Proof. By 1, \(\widetilde{M}\) is necessarily spin. Hence, to show \(w_2(TM)=w_2(\nu M)+w_1(\nu M)^2=0\), it suffices to show that the normal \(1\)-type \((\pi,w_1,w_2)\) satisfies \(w_2=w_1^2\). This holds for \(\pi\cong \mathbb{Z}\) because \(H^2(\mathbb{Z};\mathbb{Z}/2) =0\).

Now let \(\pi\) be a finite cyclic group, necessarily of even order since \(M\) and \(M'\) are stably exotic and hence nonorientable. It follows from 1 [it:main-1] that \(w_1w_2=w_1^3 \in H^3(\pi;\mathbb{Z}/2) \cong \mathbb{Z}/2\). For such \(\pi\), cupping with the nontrivial element \(w_1\in H^1(\pi;\mathbb{Z}/2)\) induces an isomorphism \(H^2(\pi;\mathbb{Z}/2)\to H^3(\pi;\mathbb{Z}/2)\), and hence \(w_1w_2=w_1^3\) implies \(w_2=w_1^2\), as desired. ◻

Proof. Since \(TW\oplus \nu_W\) is trivial for any embedding \(W\hookrightarrow \mathbb{R}^n\), the Whitney sum formula implies that \(w_1(TW)=w_1(\nu_W)\), \(w_2(TW) = w_1(\nu_W)^2 + w_2(\nu_W)\), and \(w_2(\nu_W) = w_1(TW)^2 + w_2(TW)\). ◻

Proof. Again, the proof of teichner-signature?*Proposition 1, p. 750 was given in the orientable case. In teichner-star?*Section 2 it was explained how to adapt to the nonorientable case, provided there is a vector bundle \(E \to Y\) with \(w_i(E) = w_i \in H^i(Y;\mathbb{Z}/2)\) for \(i=1,2\). We explain how to perform the necessary adaptations in the general case of \(w_1 \neq 0\), without assuming the existence of such an \(E\).

More details on Teichner’s proof of teichner-signature?*Proposition 1 were given by Orson–Powell in Orson-Powell?*Section 5, so we explain how to perform the adaptation by referencing the Orson–Powell write up. In Orson-Powell?*Lemma 5.3, replace \(\operatorname{BSCAT}\) with \(\mathop{\mathrm{BO}}\), and \(\mathop{\mathrm{Sq}}^2_{w_2(v)}\) with \(\mathop{\mathrm{Sq}}^2_{w_1,w_2}\). To justify the latter replacement, note that on the third line of the displayed equations in the proof of Orson-Powell?*Lemma 5.3, the term \(U_n \cup \rho_n^*w_1(v_n) \cup \mathop{\mathrm{Sq}}^1(\rho_n^*(X))\) from the previous line is observed to be zero, because in that paper \(w_1(v_n)=0\). In the general case \(w_1(v_n) \neq 0\) and that term survives, so we are left with \(\mathop{\mathrm{Sq}}^2_{w_1,w_2}\) as asserted.

Further, in the proof of Orson-Powell?*Lemma 5.4, for \(0 \leq q \leq 2\), the terms \(E^2_{p,q}\) of the James spectral sequence are identified with the \(E^2_{p,q}\) terms in an Atiyah–Hirzebruch spectral sequence for a Thom space \(H_p(M(v);\pi^{\mathrm{st}}_q)\), where \(v \colon B \to \mathop{\mathrm{BO}}\) is a stable vector bundle over \(B\) (corresponding to \(B_Y\) in our setting). In the case that \(v\) is a nonorientable bundle, the Thom isomorphism theorem identifies this with \(H_p(B;(\pi^{\mathrm{st}}_q)^{w_1})\), where the coefficients are now twisted by \(w_1\) (recall that after identification of \(\pi_q^{\mathrm{st}}\) with \(\Omega_q^{\mathop{\mathrm{Spin}}}\), following our convention in 1 we cease to indicate the twisting in the notation). The rest of the proof in Orson-Powell?*Lemma 5.4 and Corollary 5.7 proceeds as given in that paper, with straightforward modifications, e.g.as already noted, insert \(\mathop{\mathrm{BO}}\) in place of \(\operatorname{BSCAT}\), and \(\mathop{\mathrm{Sq}}^2_{w_1,w_2}\) in place of \(\mathop{\mathrm{Sq}}^2_{w_2(v)}\), and then adapt the displayed computation at the end of the proof of Orson-Powell?*Corollary 5.7 to include the term \(w_1(v) \cup \mathop{\mathrm{Sq}}^1(f^*(u))\) on the second line, and subsequent terms arising from that one. ◻

Proof. First we prove [item:i]. Let \((\pi,w_1,w_2)\) be the triple determining \((B,\xi)\). Take a model for \(B\pi\) such that \((w_1,w_2) \colon B\pi \to K(\mathbb{Z}/2,1) \times K(\mathbb{Z}/2,2)\) is a fibration. Then by 1, the normal 1-type \(\xi \colon B \to \mathop{\mathrm{BO}}\) is fibre homotopy equivalent to the pullback \[\label{eqn:pullback-defining-B-xi} \begin{tikzcd}[column sep = large] B_{\pi} \ar[r] \ar[d,"\xi_{\pi}"] & B\pi \ar[d,"{(w_1,w_2)}"] \\ \mathop{\mathrm{BO}}\ar[r,"P_2(-)"] & K(\mathbb{Z}/2,1) \times K(\mathbb{Z}/2,2) \end{tikzcd}\tag{1}\] of \((w_1,w_2)\) along \(\mathop{\mathrm{BO}}\to P_2(\mathop{\mathrm{BO}})\simeq K(\mathbb{Z}/2,1)\times K(\mathbb{Z}/2,2)\) from 3. We have \(\Omega_4^{\mathop{\mathrm{Spin}}}\cong 16 \mathbb{Z}\) and \(\Omega_4^{\mathop{\mathrm{TopSpin}}}\cong 8 \mathbb{Z}\) via the signature. Since we assume that \(w_1\) is nontrivial, in the smooth and topological James spectral sequences we have \[E^2_{0,4} \cong H_0(\pi;\Omega_4^{\mathop{\mathrm{Spin}}})\cong \mathbb{Z}\otimes_{\mathbb{Z}\pi}16\mathbb{Z}^{w_1}\cong \mathbb{Z}/2 \text{ and } E^{2,\mathop{\mathrm{Top}}}_{0,4} \cong H_0(\pi;\Omega_4^{\mathop{\mathrm{TopSpin}}})\cong \mathbb{Z}\otimes_{\mathbb{Z}\pi}8\mathbb{Z}^{w_1}\cong \mathbb{Z}/2.\] These groups are generated by \(K3\) and \(E_8\), respectively.

Since \(K3\) is simply-connected, every map \(K3\to B\pi\) is null homotopic and a choice of \(\xi\)-structure is the same as the choice of a spin structure. Since \(K3\) has a unique spin structure for each choice of orientation, it remains to show that the choice of orientation does not affect the element. For this we use that \([K3]\) lies in the subgroup \(E^{\infty}_{0,4} \subseteq \Omega_4(\xi)\), which, as a quotient of \(E^2_{0,4} \cong \mathbb{Z}/2\), has order at most two. Thus \([K3] = -[K3]\) and so \([K3]\) represents a unique element of \(\Omega_4(\xi)\), as claimed. This proves [item:i].

For the proof of [item:ii] and [item:iii] we first show that \([K3]\) generates the kernel of \(\Omega_4(\xi)\to\Omega_4(\xi^{\mathop{\mathrm{Top}}})\). The map \(\Omega_i^{\mathop{\mathrm{Spin}}}\to \Omega_i^{\mathop{\mathrm{TopSpin}}}\) is an isomorphism for \(0\leq i\leq 3\), and we saw above that forgetful map \(\Omega_4^{\mathop{\mathrm{Spin}}} \to \Omega_4^{\mathop{\mathrm{TopSpin}}}\) is identified, via the signature, with the inclusion \(16\mathbb{Z}\to 8\mathbb{Z}\). It follows that the map \(H_0(\pi;\Omega_4^{\mathop{\mathrm{Spin}}}) \cong \mathbb{Z}/2 \to H_0(\pi;\Omega_4^{\mathop{\mathrm{TopSpin}}}) \cong \mathbb{Z}/2\) is trivial. To see this, let \(g \in \pi\) be such that \(w_1(g) =-1\). Then \[1 \otimes 16 \mapsto 1 \otimes 8 + 1 \otimes 8 = 1 \cdot e \otimes 8 + 1 \cdot g \otimes 8 = 1 \otimes 8 + 1 \otimes (-8) =0.\] Thus we see that the kernel of \(\Omega_4(\xi)\to\Omega_4(\xi^{\mathop{\mathrm{Top}}})\) is generated by \([K3]\in E^{\infty}_{0,4} \subseteq \Omega_4(\xi)\). It follows that \([K3] =0 \in \Omega_4(\xi)\) if and only if the forgetful map is injective. This shows the first parts of both [item:ii] and [item:iii]. The second part of [item:ii] now follows directly from 1 2.

For the second part of [item:iii], first note that \([D \# K3]\) is bordant to \([K3]\), which by [item:i] is a uniquely determined element, and thus \([D \# K3]\) represents a nontrivial element of \(\Omega_4(\xi)\). On the other hand \([D \# 11(S^2 \times S^2)]=0 \in \Omega_4(\xi)\). The choice of normal 1-smoothings correspond to an action of the group of homotopy automorphisms of \(\xi\) on \(\Omega_4(\xi)\). It is easy to see that this action preserves the trivial element of \(\Omega_4(\xi)\). Hence there are no choices of normal 1-smoothings for which \([D \# K3]\) and \([D \# 11(S^2 \times S^2)]\) are bordant in \(\Omega_4(\xi)\), and hence they are not stably diffeomorphic by 1.

