm}}\nolimits(a,b),\gcd(a,b)} \oplus \underline{H}_2\), as required. ◻

Proof. If \(a=b=0\), then we have \(\mathop{\mathrm{Ker}}\nolimits[0,0] = \mathbb{Z}^2\) and \((\mathbb{Z}^2)^{\perp} = 0\). Otherwise, \(\mathop{\mathrm{Ker}}\nolimits[a,b] \cong \mathbb{Z}\), generated by \(\bigl( \begin{smallmatrix} \bar{b} \\ -\bar{a} \end{smallmatrix} \bigr)\), and \((\mathop{\mathrm{Ker}}\nolimits[a,b])^{\perp} \cong \mathbb{Z}\), generated by \(\bigl( \begin{smallmatrix} \bar{b} \\ \bar{a} \end{smallmatrix} \bigr)\). Choosing this as the single basis element of \((\mathop{\mathrm{Ker}}\nolimits[a,b])^{\perp}\), the restriction of \(\bigl[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr]\) is given by \(2\bar{a}\bar{b}\), and the restriction of \([a,b]\) is given by \(\bar{b}a + \bar{a}b = 2\mathop{\mathrm{lcm}}\nolimits(a,b)\). ◻

Proof. The hyperbolic form \(\underline{H}_2\) has only two lagrangians, \(\mathbb{Z}\times \{ 0 \}\) and \(\{ 0 \} \times \mathbb{Z}\), so for any \(I \in \mathop{\mathrm{Aut}}\nolimits(\underline{H}_2)\) we have \(I\bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \bigr), I\bigl( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \bigr) \in \{ \bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \bigr),\bigl( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \bigr),\bigl( \begin{smallmatrix} -1 \\ 0 \end{smallmatrix} \bigr),\bigl( \begin{smallmatrix} 0 \\ -1 \end{smallmatrix} \bigr)\}\). For each possible value of \(I\bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \bigr)\) there is a unique choice of \(I\bigl( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \bigr)\) such that \(I\) preserves the bilinear function of \(\underline{H}_2\). ◻

Proof. An automorphism \(I \in \mathop{\mathrm{Aut}}\nolimits(\mathbb{Z}^2)\) is an isomorphism \(\underline{E}_{c,d} \rightarrow \underline{E}_{a,b}\) if and only if it preserves the bilinear function \(\bigl[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr]\) (equivalently, \(I \in \mathop{\mathrm{Aut}}\nolimits(\underline{H}_2)\)) and \([c,d]=I^*([a,b])\) (ie.\([c,d] = [a,b] \circ I\)). Since every element of \(\mathop{\mathrm{Aut}}\nolimits(\underline{H}_2)\) is symmetric, the second condition can be replaced by \(\bigl( \begin{smallmatrix} c \\ d \end{smallmatrix} \bigr) = I \bigl( \begin{smallmatrix} a \\ b \end{smallmatrix} \bigr)\), that is, \(\bigl( \begin{smallmatrix} c \\ d \end{smallmatrix} \bigr)\) is the image of \(\bigl( \begin{smallmatrix} a \\ b \end{smallmatrix} \bigr)\) under the action of \(I\). ◻

Proof. First suppose that \(a=0\). If \(c,d \in \mathbb{Z}\), then \(\gcd(c,d)=\gcd(a,b)=b\) and \(cd=ab=0\) if and only if \(\bigl( \begin{smallmatrix} c \\ d \end{smallmatrix} \bigr) \in \{ \bigl( \begin{smallmatrix} 0 \\ b \end{smallmatrix} \bigr), \bigl( \begin{smallmatrix} b \\ 0 \end{smallmatrix} \bigr), \bigl( \begin{smallmatrix} 0 \\ -b \end{smallmatrix} \bigr), \bigl( \begin{smallmatrix} -b \\ 0 \end{smallmatrix} \bigr) \}\). By Propositions 141 and 144 \(\underline{E}\oplus \underline{H}_{2k} \cong \underline{E}_{0,b} \oplus \underline{H}_{2k}\) for some \(k \geq 0\) if and only if \(\underline{E}\cong \underline{E}_{0,b}\). Therefore \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{E}_{0,b})|=1\). Similarly \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{E}_{a,0})|=1\).