Since \(K3\) is homeomorphic to \(E_8\#E_8\#3(S^2\times S^2)\) by Freedman’s classification of simply-connected \(4\)-manifolds Freedman-82? and \(D\) is nonorientable, we have homeomorphisms \[D \# K3\cong D\#E_8\#E_8\#3(S^2\times S^2)\cong D\#E_8\#\overline{E}_8\#3(S^2\times S^2)\cong D \# 11(S^2 \times S^2)\] as claimed. ◻

Proof. Let \(\xi_{\pi} \colon B_{\pi} \to \mathop{\mathrm{BO}}\) be the fibration from 3. We consider the bordism group \(\Omega_4(\xi_{\pi})\) and the associated James spectral sequence. We show that \(E^\infty_{0,4}=\mathbb{Z}/2\).

To see this claim, note that \(w_1\) determines a map \(w_1 \colon \pi \to \mathbb{Z}/2\). Take \(w_i \neq 0 \in H^i(B\mathbb{Z}/2;\mathbb{Z}/2)\) for \(i=1,2\), and define \(\xi_{\mathbb{Z}/2} \colon B_{\mathbb{Z}/2} \to \mathop{\mathrm{BO}}\) as in 3. Then since \(w_1^2=w_2\) for both \(\mathbb{Z}/2\) and \(\pi\), it follows that \(w_1 \colon \pi \to \mathbb{Z}/2\) determines a map of fibrations \((B_{\pi},\xi_{\pi})\) to \((B_{\mathbb{Z}/2},\xi_{\mathbb{Z}/2})\), and hence determines a map of bordism groups \(\Omega_4(\xi_{\pi}) \to \Omega_4(\xi_{\mathbb{Z}/2})\). By naturality of the James spectral sequence there is an associated homomorphism between the \(E^{\infty}_{0,4}\) terms. 11 [item:i] implies that \([K3]\) determines a well-defined class in both bordism classes. In addition we saw in the proof of 11 that \([K3]\) generates both (at this point, potentially trivial) \(E^{\infty}_{0,4}\) terms, and \(\Omega_4(\xi_{\pi}) \to \Omega_4(\xi_{\mathbb{Z}/2})\) sends \([K3] \mapsto [K3]\). Thus if \([K3]\) is nonzero in \(E^{\infty}_{0,4} \subseteq \Omega_4(\xi_{\mathbb{Z}/2})\), then certainly \([K3]\) is nonzero in \(E^{\infty}_{0,4} \subseteq \Omega_4(\xi_{\pi})\). This completes the proof of the claim.

Thus we assume that \(\pi = \mathbb{Z}/2\), and we show that \([K3] \neq 0 \in \Omega_4(\xi_{\mathbb{Z}/2})\). Since \(\Omega_0^{\mathop{\mathrm{Spin}}} \cong \mathbb{Z}\), \(\Omega_1^{\mathop{\mathrm{Spin}}} = \Omega_2^{\mathop{\mathrm{Spin}}}=\mathbb{Z}/2\), and \(\Omega_3^{\mathop{\mathrm{Spin}}}=0\), we have for \(E^2_{p,q} \cong H_p(B\mathbb{Z}/2;\Omega_q^{\mathop{\mathrm{Spin}}})\) that \[E^2_{0,4}=E^2_{2,2}=E^2_{3,1}=E^2_{4,0}=\mathbb{Z}/2 \text{ and } E^2_{1,3}=0.\] On the other hand, \(\Omega_4(\xi_{\mathbb{Z}/2})=\Omega_4^{\mathop{\mathrm{Pin}}^+}\cong\mathbb{Z}/16\) by and as we recalled in 8. Hence for order reasons \(E^\infty_{0,4}\) must equal \(\mathbb{Z}/2\) (as must \(E^\infty_{2,2}\), \(E^\infty_{3,1}\), and \(E^\infty_{4,0}\)). Since \([K3]\) generates \(E^{\infty}_{0,4}\), it follows that \([K3] \neq 0 \in \Omega_4(\xi_{\mathbb{Z}/2})\).

By the claim, \([K3] \neq 0 \in \Omega_4(\xi_{\pi})\). Then 11 [item:iii] completes the proof of 3. ◻

Proof of 3. Let \(K\) be a presentation \(2\)-complex for a presentation of \(G\) realising \(\mathrm{def}(G)\). Then the Euler characteristic of \(K\) is \(1-\mathrm{def}(G)\). By Wall thickenings?*Proposition 5.1, there exists a \(5\)-dimensional thickening \(W\) of \(K\) with \(w_i(\nu W)=w_i\) for \(i=1,2\). Let \(N:=\partial W\). Then \(N\) has normal \(1\)-type \((G,w_1,w_2)\). Since \(W \cup_N W\) is odd-dimensional, \(0= \chi(W \cup_N W) = 2\chi(W) -\chi(N)\), and hence \(N\) has Euler characteristic \(\chi(N) = 2(1-\mathrm{def}(G))= 2-2\mathrm{def}(G)\). Let \(M:=N\#(S^2\times S^2)\), which thus has Euler characteristic \(4-2\mathrm{def}(G)\).

To construct \(M'\), we will use a manifold \(A\) constructed by Akbulut in akbulut? that is homeomorphic but not stably diffeomorphic to \((S^1\widetilde{\times} S^3)\#(S^2\times S^2)\). The bordism group of the normal 1-type of \(A\) is \(\mathbb{Z}/2\), since \(E^2_{p,q} = H_p(S^1;\Omega_q^{\mathop{\mathrm{Spin}}})=0\) for \((p,q) = (1,3)\) and \(p>1\). Since \(A\) is not stably diffeomorphic to \((S^1\widetilde{\times} S^3)\#(S^2\times S^2)\), which represents the trivial element, \(A\) must be stably diffeomorphic to \((S^1\widetilde{\times} S^3)\#K3\). Choose an element \(g\) of \(G\) with \(w_1(g)=-1\) and consider \(A\) as an element of \(\Omega_4(\xi(G,w_1,w_2))\) by sending a generator of \(\pi_1(A)\cong \mathbb{Z}\) to \(g\). Perform a 1-surgery on \(N\#A\) that identifies \(g\) with the generator of \(\pi_1(A)\), to obtain the manifold \(M'\) whose fundamental group is again \(G\). Since \(A\) has Euler characteristic \(2\), \(N\#A\) has Euler characteristic \(2-2\mathrm{def}(G)\). The surgery changes the Euler characteristic by \(\chi(D^2 \times S^2) = 2\), so that \(M'\) has Euler characteristic \(4-2\mathrm{def}(G)\) as claimed. This shows [it:smallex1].

To show [it:smallex2], start with \(N\# (S^1\widetilde{\times} S^3)\#(S^2\times S^2)\) and perform a surgery that identifies \(g\) with the generator of \(\pi_1(S^1)\). This returns \(N\#(S^2\times S^2)=M\). Since \(A\) and \((S^1\widetilde{\times} S^3)\#(S^2\times S^2)\) are homeomorphic, it follows that \(M\) and \(M'\) are homeomorphic. Since \(A\) is stably diffeomorphic to \((S^1\widetilde{\times} S^3)\#K3\), it similarly follows that \(M'\) is stably diffeomorphic to \(N \# K3\). By 11 [item:iii], the manifolds \(N\#K3\) and \(N\#11(S^2\times S^2)\) are not stably diffeomorphic and hence \(M\) and \(M'\) are not stably diffeomorphic. This completes the proof of [it:smallex2].

Items [it:smallex4]–[it:smallex5] follow from similar properties of \(A\); that is, \(A\) is obtained from \((S^1\widetilde{\times} S^3)\#(S^2\times S^2)\) by a Gluck twist akbulut? and the orientation double covers of \(A\) and \((S^1\widetilde{\times} S^3)\#(S^2\times S^2)\) are diffeomorphic Torres-JKTR?. See torres?*Section 3.4 for a more detailed argument. ◻

Proof of 1 assuming 4.. Since [it:main-2] was already proved in 3, we only show [it:main-1] and [it:main-3]. If \(w_1^3\neq w_1w_2\), then \(d_3^{\pi}\) is nontrivial by 4. Hence \([K3]=0\in \Omega_4(\xi_{\pi})\) and thus no stably exotic \(4\)-manifolds with normal \(1\)-type \((\pi,w_1,w_2)\) exist by 11 [item:ii]. This implies [it:main-1].

Now assume that \(w_1^3=w_1w_2\) and that \(\pi\) has cohomological dimension at most \(3\). The differential \(d_3^{\pi}\) is trivial by 4. Since \(\pi\) has cohomological dimension at most \(3\), all higher differentials are trivial as well. It follows that \([K3]\neq 0\in \Omega_4(\xi_{\pi})\). The existence of stably exotic \(4\)-manifolds with normal \(1\)-type \((\pi,w_1,w_2)\) now follows from 11 [item:iii]. This completes the proof of [it:main-3]. ◻

Proof. We consider the map \(\Theta\colon \mathbb{R}P^\infty\to X\), with \(u_1:=\Theta^*v_1\neq 0\) the generator of \(H^1(\mathbb{R}P^\infty;\mathbb{Z}/2)\) and \(u_2:=\Theta^*v_2=0 \in H^2(\mathbb{R}P^\infty;\mathbb{Z}/2)\). Since \(X= K_1 \times K_2\) is a product of Eilenberg-Maclane spaces, this determines \(\Theta\) up to homotopy. We consider the fibration \(\xi_{\mathbb{R}P^\infty} \colon B_{\mathbb{R}P^\infty} \to \mathop{\mathrm{BO}}\) determined as in 3 by \(\mathbb{R}P^\infty\), \(u_1\), and \(u_2\). We have an induced map \(\Theta_* \colon \Omega_4(\xi_{\mathbb{R}P^\infty}) \to \Omega_4(\xi_X)\).