Next note that for \(c,d \in \mathbb{Z}\) we have \(|cd|=\gcd(c,d)^2\) if and only if \(|c|=|d|=\gcd(c,d)\). So if \(|a|=|b|\), then by Proposition 141 \(\underline{E}\oplus \underline{H}_{2k} \cong \underline{E}_{a,b} \oplus \underline{H}_{2k}\) for some \(k \geq 0\) if and only if \(\underline{E}\cong \underline{E}_{a,b}\) or \(\underline{E}\cong \underline{E}_{-a,-b}\). In both cases \(\underline{E}\cong \underline{E}_{a,b}\) by Proposition 144, showing that \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{E}_{a,b})|=1\).

If \(ab \neq 0\) and \(|a| \neq |b|\), then define \(\bar{a} = \frac{a}{\gcd(a,b)}\) and \(\bar{b} = \frac{b}{\gcd(a,b)}\). By Proposition 141 \(\underline{E}\oplus \underline{H}_{2k} \cong \underline{E}_{a,b} \oplus \underline{H}_{2k}\) for some \(k \geq 0\) if and only if \(\underline{E}\cong \underline{E}_{\bar{c}\gcd(a,b),\bar{d}\gcd(a,b)}\) for some \(\bar{c}, \bar{d} \in \mathbb{Z}\) such that \(\gcd(\bar{c},\bar{d})=1\) and \(\bar{c}\bar{d} = \bar{a}\bar{b}\). Equivalently, \(\underline{E}\cong \underline{E}_{\bar{c}\gcd(a,b),\mathop{\mathrm{lcm}}\nolimits(a,b)/\bar{c}}\) for some \(\bar{c} \in \mathbb{Z}\setminus \{ 0 \}\) such that \(\bar{c}\) divides \(\bar{a}\bar{b}\) and \(\gcd(\bar{c},\frac{\bar{a}\bar{b}}{\bar{c}})=1\). If the prime factorisation of \(|\bar{a}\bar{b}| = \frac{|ab|}{\gcd(a,b)^2}\) is \(|\bar{a}\bar{b}|=p_1^{k_1} \ldots p_r^{k_r}\), then \(\bar{c}\) satisfies the conditions if and only if \(\bar{c} = \epsilon \prod_{i \in I} p_i^{k_i}\) for some \(\epsilon \in \{ \pm 1 \}\) and \(I \subseteq \{ 1, \ldots , r \}\). So there are \(2^{r+1}\) possible values of \(\bar{c}\). By our assumptions \(|\bar{a}\bar{b}|>1\), so \(|\bar{c}| \neq \left| \frac{\bar{a}\bar{b}}{\bar{c}} \right|\) if \(\bar{c}\) satisfies the conditions, hence \(|\bar{c}\gcd(a,b)| \neq \left| \frac{\mathop{\mathrm{lcm}}\nolimits(a,b)}{\bar{c}} \right|\). Therefore \(\mathop{\mathrm{Aut}}\nolimits(\underline{H}_2)\) acts freely on the set of possible pairs \(\bigl( \begin{smallmatrix} \bar{c}\gcd(a,b) \\ \mathop{\mathrm{lcm}}\nolimits(a,b)/\bar{c} \end{smallmatrix} \bigr)\). So it follows from Proposition 144 that \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{E}_{a,b})|=\frac{2^{r+1}}{|\mathop{\mathrm{Aut}}\nolimits(\underline{H}_2)|} = 2^{r-1}\). ◻

Proof. Let \(R = \mathop{\mathrm{Tor}}\nolimits M\) and \(\underline{R}= (R, 0, 0)\), then \(\underline{M}\cong \underline{\bar{M}}\oplus \underline{R}\). If \(\underline{M}{}'\oplus \underline{H}_{2k} \cong \underline{\bar{M}}\oplus \underline{H}_{2k}\) for some \(k \geq 0\), then \((\underline{M}{}'\oplus \underline{R}) \oplus \underline{H}_{2k} \cong \underline{M}\oplus \underline{H}_{2k}\), and if \(\underline{M}{}'\cong \underline{M}{}''\), then \(\underline{M}{}'\oplus \underline{R}\cong \underline{M}{}''\oplus \underline{R}\). Therefore the assignment \(\underline{M}{}'\mapsto \underline{M}{}'\oplus \underline{R}\) induces a well-defined map \(F : \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{\bar{M}}) \rightarrow \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{M})\).