A 4-manifold or a 5-dimensional bordism \(W\) admits the structure of a normal bordism over \(\xi_{\mathbb{R}P^\infty}\) if and only if \(w_2(\nu_W) =0\), which holds if and only if \(\nu_W\) admits a \(\mathop{\mathrm{Pin}}^+\) structure, which in turn holds if and only if \(TW\) admits a \(\mathop{\mathrm{Pin}}^-\) structure by 7. We deduce that \(\Omega_4(\xi_{\mathbb{R}P^\infty})\cong\Omega_4^{\mathop{\mathrm{Pin}}^-}\), and then we recall that \(\Omega_4^{\mathop{\mathrm{Pin}}^-}\) vanishes by 8. In particular, \([K3]\) is trivial in \(\Omega_4(\xi_{\mathbb{R}P^\infty}) \cong \Omega_4^{\mathop{\mathrm{Pin}}^-}=0\).

We now deduce that the differential \(d_3^{\mathbb{R}P^{\infty}} \colon H_3(\mathbb{R}P^\infty;\mathbb{Z}/2)/\operatorname{Im}d_2^{\mathbb{R}P^{\infty}} \to H_0(\mathbb{R}P^\infty;\mathbb{Z}^{u_1}) \cong \mathbb{Z}/2\) is nontrivial. To do so, we show that \(E^2_{2,3}=E^3_{4,1}=E^2_{5,0}=0\) in the James spectral sequence computing \(\Omega_*(\xi_{\mathbb{R}P^\infty})\), and hence only \(d_3^{\mathbb{R}P^\infty}\) can possibly kill the \(E^*_{0,4}\) term; since \([K3]=0\) this term must be killed by one of \(d_3^{\mathbb{R}P^{\infty}}\), \(d_4^{\mathbb{R}P^{\infty}}\), or \(d_5^{\mathbb{R}P^{\infty}}\).

First, \(E^2_{2,3} = H_2(\mathbb{R}P^\infty;\Omega_3^{\mathop{\mathrm{Spin}}})=0\) because \(\Omega_3^{\mathop{\mathrm{Spin}}} =0\). Next, since \(u_2=0\), we have \[(\mathop{\mathrm{Sq}}^2+u_1\mathop{\mathrm{Sq}}^1)(u_1^2)=\mathop{\mathrm{Sq}}^2(u_1^2) + u_1 (\mathop{\mathrm{Sq}}^1u_1) u_1 + u_1^2(\mathop{\mathrm{Sq}}^1u_1) =u_1^4.\] Hence using 10 [prop:differentials-item-i] we can compute \(d_2^{\mathbb{R}P^{\infty}} \colon E_{4,1}^2 \to E^2_{2,2}\). We have \(E_{4,1}^2 \cong H_4(\mathbb{R}P^\infty;\Omega_1^{\mathop{\mathrm{Spin}}}) \cong \mathbb{Z}/2\), with generator dual to \(u_1^4\), and \(E_{2,2}^2 \cong H_2(\mathbb{R}P^\infty;\Omega_2^{\mathop{\mathrm{Spin}}}) \cong \mathbb{Z}/2\), with generator dual to \(u_1^2\). Thus by 10 [prop:differentials-item-i], \(d_2^{\mathbb{R}P^{\infty}}\) is an isomorphism and so \(E^3_{4,1}=0\). We also have \(E^2_{5,0}=H_5(\mathbb{R}P^\infty;\mathbb{Z}^-)=0\). Thus we have shown that \(E^2_{2,3}=E^3_{4,1}=E^2_{5,0}=0\), as asserted. It follows that the only possibly nontrivial differential that can hit \(E^k_{0,4}\), for some \(k \geq 2\), is the differential \(d_3^{\mathbb{R}P^{\infty}} \colon E^3_{3,2}\to E^3_{0,4}=\mathbb{Z}/2\). Since we showed that \(K3\) is null-bordant in \(\Omega_4(\xi_{\mathbb{R}P^{\infty}})\), this differential has to be nontrivial. By naturality of the James spectral sequence, the square \[\begin{tikzcd} H_3(\mathbb{R}P^\infty;\mathbb{Z}/2)/\operatorname{Im}d_2^{\mathbb{R}P^{\infty}}\ar[d,"d_3^{\mathbb{R}P^{\infty}}","\cong"']\ar[r,"\Theta_*"']&H_3(X;\mathbb{Z}/2)/\operatorname{Im}d_2^{X}\ar[d,"d_3^X"]\\ H_0(\mathbb{R}P^\infty;\mathbb{Z}^{u_1})\ar[r,"\cong",,"\Theta_*"'] & H_0(X;\mathbb{Z}^{v_1}) \end{tikzcd}\] commutes. Since \(d_3^{\mathbb{R}P^{\infty}}\) is nontrivial (in fact it is an isomorphism because both domain and codomain are order two), and the bottom horizontal map is an isomorphism (again between groups of order two), it follows that \(d_3^X\) is nontrivial, as required. ◻

Proof. By the Künneth theorem \[H_3(X;\mathbb{Z}/2)\cong H_3(K_2;\mathbb{Z}/2)\oplus \big(H_2(K_2;\mathbb{Z}/2)\otimes H_1(K_1;\mathbb{Z}/2)\big) \oplus H_3(K_1;\mathbb{Z}/2)\cong H_3(K_2;\mathbb{Z}/2)\oplus(\mathbb{Z}/2)^2.\] By naturality of the James spectral sequences, comparing those for \((B_{K_2},\xi_{K_2})\) and \((B_X,\xi_X)\) via the inclusion of the second factor \(\{*\}\times K_2 \hookrightarrow X\), there is a commutative diagram \[\begin{tikzcd} H_3(K_2;\mathbb{Z}/2)/\operatorname{Im}d_2^{K_2}\ar[d,"d_3^{K_2}"]\ar[r]&H_3(X;\mathbb{Z}/2)/\operatorname{Im}d_2^{X}\ar[d,"d_3^{X}"]\\ H_0(K_2;\mathbb{Z})\ar[r]&H_0(X;\mathbb{Z}^{v_1}). \end{tikzcd}\] Since \(H_0(K_2;\mathbb{Z})\cong \mathbb{Z}\), and \(H_3(K_2;\mathbb{Z}/2)\) is 2-torsion, \(d_3^{K_2}\) is trivial and hence \(d_3^X\) is trivial on the image of \(H_3(K_2;\mathbb{Z}/2)\). The quotient \(H_3(X;\mathbb{Z}/2)/H_3(K_2;\mathbb{Z}/2)\cong (\mathbb{Z}/2)^2\) is generated by the two classes dual to \(v_1v_2\) and \(v_1^3\).

Consider the map \(\Theta'\colon \mathbb{R}P^\infty\to X\) with \((\Theta')^*v_1\neq 0\) and \((\Theta')^*(v_2) = (\Theta')^*(v_1^2)\), i.e.\((\Theta')^*(v_2+v_1^2)=0\). Again since \(X\) is a product of Eilenberg-Maclane spaces, this determines \(\Theta'\) up to homotopy. Write \(u'_1:=(\Theta')^*v_1 \in H^1(\mathbb{R}P^\infty;\mathbb{Z}/2)\) and \(u'_2:=(\Theta')^*v_2 \in H^2(\mathbb{R}P^\infty;\mathbb{Z}/2)\). This gives rise to a fibration \(\xi'_{\mathbb{R}P^\infty} \colon B'_{\mathbb{R}P^\infty} \to \mathop{\mathrm{BO}}\) as in 3, with a corresponding bordism group \(\Omega_4(\xi'_{\mathbb{R}P^\infty})\) and a map \(\Omega_4(\xi'_{\mathbb{R}P^\infty}) \to \Omega_4(\xi_X)\). Since \(u_2'=(u_1')^2\), it follows from the proof of 3 that \([K3]\) is nontrivial in \(\Omega_4(\xi'_{\mathbb{R}P^\infty})\). In particular the differential \(d_3^{\mathbb{R}P^{\infty}} \colon H_3(\mathbb{R}P^{\infty};\mathbb{Z}/2)/\operatorname{Im}d_2^{\mathbb{R}P^{\infty}} \to H_0(\mathbb{R}P^{\infty};\mathbb{Z}^{u_1'})\) must be trivial. From the square \[\begin{tikzcd} H_3(\mathbb{R}P^\infty;\mathbb{Z}/2)/\operatorname{Im}d_2^{\mathbb{R}P^{\infty}}\ar[d,"d_3^{\mathbb{R}P^{\infty}}","0"']\ar[r,"\Theta_*'"'] & H_3(X;\mathbb{Z}/2)/\operatorname{Im}d_2^{X}\ar[d,"d_3^X"]\\ H_0(\mathbb{R}P^\infty;\mathbb{Z}^{u_1'})\ar[r,"\cong","\Theta_*'"'] & H_0(X;\mathbb{Z}^{v_1}), \end{tikzcd}\] we deduce that the right-down composition is trivial. Taking \(\mathop{\mathrm{Hom}}_{\mathbb{Z}/2}(-,\mathbb{Z}/2)\) duals, and using the map \(\mathop{\mathrm{ev}}\colon H^3(X;\mathbb{Z}/2) \xrightarrow{\cong} H_3(X;\mathbb{Z}/2)^*\) from universal coefficients, we see that the composition \[\mathbb{Z}/2 \cong H_0(X;\mathbb{Z}^{v_1})^* \xrightarrow{\mathop{\mathrm{ev}}^{-1} \circ (d_3^X)^*} H^3(X;\mathbb{Z}/2) \xrightarrow{(\Theta')^*} H^3(\mathbb{R}P^\infty;\mathbb{Z}/2)\] is trivial. Since \((\Theta')^*(v_1^3) = (u'_1)^3\) and \((\Theta')^*(v_1v_2) = u_1'u_2' = (u'_1)^3\) are nontrivial in \(H^3(\mathbb{R}P^{\infty};\mathbb{Z}/2)\), the differential \(d_3^X\) is neither dual to \(1 \mapsto v_1^3\) nor to \(1 \mapsto v_1v_2\). Since \(d_3^X\) is nontrivial by 2, there is one remaining option, and \(d_3^X\) must be dual to \(1 \mapsto v_1^3+v_1v_2\). ◻