Next consider an isomorphism \(\underline{M}{}'\oplus \underline{H}_{2k} \cong \underline{M}\oplus \underline{H}_{2k}\) for some \(k \geq 0\) and \(\underline{M}{}'= (M', \lambda', \mu')\). It restricts to an isomorphism \(\mathop{\mathrm{Tor}}\nolimits(M' \oplus \mathbb{Z}^{2k}) \cong \mathop{\mathrm{Tor}}\nolimits M' \rightarrow \mathop{\mathrm{Tor}}\nolimits(M \oplus \mathbb{Z}^{2k}) \cong R\), and since it is a morphism of extended quadratic forms, \(\mu' \big| _{\mathop{\mathrm{Tor}}\nolimits M'} = 0\). So \(\underline{\bar{M}}{}'\) is defined, \(\underline{M}{}'\cong \underline{\bar{M}}{}'\oplus \underline{R}\), and there is an induced isomorphism \(\underline{\bar{M}}{}'\oplus \underline{H}_{2k} \cong \underline{\bar{M}}\oplus \underline{H}_{2k}\). If \(\underline{M}{}'\cong \underline{M}{}''\), then \(\underline{\bar{M}}{}'\cong \underline{\bar{M}}{}''\), so the assignment \(\underline{M}{}'\mapsto \underline{\bar{M}}{}'\) induces a well-defined map \(G : \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{M}) \rightarrow \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{\bar{M}})\), which is the inverse of \(F\). ◻

Proof. If \((M',\lambda',\mu') \oplus \underline{H}_{2k} \cong (M,\lambda,\mu) \oplus \underline{H}_{2k}\) for some \(k \geq 0\), then \((M',\lambda',\iota \circ \mu') \oplus \underline{H}_{2k} \cong (M,\lambda,\iota \circ \mu) \oplus \underline{H}_{2k}\), and if \((M',\lambda',\mu') \cong (M'',\lambda'',\mu'')\), then \((M',\lambda',\iota \circ \mu') \cong (M'',\lambda'',\iota \circ \mu'')\). Therefore the assignment \((M',\lambda',\mu') \mapsto (M',\lambda',\iota \circ \mu')\) induces a well-defined map \(\iota_* : \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,\lambda,\mu) \rightarrow \mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M,\lambda,\iota \circ \mu)\).

If \((M',\lambda',\mu') \oplus \underline{H}_{2k} \cong (M,\lambda,\iota \circ \mu) \oplus \underline{H}_{2k}\) for some \(k \geq 0\), then \(\mathop{\mathrm{Im}}\nolimits(\mu') = \mathop{\mathrm{Im}}\nolimits(\iota \circ \mu) \leq \mathop{\mathrm{Im}}\nolimits(\iota) \leq Q_2\). So if \(\iota\) is injective and \(\iota^{-1} : \mathop{\mathrm{Im}}\nolimits(\iota) \rightarrow Q_1\) denotes its inverse, then \((M',\lambda',\mu') = (M',\lambda',\iota \circ \iota^{-1} \circ \mu')\), showing that \(\iota_*\) is surjective. And if \((M',\lambda',\iota \circ \mu') \cong (M'',\lambda'',\iota \circ \mu'')\), then \((M',\lambda',\mu') = (M',\lambda',\iota^{-1} \circ \iota \circ \mu') \cong (M'',\lambda'',\iota^{-1} \circ \iota \circ \mu'') = (M'',\lambda'',\mu'')\), so \(\iota_*\) is injective. ◻