Proof of 4. One more application of naturality yields the square: \[\begin{tikzcd}[column sep = large] H_3(\pi;\mathbb{Z}/2)/\operatorname{Im}d_2^\pi\ar[d,"d_3^{\pi}"]\ar[r,"{(w_1,w_2)_*}"] & H_3(X;\mathbb{Z}/2)/\operatorname{Im}d_2^X\ar[d,"d_3^X"]\\ H_0(\pi;\mathbb{Z}^{w_1})\ar[r,"\cong"] & H_0(X;\mathbb{Z}^{v_1}). \end{tikzcd}\] Since \(d_3^X\) is dual to \(v_1^3+v_1v_2\) by 3, and the map \(B\pi \to X = K(\mathbb{Z}/2,1) \times K(\mathbb{Z}/2,2)\) pulls the universal classes \(v_1 \in H^1(X;\mathbb{Z}/2)\) and \(v_2 \in H^2(X;\mathbb{Z}/2)\) back to \(w_1\) and \(w_2\) respectively, it follows that \(d_3^\pi\) is dual to \(1 \mapsto w_1^3 + w_1w_2\). ◻

Proof. The long exact sequence of homotopy groups for the fibration sequence 2 defining \(F\) yields \[0 \to \pi_1(F) \xrightarrow{u_*} \pi_1(X) \to 0,\] which shows that \(u\) induces an isomorphism \(\pi_1(F)\cong \pi_1(X)\cong \mathbb{Z}/2\).

Thus \(v_1:=u^*\iota_1\in H^1(F;\mathbb{Z}/2)\) is nontrivial. Since there is a space \(Y\) with cohomology classes \(a_i\in H^i(Y;\mathbb{Z}/2)\) for \(i=1,2\), such that \(a_1^3=a_1a_2\) and \(a_2\neq a_1^2\) (see 3 for an example) and \(F\) is the universal space with cohomology classes \(v_1,v_2\) such that \(v_1^3=v_1v_2\), we must have that \(v_2\neq v_1^2\).

The long exact sequence of homotopy groups for the fibration sequence defining \(F\) also shows that there is a short exact sequence \[\label{eqn:ses-for-pi2-F} 0\to \pi_3(K(\mathbb{Z}/2,3)) \cong \mathbb{Z}/2\to \pi_2(F)\to \pi_2(X) \cong \mathbb{Z}/2\to 0.\tag{4}\] Let \(i_2\colon K_2\to X\) be the factor inclusion. Then \(i_2^*(\iota_1)=0\), so \(i_2^*(\iota_1^3+\iota_1\iota_2)=0\), and hence \(i_2\) admits a lift to \(F\). This implies that the sequence 4 splits, and hence \[\pi_2(F)\cong \mathbb{Z}/2\oplus\mathbb{Z}/2\] as an abelian group. Similarly, let \(s\colon K_1\to X\) be determined by \(s^*\iota_1=x\) and \(s^*\iota_2=x^2\) where, \(x\) is the generator of \(H^1(K_1;\mathbb{Z}/2)\). Then \(s^*(\iota_1^3+\iota_1\iota_2)=x^3+x^3=0\) and hence \(s\) admits a lift to \(F\). This gives a splitting of the 2-connected map \(c\colon F\to K_1\) and shows that \(F\) has trivial \(k\)-invariant.

It remains to determine the \(\mathbb{Z}[\pi_1(F)]\)-module structure of \(\pi_2(F)\). The only possibilities are the trivial action and \(\pi_2(F) \cong (\mathbb{Z}/2)[\mathbb{Z}/2]\). Suppose for a contradiction that the \(\pi_1\)-action on \(\pi_2\) is trivial. Then we would have \(F\simeq K_1 \times K_2\times K_2\). In the \(\mathbb{Z}/2\)-cohomology ring of \(K_1 \times K_2\times K_2\), multiplication with the unique nontrivial class in degree 1 is injective, by the Künneth theorem and because \(H^*(K_1;\mathbb{Z}/2) \cong (\mathbb{Z}/2)[x]\). Hence \(v_1^2 \neq v_2\) implies \(v_1^3\neq v_1v_2\), contradicting that \(v_1^3= v_1v_2\) by construction. We see that \(\pi_2(F)\cong (\mathbb{Z}/2)[\mathbb{Z}/2]\) as claimed. ◻

Proof. The space \(K_2\times K_2 \times S^{\infty}\) has the same homotopy groups as \(\widetilde{F}\). Moreover the fixed point sets of the \(\mathbb{Z}/2\) action on both \(\widetilde{F}\) and \(K_2\times K_2 \times S^{\infty}\) are empty. Hence the spaces are equivariantly homotopy equivalent by the equivariant Whitehead theorem; see e.g. matumoto?*Theorem 5.3. ◻

Proof. By 13, \(\pi_1(F)\cong \mathbb{Z}/2\) and thus \(H^1(F;\mathbb{Z}/2)\cong \mathbb{Z}/2\). Again by 13, \(v_1\neq0\) and \(v_2\neq v_1^2\).

Looking at the Leray–Serre spectral sequence for \(\widetilde{F} \to F \xrightarrow{c} K_1\) with \(E^2_{p,q} \cong H_p(K_1;H_q(\widetilde{F};\mathbb{Z}/2))\) and converging to \(H_{p+q}(F;\mathbb{Z}/2)\), analysing the \(p+q=2\) anti-diagonal yields an exact sequence \[H_0(K_1;H_2(\widetilde{F};\mathbb{Z}/2))\to H_2(F;\mathbb{Z}/2)\to H_2(K_1;\mathbb{Z}/2)\to 0.\] By 2, \[H_0(K_1;H_2(\widetilde{F};\mathbb{Z}/2))\cong \mathbb{Z}/2\otimes_{(\mathbb{Z}/2)[\mathbb{Z}/2]}(\mathbb{Z}/2)[\mathbb{Z}/2]\cong \mathbb{Z}/2.\] Since the \(k\)-invariant of \(F\) is trivial, there exists a section \(s\colon K_1\to F\) and thus there are no differentials into \(H_0(K_1;H_2(\widetilde{F};\mathbb{Z}/2))\). Hence we have a short exact sequence \(0 \to \mathbb{Z}/2 \to H_2(F;\mathbb{Z}/2) \to \mathbb{Z}/2 \to 0\) and so \[H^2(F;\mathbb{Z}/2)\cong H_2(F;\mathbb{Z}/2)\cong (\mathbb{Z}/2)^2,\] as claimed.

Since \(s\colon K_1\to F\) induces an isomorphism on fundamental groups, it induces an isomorphism on \(H^1(-;\mathbb{Z}/2)\), and hence \(s^*(v_1) \neq 0 \in H^1(K_1;\mathbb{Z}/2)\).

We saw that \(v_2 \neq v_1^2\) in the proof of 13, so it remains to show that \(s^*v_2=s^*v_1^2 \in H^2(K_1;\mathbb{Z}/2)\) for every section \(s\). So, fix a section \(s \colon K_1\to F\). Since \(v_1^3 = v_1v_2 \in H^3(F;\mathbb{Z}/2)\) we have \(s^*(v_1^3)=s^*(v_1v_2)\), and therefore \[s^*(v_1)\big(s^*(v_1^2) + s^*(v_2)\big)=0 \in H^3(K_1;\mathbb{Z}/2).\] Since \(s^*v_1\neq 0 \in H^1(K_1;\mathbb{Z}/2)\) and \(H^*(K_1;\mathbb{Z}/2)\) is a polynomial ring in a single variable (\(K_1 \simeq \mathbb{R}P^\infty\)), this implies that \(s^*v_1^2 + s^*v_2=0\), and so \(s^*v_1^2 = s^*v_2 \in H^2(K_1;\mathbb{Z}/2)\), as claimed.

The uniqueness of \(v_2\) satisfying this follows from the fact that \(H^2(F;\mathbb{Z}/2)\) only has four elements: the element in one summand of \(H^2(F;\mathbb{Z}/2) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2\) is determined by \(s^*v_2 = s^*v_1^2\). This leaves two possibilities remaining, but one of them is \(v_1^2\), and so the fact that \(v_2 \neq v_1^2\) leaves a single element. ◻

Proof. Consider the exact sequence \[\label{eqn:leray-serre-cohomology-for-F} H^2(K_1;\mathbb{Z}/2)\xrightarrow{c^*} H^2(F;\mathbb{Z}/2)\xrightarrow{p^*} H^0(K_1;H^2(\widetilde{F};\mathbb{Z}/2)),\tag{5}\] which again follows from the Leray–Serre spectral sequence for \(\widetilde{F} \xrightarrow{p} F \xrightarrow{c} K_1\), this time the cohomology version. Note that \[H^0\big(K_1;H^2(\widetilde{F};\mathbb{Z}/2)\big) \cong H^0\big(\mathbb{Z}/2;H^2(K_2;\mathbb{Z}/2\big) \oplus H^2\big(K_2;\mathbb{Z}/2)\big)\] is isomorphic to the fixed subgroup of the coefficients \(H^2(K_2;\mathbb{Z}/2) \oplus H^2(K_2;\mathbb{Z}/2)\) under the \(\mathbb{Z}/2\)-action \((x,y) \mapsto (y,x)\) from 2.

Since neither \(v_2=0\) nor \(v_2=v_1^2\) by 4, \(v_2\) is not pulled back from \(K_1\) under \(c\). Hence by exactness of 5 , it follows that \(p^*v_2\neq 0\). We must therefore have \[p^*v_2=(\iota_2,\iota_2)\in H^2(K_2;\mathbb{Z}/2) \oplus H^2(K_2;\mathbb{Z}/2),\] because this is the unique nontrivial fixed point of the \(\mathbb{Z}/2\)-action. ◻

Proof. For the statement of [item:prop-K2-low-degrees-1] we refer to Hatcher-alg-top?*pp. 499-500 or Mosher-Tangora?*p. 27. Admissible means that \(i_j\geq 2i_{j+1}\) for each \(j\), the degree \(d(I)\) is \(\sum i_j\), and the excess is \(2i_1 -d(I)\).

For [item:prop-K2-low-degrees-2], the first and second homology groups are straightforward. For degrees higher than two we deduce the homology from the cohomology given in [item:prop-K2-low-degrees-1]. The only admissible sequence of degree one yields \(\mathop{\mathrm{Sq}}^1(\iota_2)\) and so \(H^3(K_2;\mathbb{Z}/2) \cong \mathbb{Z}/2\). There are no degree two admissible sequences of excess less than two, and so \(H^4(K_2;\mathbb{Z}/2) \cong \mathbb{Z}/2\) generated by \(\iota_2^2\). Finally, from degree three admissible sequences of excess less than two, we obtain \(\mathop{\mathrm{Sq}}^2\mathop{\mathrm{Sq}}^1(\iota_2)\), and so \(H^5(K_2;\mathbb{Z}/2) \cong (\mathbb{Z}/2)^2\) generated by \(\{\iota_2\mathop{\mathrm{Sq}}^1\iota_2, \mathop{\mathrm{Sq}}^2\mathop{\mathrm{Sq}}^1(\iota_2)\}\).

For the integral homology in [item:prop-K2-low-degrees-3], again the first and second homology groups are straightforward, and for the third and fourth homology we refer to clement?*Appendix C. ◻

Proof. To prove part [it:wtF-i], we consider the pullback square in 3, and augment it with another pullback square: \[\begin{tikzcd}[row sep = small] B_{Y_i} \ar[r] \ar[d] & K_2 \ar[d] \ar[r] & * \ar[d] \\ \mathop{\mathrm{BO}}\ar[r,"w_1 \times w_2"] & K_1 \times K_2 \ar[r] & K_1. \end{tikzcd}\] The concatenation of two pullback squares is again a pullback. Hence \(B_{Y_i}\) is the pullback of the outer rectangle, which we recognise as \(\mathop{\mathrm{BSO}}\). So \(\Omega_4(\xi_{Y_i})\cong\Omega_4^{SO}\cong \mathbb{Z}\) as desired. This proves [it:wtF-i], but for later reference we show the entries \(E^2_{p,q} \cong H_p(Y_i;\Omega_q^{\mathop{\mathrm{Spin}}})\) with \(p+q=4\) in the James spectral sequence (9) for \(\Omega_4(\xi_{Y_i})\). The coefficients are untwisted due to the \(0\) in \((Y_i,0,x_i)\), and for the computations of the homology groups we apply 14.

The term \(E^2_{0,4} \cong H_0(K_2;\Omega_4^{\mathop{\mathrm{Spin}}})\) (with untwisted coefficients) is isomorphic to \(16\mathbb{Z}\) via the signature. Since \(\Omega_4^{\mathop{\mathrm{SO}}} \cong \mathbb{Z}\), generated by \(\mathbb{C}P^2\), and again detected by the signature, we must have that all terms on the \(E^2\) page with \(p+q=4\) survive to the \(E^\infty\) page and give the iterated graded groups of the filtration \(16\mathbb{Z}\leq 8\mathbb{Z}\leq 4\mathbb{Z}\leq \mathbb{Z}\).

Next we show [it:wtF-ii]. For this we consider the James spectral sequence for the computation of \(\Omega_4(\xi_Y)\). First note that by the Künneth theorem for \(k\leq 3\) we have \(H_k(Y;\mathbb{Z}/2)\cong H_k(Y_1;\mathbb{Z}/2)\oplus H_k(Y_2;\mathbb{Z}/2)\), which vanishes for \(k=1\) and is \(\mathbb{Z}/2 \oplus \mathbb{Z}/2\) for \(k=2,3\) by 14. Furthermore again by the Künneth theorem and 14, \[H_4(Y;\mathbb{Z})\cong H_4(Y_1;\mathbb{Z})\oplus H_4(Y_2;\mathbb{Z})\oplus H_2(Y_1;\mathbb{Z})\otimes H_2(Y_2;\mathbb{Z}) \cong \mathbb{Z}/4 \oplus \mathbb{Z}/4 \oplus \mathbb{Z}/2.\] We display the relevant groups in the \(E^2_{p,q} \cong H_p(Y;\Omega_q^{\mathop{\mathrm{Spin}}})\) page next, together with the differentials that we will soon have to compute. There is again no twisting in the coefficients.

We compute the differential \[d_2 \colon E^2_{4,0} \cong \mathbb{Z}/4 \oplus \mathbb{Z}/4 \oplus \mathbb{Z}/2 \to E^2_{2,1} \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2.\] By 10 [prop:differentials-item-ii], this is given as the composition \[H_4(Y;\mathbb{Z}) \to H_4(Y;\mathbb{Z}/2) \to H_2(Y;\mathbb{Z}/2)\] of reduction of coefficients modulo two, followed by the dual to the map \(\mathop{\mathrm{Sq}}^2 + (x_1 +x_2)\cup- \colon H^2(Y;\mathbb{Z}/2) \to H^4(Y;\mathbb{Z}/2)\). By the Künneth theorem and 14, we have that \(H_4(Y;\mathbb{Z}/2) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/2\), and the reduction modulo two map \[H_4(Y;\mathbb{Z}) \cong \mathbb{Z}/4 \oplus \mathbb{Z}/4 \oplus \mathbb{Z}/2 \to H_4(Y;\mathbb{Z}/2) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/2\] is the projection. To compute the second map in the composition, note that for both \(x_i\), we have \(\mathop{\mathrm{Sq}}^2(x_i)+(x_1+x_2)x_i=x_1x_2\in H^4(Y;\mathbb{Z}/2)\). Combining the two maps, we see that the kernel of \(d_2\colon H_4(Y;\mathbb{Z})\to H_2(Y;\mathbb{Z}/2)\) is \(H_4(Y_1;\mathbb{Z})\oplus H_4(Y_2;\mathbb{Z})\cong (\mathbb{Z}/4)^2\) and that the \(\mathbb{Z}/2\) summand maps to \((1,1) \in \mathbb{Z}/2 \oplus \mathbb{Z}/2\).

We deduce that \(\Omega_4(\xi_{Y_1})\oplus \Omega_4(\xi_{Y_2})\cong \mathbb{Z}\oplus \mathbb{Z}\) surjects onto \(\Omega_4(\xi_{Y})\). To see this first note that the factor inclusion \(Y_i \to Y\) does indeed induce a map \(\Omega_4(\xi_{Y_i}) \to \Omega_4(\xi_{Y})\). In addition, other than the one we just analysed, there are no nontrivial differentials with domain the terms \(E^k_{0,4}\), \(E^k_{2,2}\), \(E^k_{3,1}\), or \(E^k_{4,0}\), for any \(k \geq 2\). The corresponding \(E^{\infty}\) pages are therefore quotients of two copies of the terms on the \(p+q=4\) line of the first displayed \(E^2\) page above (the one from the James spectral sequence for \(\Omega_4(\xi_{Y_i})\)).

The element \(16\in 16\mathbb{Z}\leq \mathbb{Z}\cong \Omega_4(\xi_{Y_i})\) is represented by \(K3\) for both \(i\). Hence considering its image in \(\Omega_4(\xi_{Y})\) we see that \((16,0)=(0,16) \in \Omega_4(\xi_{Y})\). Hence certainly \((16,-16)\) maps to zero in \(\Omega_4(\xi_{Y})\). Since the signature gives a nontrivial homomorphism \(\Omega_4(\xi_{Y}) \to \mathbb{Z}\), and there is one relation already, we know \(\Omega_4(\xi_{Y})\) has rank one. It follows that there exists some positive integer \(\ell\) dividing \(16\) such that \(\Omega_4(\xi_Y)\cong \mathbb{Z}^2/\{(n\ell,-n\ell)\mid n\in \mathbb{Z}\}\).

Since for both \(x_i\), we have \(\mathop{\mathrm{Sq}}^2(x_i)+(x_1+x_2)x_i=x_1x_2\in H^4(Y;\mathbb{Z}/2)\), the element \((1,1)\in H_2(Y;\mathbb{Z}/2) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2\) lies in the image of \[d_2\colon E^2_{4,1} \cong H_4(Y;\mathbb{Z}/2)\to E^2_{2,2} \cong H_2(Y;\mathbb{Z}/2)\] by 10 [prop:differentials-item-i]; this is essentially the same computation as that performed above to compute part of \(d_2 \colon E^2_{4,0} \to E^2_{2,1}\). So \(\ell \leq 8\). There are no nontrivial differentials into \(H_4(Y;\mathbb{Z})\), and no further differentials out of it other than the one we computed. Thus \(\ell \geq 4\). Hence \(\ell=4\) or \(\ell=8\), depending on whether the image of \[d_2\colon E^2_{5,0} \cong H_5(Y;\mathbb{Z})\to E^2_{3,1} \cong H_3(Y;\mathbb{Z}/2)\] is nontrivial or trivial respectively. Let us compute this differential. First note that \[H_5(Y;\mathbb{Z})\cong H_5(Y_1;\mathbb{Z})\oplus H_5(Y_2;\mathbb{Z})\oplus \mathop{\mathrm{Tor}}_1(H_2(Y_1;\mathbb{Z}),H_2(Y_2;\mathbb{Z}))\] by the Künneth theorem. As shown in the proof of teichner-phd?*Theorem 3.3.2 (2), all differentials vanishes on the \(H_5(Y_i;\mathbb{Z})\)-summands. It remains to compute the differential on \(\mathop{\mathrm{Tor}}_1(H_2(Y_1;\mathbb{Z}),H_2(Y_2;\mathbb{Z})) \cong \mathbb{Z}/2\).