Proof. We need to show that if \(\underline{M}\oplus \underline{H}_{2k} \cong \underline{M}{}'\oplus \underline{H}_{2k}\) for some \(\underline{M}{}'= (M', \lambda', \mu')\) and \(k \geq 0\), then \(\underline{M}\cong \underline{M}{}'\). The isomorphism \(\underline{M}\oplus \underline{H}_{2k} \cong \underline{M}{}'\oplus \underline{H}_{2k}\) restricts to an isomorphism between \(\mathop{\mathrm{Ker}}\nolimits(\mu \oplus 0) = 0 \oplus \mathbb{Z}^{2k}\) and \(\mathop{\mathrm{Ker}}\nolimits(\mu' \oplus 0) = (\mathop{\mathrm{Ker}}\nolimits\mu') \oplus \mathbb{Z}^{2k}\), which is only possible if \(\mathop{\mathrm{Ker}}\nolimits\mu' \cong 0\), ie.\(\mu'\) is injective. Then by Lemma 139 a) and d) we have \(\underline{M}\cong \kappa(\underline{M}) \cong \kappa(\underline{M}{}') \cong \underline{M}{}'\). ◻

Proof. Since \(Q\) is free, we have \(\mu \big| _{\mathop{\mathrm{Tor}}\nolimits M} = 0\), and by Lemma 147 \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{M})| = |\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{\bar{M}})|\) (where \(\underline{\bar{M}}= (\bar{M}, \bar{\lambda}, \bar{\mu})\) is defined as in Definition 58), so we need to prove the corresponding statement for \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{\bar{M}})|\). By assumption there is an isomorphism \(\bar{M} \cong \mathbb{Z}^2\) under which \(\bar{\lambda}\) corresponds to \(\bigl[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr]\), and \(\bar{\mu}\) is surjective (hence \(\mathop{\mathrm{rk}}\nolimits Q \leq 2\)).

If \(Q \cong 0\), then \(\underline{\bar{M}}\cong (\mathbb{Z}^2, \bigl[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr], 0 )\). By applying Lemma 148 to the inclusion \(0 \rightarrow \mathbb{Z}\) and Proposition 145, we have \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{\bar{M}})| = |\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{E}_{0,0})| = 1\).

If \(Q \cong \mathbb{Z}^2\), then since \(\bar{\mu}\) is surjective, it is also injective, so by Lemma 149 \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{\bar{M}})| = 1\).

If \(Q \cong \mathbb{Z}\), then the basis \([x],[y]\) determines an isomorphism \(\underline{\bar{M}}\cong \underline{E}_{a,b}\). Since \(\bar{\mu}\) is surjective, \(\gcd(a,b)=1\), which also implies that \(|a|=|b|\) holds if and only if \(|ab|=1\). Therefore the remaining cases follow from the corresponding parts of Proposition 145. ◻

Proof. Since \(\mu \big| _{\mathop{\mathrm{Tor}}\nolimits M} = 0\), by the proof of Lemma 147 it is enough to consider the case when \(\underline{M}\) is free. If \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{M})|=1\), then any extended quadratic form that represents an element of \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(\underline{M})\) is isomorphic to \(\underline{N}\). Otherwise \(\mathop{\mathrm{rk}}\nolimits Q = 1\), and if \(Q\) is identified with \(\mathbb{Z}\), then \(\underline{M}\cong \underline{E}_{a,b}\) for some \(a,b \in \mathbb{Z}\). By Proposition 141 \(\underline{N}\cong \underline{E}_{c,d}\) for some \(c,d \in \mathbb{Z}\). The only non-trivial automorphism of \(\mathbb{Z}\) is \(-\mathop{\mathrm{id}}\nolimits_{\mathbb{Z}}\), and if \(h = -\mathop{\mathrm{id}}\nolimits_{\mathbb{Z}}\), then \((N, \lambda_N, h \circ \mu_N) \cong \underline{E}_{-c,-d}\). By Proposition 144 \(\underline{E}_{-c,-d} \cong \underline{E}_{c,d}\). ◻

Proof of Theorem 5. By Theorem 11 \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M)| = |\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f)) / A^{(B,\xi)}_{[M,f]}|\). By Proposition 151 any automorphism of \(H_q(B)\) that acts on \(\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))\), in particular any element of \(A^{(B,\xi)}_{[M,f]}\), acts trivially. Therefore \(|\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(M)| = |\mathcal{S}^{\mathop{\mathrm{st}}\nolimits}(E_q(M,f))|\), which was determined in Theorem 150. ◻