We derive bases for the \(\mathbb{Z}/2\)-coefficient cohomology and homology of \(Y\) using the Künneth theorem and 14. For \(H^3(Y;\mathbb{Z}/2)\), we consider the basis \(\{\mathop{\mathrm{Sq}}^1(x_1),\mathop{\mathrm{Sq}}^1(x_2)\}\) and for \(H^5(Y;\mathbb{Z}/2)\) we consider the basis \[\{\mathop{\mathrm{Sq}}^2\mathop{\mathrm{Sq}}^1(x_1),x_1\mathop{\mathrm{Sq}}^1(x_1),x_2\mathop{\mathrm{Sq}}^1(x_1),x_1\mathop{\mathrm{Sq}}^1(x_2), x_2\mathop{\mathrm{Sq}}^1(x_2),\mathop{\mathrm{Sq}}^2\mathop{\mathrm{Sq}}^1(x_2)\}.\] For \(H_3(Y;\mathbb{Z}/2)\) and \(H_5(Y;\mathbb{Z}/2)\) we consider the dual bases, denoted e.g.\(\{\mathop{\mathrm{Sq}}^1(x_1)^{\wedge},\mathop{\mathrm{Sq}}^1(x_2)^\wedge\}\).

Again by 10 [prop:differentials-item-ii], we need to understand the effect of the reduction modulo two map \(H_5(Y;\mathbb{Z}) \to H_5(Y;\mathbb{Z}/2)\) on the \(\mathop{\mathrm{Tor}}\) summand, followed by the dual to \(\mathop{\mathrm{Sq}}^2(-) + (x_1 + x_2)\cup -\). Since the generator of \(\mathbb{Z}/2\cong \mathop{\mathrm{Tor}}_1(H_2(Y_1;\mathbb{Z}),H_2(Y_2;\mathbb{Z}))\leq H_5(Y;\mathbb{Z})\) lies in the image of induced map on homology from \(Y_1^{(3)}\times Y_2^{(3)}\), and is invariant under the action of the automorphism interchanging \(Y_1\) and \(Y_2\), it follows that its image in \(H_5(Y;\mathbb{Z}/2)\) under reduction modulo two is \((x_2\mathop{\mathrm{Sq}}^1(x_1))^{\wedge}+(x_1\mathop{\mathrm{Sq}}^1(x_2))^{\wedge}\).

For the generators \(\mathop{\mathrm{Sq}}^1(x_i)\) of \(H^3(Y;\mathbb{Z}/2)\), we have \[\mathop{\mathrm{Sq}}^2\mathop{\mathrm{Sq}}^1(x_i)+(x_1+x_2)\mathop{\mathrm{Sq}}^1(x_i)=\mathop{\mathrm{Sq}}^2\mathop{\mathrm{Sq}}^1(x_i)+x_1\mathop{\mathrm{Sq}}^1(x_i)+x_2\mathop{\mathrm{Sq}}^1(x_i)\in H^5(Y;\mathbb{Z}/2).\] Computing the dual of this map and applying it to \((x_2\mathop{\mathrm{Sq}}^1(x_1))^{\wedge}+(x_1\mathop{\mathrm{Sq}}^1(x_2))^{\wedge}\) shows that the image of \(d_2\colon H_5(Y;\mathbb{Z})\to H_3(Y;\mathbb{Z}/2) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2\) is the subgroup generated by \(\mathop{\mathrm{Sq}}^1(x_1)^{\wedge}+\mathop{\mathrm{Sq}}^1(x_2)^{\wedge}\). Hence this differential is nontrivial, so \(\ell=4\), and [it:wtF-ii] follows.

Part [it:wtF-iii] follows from 5.

Now we show [it:wtF-iv]. Since \(H_2(\widetilde{F}^{(2)};\mathbb{Z}/2)\to H_2(\widetilde{F};\mathbb{Z}/2)\) is surjective, the image of \(\Omega_4(\xi_{\widetilde{F}^{(2)}})\) in \(\Omega_4(\xi_{\widetilde{F}})\) is the filtration step \(F_{2,2}\) in the spectral sequence for the latter group. By the proof of [it:wtF-ii], the filtration arising from the spectral sequence is \[\label{eqn:filtration-groups} 0 \leq 16\mathbb{Z}\cong \frac{16\mathbb{Z}\oplus 16\mathbb{Z}}{(16,-16)} \leq \frac{8\mathbb{Z}\oplus 8\mathbb{Z}}{(8,-8)} \leq \frac{4\mathbb{Z}\oplus 4\mathbb{Z}}{(4,-4)} \leq \frac{\mathbb{Z}\oplus \mathbb{Z}}{(4,-4)} = \Omega_4(\xi_{\widetilde{F}})\tag{6}\] Hence in particular \(F_{2,2} \cong (8\mathbb{Z}\oplus8\mathbb{Z})/(8,-8)\), which proves [it:wtF-iv].

Finally, we showed in the proof of [it:wtF-ii] that the kernel of \(d_2\colon H_5(Y;\mathbb{Z})\to H_2(Y;\mathbb{Z}/2)\) is \(H_5(Y_1;\mathbb{Z})\oplus H_5(Y_2;\mathbb{Z})\). As mentioned above, it was shown in the proof of teichner-phd?*Theorem 3.3.2 (2) that all differentials, including the \(d_3\) differentials, vanish on the \(H_5(Y_i;\mathbb{Z})\)-summands. This shows [it:wtF-v]. ◻

Proof. Choose a CW structure on \(F\) and let \(\widetilde{F}^{(2)}\) and \(\widetilde{F}^{(4)}\) denote the corresponding \(2\)- and \(4\)-skeleta of the universal cover, respectively. Let \(k\colon \widetilde{F}^{(2)} \to \widetilde{F}^{(4)}\) denote the inclusion map. Consider the map \(j_* \colon \Omega_4(\xi_{\widetilde{F}^{(4)}})\to \Omega_4(\xi_{\widetilde{F}})\), and let \(j_*|\) denote its restriction to the image of \(\Omega_4(\xi_{\widetilde{F}^{(2)}})\) in domain and codomain. It follows from the vanishing of \(d_3\colon E_{5,0}^3\to E_{2,2}^3\), shown in 6 [it:wtF-v], that \(j_*|\) is an isomorphism. By 6 [it:wtF-iv], \[\operatorname{Im}\Omega_4(\xi_{\widetilde{F}^{(2)}}) \cong (8\mathbb{Z}\oplus8\mathbb{Z})/(8,-8).\]

The class \([K3]\) lies in the image of \(\Omega_4(\xi_{\widetilde{F}^{(2)}})\) and is represented by \((16,0)\). By 2 6, the action of \(\mathbb{Z}/2=\pi_1(F)\) on \(\Omega_4(\xi_{\widetilde{F}}) \cong \Omega_4(\xi_{Y})\cong \mathbb{Z}^2/(4,-4)\) interchanges the summands. Moreover, since \(v_1\) is nontrivial on the generator of \(\pi_1(F)\), acting by this generator changes the orientation. So altogether the action sends \((z,z')\mapsto (-z',-z)\). The deck transformation of \(\widetilde{F}\) restricts to \(\widetilde{F}^{(2)}\) and \(\widetilde{F}^{(4)}\). Hence on the image \((8\mathbb{Z}\oplus8\mathbb{Z})/(8,-8)\) of \(\Omega_4(\xi_{\widetilde{F}^{(2)}})\) in \(\Omega_4(\xi_{\widetilde{F}^{(4)}})\) the deck transformation acts by sending \((-8,0)\) to \((0,8)\).

We consider the composition \(\widetilde{F}^{(2)}\xrightarrow{k} \widetilde{F}^{(4)}\xrightarrow{p} F^{(4)}\) and the induced maps on bordism groups \[\Omega_4(\xi_{\widetilde{F}^{(2)}})\xrightarrow{k_*} \Omega_4(\xi_{\widetilde{F}^{(4)}})\xrightarrow{p_*} \Omega_4(\xi_{F^{(4)}}).\] In particular this composition factors through \(p_*| \colon \operatorname{Im}k_* \to \Omega_4(\xi_{F^{(4)}})\). Since deck transformations commute with \(p \colon \widetilde{F}^{(4)} \to F^{(4)}\), \(p_*|\) factors through the quotient of \(\operatorname{Im}k_*\) by the \(\mathbb{Z}/2\) deck transformation action, yielding \[(8\mathbb{Z}\oplus8\mathbb{Z})/(8,-8) \cong \operatorname{Im}k_* \to \mathbb{Z}\otimes_{\mathbb{Z}[\mathbb{Z}/2]} \operatorname{Im}k_* \to \Omega_4(\xi_{F^{(4)}}).\] The class \([K3]\) is represented by \((16,0) \in (8\mathbb{Z}\oplus8\mathbb{Z})/(8,-8)\), but in the quotient of \((8\mathbb{Z}\oplus8\mathbb{Z})/(8,-8)\) by the action sending \((-8,0)\) to \((0,8)\) we have \[(16,0)=(8,8)=(8,0)+(0,8)=(8,0)+(-8,0)=(0,0),\] where the first equality is given by subtracting the trivial element \((8,-8)\), and the penultimate equality uses the deck transformation. Hence \([K3]\) vanishes in \(\Omega_4(\xi_{F^{(4)}})\), as claimed. ◻

Proof. By 4, the differential \(d_3 \colon H_3(F^{(4)};\mathbb{Z}/2) \to H_0(F^{(4)};\Omega_4^{\mathop{\mathrm{Spin}}})=E^3_{0,4}\cong \mathbb{Z}/2\) in the James spectral sequence for \(F^{(4)}\) is dual to \(j^*v_1^3+j^*v_1j^*v_2=j^*(v_1^3+v_1v_2)=0\). Hence \(E_{0,4}^4 \cong \mathbb{Z}/2\). By 7, \([K3]\) vanishes in \(\Omega_4(\xi_{F^{(4)}})\) and hence \(E_{0,4}^\infty=0\). Since \(H_5(F^{(4)};\mathbb{Z})=0\), it follows that the differential \(d_4 \colon H_4(F^{(4)};\mathbb{Z}/2) \to H_0(F^{(4)};\Omega_4^{\mathop{\mathrm{Spin}}})\) must be nontrivial. Since \(j_*\colon H_4(F^{(4)};\mathbb{Z}/2)\to H_4(F;\mathbb{Z}/2)\) is surjective, the differential \(d_4 \colon H_4(F;\mathbb{Z}/2) \to H_0(F;\Omega_4^{\mathop{\mathrm{Spin}}})\) is also nontrivial by naturality of the James spectral sequence. ◻

Proof. By 4, \(H^2(F;\mathbb{Z}/2)\) is generated by \(v_1^2\) and \(v_2\). Since \(v_1^3=v_1v_2\), \[v_1^4=\mathop{\mathrm{Sq}}^1(v_1^3)=\mathop{\mathrm{Sq}}^1(v_1v_2)=v_1^2v_2+v_1\mathop{\mathrm{Sq}}^1(v_2)=v_1^4+v_1\mathop{\mathrm{Sq}}^1(v_2).\] Thus \(v_1\mathop{\mathrm{Sq}}^1(v_2)=0\). Using this and again that \(v_1^3=v_1v_2\), we have \[\mathop{\mathrm{Sq}}^2(v_1^2)+v_1\mathop{\mathrm{Sq}}^1(v_1^2)+v_2v_1^2=v_1^4+0+v_1^2v_2= v_1(v_1^3 + v_1v_2) =0,\] and \[\mathop{\mathrm{Sq}}^2(v_2)+v_1\mathop{\mathrm{Sq}}^1(v_2)+v_2^2=v_1\mathop{\mathrm{Sq}}^1(v_2)=0.\qedhere\] ◻

Proof. The lift \(f\) induces a map \(B_V \to B_F\) over \(\mathop{\mathrm{BO}}\) and a corresponding map of bordism groups \(\Omega_4(\xi_V) \to \Omega_4(\xi_F)\).

By 10 and 8, the \(d_2\) differential with domain \(H_4(F;\Omega_1^{\mathop{\mathrm{Spin}}})\) is trivial. The differential with domain \(H_4(V;\Omega_1^{\mathop{\mathrm{Spin}}})\) is given by \(\mathop{\mathrm{Sq}}_2^{w_1,w_2}\). Hence the \(E^4_{4,1}\) term for \(F\) is \(H_4(F;\mathbb{Z}/2)/\operatorname{Im}d_2^F\) and the \(E^4_{4,1}\) term for \(V\) is \(\ker(\mathop{\mathrm{Sq}}_2^{w_1,w_2})/\operatorname{Im}d_2^V\) (as before recall that \(\Omega_3^{\mathop{\mathrm{Spin}}}=0\) so in both cases \(d_3=0\)). By naturality of the James spectral sequence, there is a commutative diagram \[\begin{tikzcd} \ker(\mathop{\mathrm{Sq}}_2^{w_1,w_2})\ar[r, two heads]\ar[d,"f_*"]& \ker(\mathop{\mathrm{Sq}}_2^{w_1,w_2})/\operatorname{Im}d_2^V\ar[r,"{d_4^V}"]\ar[d,"f_*"] & H_0(V;\Omega_4^{\mathop{\mathrm{Spin}}})\ar[d,two heads]\\ H_4(F;\mathbb{Z}/2)\ar[r, two heads] \ar[rr,"\mathfrak{o}",bend right=15] & H_4(F;\mathbb{Z}/2)/\operatorname{Im}d_2^F\ar[r,"{d_4^F}"]&H_0(F;\Omega_4^{\mathop{\mathrm{Spin}}})\cong \mathbb{Z}/2. \end{tikzcd}\] By 4, the bottom composition is dual to \(\mathfrak{o}\). Hence the down-right-right composition is dual to \([f^*\mathfrak{o}]\). Thus the right-right-down composition is also dual to \([f^*\mathfrak{o}]\), as claimed. Since the differential \(d_4^V\) is independent of the choice of lift \(f\), so is the class \([f^*\mathfrak{o}]\). ◻

Proof. From 2, 14, and the computations with the Künneth theorem in the proof of 6, it follows that \[H^2(\widetilde{F};\mathbb{Z}/2)\cong (\mathbb{Z}/2)[\mathbb{Z}/2],\, H^3(\widetilde{F};\mathbb{Z}/2)\cong (\mathbb{Z}/2)[\mathbb{Z}/2], \text{ and }H^4(\widetilde{F};\mathbb{Z}/2)\cong (\mathbb{Z}/2)[\mathbb{Z}/2]\oplus \mathbb{Z}/2.\] Since \(H^p(K_1;(\mathbb{Z}/2)[\mathbb{Z}/2]) \cong H^p(S^{\infty};\mathbb{Z}/2)=0\) for \(p>0\), the Leray–Serre spectral sequence for \(\widetilde{F}\to F\to K_1\) shows that there is a short exact sequence \[\label{eqn:ses-for-h4-F-Z2} 0\to H^4(K_1;\mathbb{Z}/2)\xrightarrow{c^*} H^4(F;\mathbb{Z}/2)\xrightarrow{p^*}H^0(K_1;H^4(\widetilde{F};\mathbb{Z}/2))\to 0.\tag{7}\] We have \(H^0(K_1;H^4(\widetilde{F};\mathbb{Z}/2))\cong H^4(\widetilde{F};\mathbb{Z}/2)^{\mathbb{Z}/2}\cong (\mathbb{Z}/2)^2\) generated by \(x_1x_2\) and \(x_1^2+x_2^2\). Here recall that the \(x_i\) denote the generators of \(H^2(Y;\mathbb{Z}/2) = H^2(\widetilde{F};\mathbb{Z}/2)\). Furthermore, a section \(s\colon K_1\to F\) of \(c\) yields a splitting of the short sequence above. Thus a basis of \(H^4(F;\mathbb{Z}/2)\) is given by \(b:=c^*(\iota_1^4)\), the image of \(H^4(K_1;\mathbb{Z}/2)\), and \(a_1\) and \(a_2\), where \(a_1\) and \(a_2\) are determined by \(p^*(a_1)=x_1^2+x_2^2\), \(p^*(a_2)=x_1x_2\), and the condition that \(s^*(a_i)=0\). In particular \(\mathfrak{o}\) is a \(\mathbb{Z}/2\)-linear combination of \(a_1\), \(a_2\), and \(b\). It also follows from the short exact sequence 7 that the conditions \(p^*\mathfrak{o}=x_1x_2\) and \(s^*\mathfrak{o}=0\) determine a unique class.

Let \(s\colon K_1\to F\) be a section. We now show that \(s^*\mathfrak{o}=0\). We compare the James spectral sequence for \(\Omega_4(\xi_F)\) with the spectral sequence for \(\Omega_4(\xi_{K_1})\), where \(\xi_{K_1}\) is the fibration determined by \((K_1,s^*v_1,s^*v_2)\) as in 3, and the \(v_i\) are as in 3 . Since \(s\) is a section, naturality of the James spectral sequences implies that the \(d_2\) differential with domain \(H_4(K_1;\Omega_1^{\mathop{\mathrm{Spin}}})\) is again trivial. Therefore \(E^4_{4,1}\cong E^2_{4,1} \cong H_4(K_1;\mathbb{Z}/2)\) (there is no \(d_3\) either, because \(\Omega_3^{\mathop{\mathrm{Spin}}}=0\)). Also recall from 4 that \(s^*v_1^2=(s^*v_1)^2 =s^*v_2\). We can therefore apply 3, which implies that the differential \(d_4^{K_1}\colon E^4_{4,1}\cong H_4(K_1;\mathbb{Z}/2)\to E^4_{0,4}\) is trivial. By naturality of the James spectral sequence the following diagram commutes, \[\begin{tikzcd}[column sep=large] H_4(K_1;\mathbb{Z}/2)\ar[r,"\cong"]\ar[d,"s_*"]&E_{4,1}^4\ar[r,"d_4^{K_1}=0"]\ar[d,"s_*"]&E^4_{0,4}\cong \mathbb{Z}/2\ar[d,"\cong","s_*"']\\ H_4(F;\mathbb{Z}/2)\ar[r,two heads]\ar[rr,"\mathfrak{o}",bend right=15]&E_{4,1}^4\ar[r,"d_4^F"]&E^4_{0,4}\cong \mathbb{Z}/2 \end{tikzcd}\] This implies that \(s^*\mathfrak{o}=0\), as desired.

It remains to show that \(p^*\mathfrak{o}=x_1x_2\), i.e.that \(\mathfrak{o}=a_2\). Since \(s\) is a section of \(c\), \(s^*b = s^*c^*(\iota_1^4) = \iota_1^4 \neq 0\). Thus since \(s^*\mathfrak{o}=0\), we see that \(\mathfrak{o}=\varepsilon_1a_1+\varepsilon_2a_2\) for some \(\varepsilon_i\in \mathbb{Z}/2\). Since the \(d_4\) differential is trivial for \(\Omega_4(\xi_{\widetilde{F}})\) (the domain is 2-torsion and the codomain is \(16\mathbb{Z}\)), \(p^*\mathfrak{o}=p^*(\varepsilon_1a_1+\varepsilon_2a_2)\) has to be trivial in \(H^*(\widetilde{F};\mathbb{Z}/2)/\operatorname{Im}(\mathop{\mathrm{Sq}}^2_ {w_1,w_2})\), by 15 applied with \(f=p\). That is, \(p^*(\varepsilon_1a_1+\varepsilon_2a_2)\) lies in the image of the map \(\mathop{\mathrm{Sq}}^2_{w_1,w_2}\colon H^2(\widetilde{F};\mathbb{Z}/2)\to H^4(\widetilde{F};\mathbb{Z}/2)\). This image is \(\{0,x_1x_2\}\) and hence \(\varepsilon_1=0\). Since \(\mathfrak{o}\neq 0\) by 3, \(\mathfrak{o}=a_2\) as claimed. ◻

Proof of 2. Item [it:B-1] was shown in 13 and [it:B-2] follows from 9.

Now we show [it:B-4]. Let \((\pi,w_1,w_2)\) be a normal \(1\)-type. If \(w_1^3=w_1w_2\), there is a lift \(f\colon B\pi \to F\) of \((w_1,w_2)\) along \(u\). If \[0=[f^*\mathfrak{o}]\in H^4(\pi;\mathbb{Z}/2)/\operatorname{Im}\big(\mathop{\mathrm{Sq}}^2_{w_1,w_2} \colon H^2(\pi,\mathbb{Z}/2)\to H^4(\pi;\mathbb{Z}/2)\big),\] then the composition \[\label{eqn:composition-with-d4} \ker(\mathop{\mathrm{Sq}}_2^{w_1,w_2})\twoheadrightarrow \ker(\mathop{\mathrm{Sq}}_2^{w_1,w_2})/\operatorname{Im}d_2^\pi\xrightarrow{d_4^\pi} \mathbb{Z}/2\tag{8}\] vanishes, by 15 applied with \(V=B\pi\). Hence \(d_4^\pi=0\) because the first map is surjective. It follows that \([K3]\) survives to the \(E^5\)-page. By hypothesis \(H_5(\pi;\mathbb{Z})=0\), and so in fact \([K3] \neq 0\in \Omega_4(\xi_{\pi})\). The existence of stably exotic \(4\)-manifolds with normal \(1\)-type \((\pi,w_1,w_2)\) now follows from 11 [item:iii]. This completes the proof of [it:B-4].

Lastly, we show [it:B-3]. Let \((\pi,w_1,w_2)\) be a normal \(1\)-type. If \(w_1^3\neq w_1w_2\), no stably exotic \(4\)-manifolds with normal \(1\)-type \((\pi,w_1,w_2)\) exist by 1 [it:main-1]. So we assume \(w_1^3=w_1w_2\). As above, there is a lift \(f\colon B\pi \to F\) of \((w_1,w_2)\) along \(u\). If \[0\neq [f^*\mathfrak{o}]\in H^4(\pi;\mathbb{Z}/2)/\operatorname{Im}\mathop{\mathrm{Sq}}^2_{w_1,w_2},\] then the composition 8 is nonzero by 15, again applied with \(V=B\pi\). Hence \(d_4^\pi \neq 0\), and so \(0=[K3]\in \Omega_4(\xi_{\pi})\). Thus no stably exotic \(4\)-manifolds with normal \(1\)-type \((\pi,w_1,w_2)\) exist by 11 [item:ii]. This implies [it:B-3]. ◻

Proof. First we construct a map \(f\colon B\pi\to F\) as follows. Start with a map \(\widehat f\colon T^4\to \widetilde{F}\simeq K_2\times K_2\) given by \((t_1t_2,t_3t_4)\), where \(t_k\in H^1(\mathbb{Z}^4,\mathbb{Z}/2)\) corresponds to the generator \(a_k\) of \(\mathbb{Z}^4\). This map is equivariant with respect to the \(\mathbb{Z}/2\)-actions by 2 and hence descends to a map \(f\colon B\pi\to F\). Let \(w_i:=f^*v_i\) for \(i=1,2\), where \(v_1\) and \(v_2\) are as in 3 . Then \(w_1^3=w_1w_2\) holds by construction of \(F\).

By 2 [it:B-3], it remains to show that \[[f^*\mathfrak{o}]\neq 0\in H^4(\pi;\mathbb{Z}/2)/\operatorname{Im}\mathop{\mathrm{Sq}}^2_{w_1,w_2},\] i.e.that \(f^*\mathfrak{o}\) does not lie in the image of \[\mathop{\mathrm{Sq}}^2_{w_1,w_2} = \mathop{\mathrm{Sq}}^2(-)+w_1\mathop{\mathrm{Sq}}^1(-)+w_2 \cup - \colon H^2(\pi,\mathbb{Z}/2)\to H^4(\pi;\mathbb{Z}/2).\] To begin the proof of this, note that by construction \(\widehat{f}^*x_1=t_1t_2\) and \(\widehat{f}^*x_2=t_3t_4\). By 2 [it:B-2], \(p^*\mathfrak{o}=x_1x_2\). Thus \[\label{eqn:wh-f-p-o-nontrivial} \widehat{f}^*p^*\mathfrak{o}=\widehat f^*(x_1x_2)=t_1t_2t_3t_4\neq 0\in H^4(\mathbb{Z}^4;\mathbb{Z}/2).\tag{9}\] Let \(j\colon T^4\to B\pi\) be determined by the inclusion of \(\mathbb{Z}^4\). Note that \(f \circ j = p \circ \widehat{f} \colon T^4 \to F\). Since \(j^*f^*\mathfrak{o}=\widehat f^*p^*\mathfrak{o}\neq 0\) by 9 , if we show that \(j^*\circ \mathop{\mathrm{Sq}}^2_{w_1,w_2}=0\) it will follow that \(f^*\mathfrak{o}\) does not lie in the image of \(\mathop{\mathrm{Sq}}^2_{w_1,w_2}\). Since \(j^*w_1=0\), \(j^*w_2=\widehat f^*p^*v_2=\widehat f^*(x_1+x_2)=t_1t_2+t_3t_4\) and \(\mathop{\mathrm{Sq}}^2\) vanishes on \(T^4\), the map \(j^*\circ \mathop{\mathrm{Sq}}^2_{w_1,w_2}\) is given by \[j^*(w_2 \cup -) = j^*w_2 \cup j^*(-) = (t_1t_2+t_3t_4) \cup j^*(-).\] Furthermore the image of \(j^*\colon H^2(\pi;\mathbb{Z}/2)\to H^2(\mathbb{Z}^4;\mathbb{Z}/2)\) consists of the elements invariant under the \(\mathbb{Z}/2\)-action. Hence a basis for \(\operatorname{Im}j^*\) is given by \(\{t_1t_3,t_2t_4,t_1t_2+t_3t_4,t_1t_4+t_2t_3\}\). It is straightforward to see that the cup product of each of these elements with \(t_1t_2+t_3t_4\) is trivial. Hence \(j^*\circ \mathop{\mathrm{Sq}}^2_{w_1,w_2}=0\), as needed.

Thus by 2 [it:B-3], there are no stably exotic \(4\)-manifolds with normal \(1\)-type \((\pi,w_1,w_2)\). ◻

Proof. We first compute the map on \(\mathbb{Z}/2\)-homology. Consider the Leray–Serre spectral sequence for the fibration \(K(\mathbb{Z}/2,3)\to \widetilde{G}\to \widetilde{F}\) obtained by looping the fibration 10 . Since \(H_q(K(\mathbb{Z}/2,3);\mathbb{Z}/2)=0\) for \(q=1,2\), the map \(\widetilde{y}_*\colon H_i(\widetilde{G};\mathbb{Z}/2)\to H_i(\widetilde{F};\mathbb{Z}/2)\) is an isomorphism for \(i\leq 2\) and there is an exact sequence \[\label{eqn:an-exact-sequence-with-wtF-and-wtF-prime} \cdots\to H_4(\widetilde{G};\mathbb{Z}/2)\to H_4(\widetilde{F};\mathbb{Z}/2)\xrightarrow{d_4} H_0(\widetilde{F};H_3(K(\mathbb{Z}/2,3);\mathbb{Z}/2))\to H_3(\widetilde{G};\mathbb{Z}/2)\xrightarrow{\widetilde{y}_*} H_3(\widetilde{F};\mathbb{Z}/2)\to 0.\tag{11}\] We will show that \(d_4\) is nontrivial. The concatenation of the two pullback squares \[\begin{tikzcd} \widetilde{G} \ar[r] \ar[d] & G \ar[r] \ar[d] & \{*\} \ar[d] \\ \widetilde{F} \ar[r,"p"] & F \ar[r,"\mathfrak{o}"] & K(\mathbb{Z}/2,4) \end{tikzcd}\] is again a pullback square, showing that \(\widetilde{G}\) is the homotopy fibre of \(p^*\mathfrak{o} \colon \widetilde{F}\to K(\mathbb{Z}/2,4)\). It follows that the composition \(H_4(\widetilde{G};\mathbb{Z}/2)\to H_4(\widetilde{F};\mathbb{Z}/2)\to H_4(K(\mathbb{Z}/2,4);\mathbb{Z}/2)\) is trivial. Since \(p^*\mathfrak{o}=x_1x_2\neq 0\) by 9, the map \(H_4(\widetilde{F};\mathbb{Z}/2)\to H_4(K(\mathbb{Z}/2,4);\mathbb{Z}/2)\) is nontrivial. Thus \(H_4(\widetilde{G};\mathbb{Z}/2)\to H_4(\widetilde{F};\mathbb{Z}/2)\) is not surjective, and so by exactness of 11 the map \(d_4\) is nontrivial, as desired. However \(H_3(K(\mathbb{Z}/2,3);\mathbb{Z}/2)\cong \mathbb{Z}/2\), and so the codomain of \(d_4\) is also \(\mathbb{Z}/2\), and it follows that \(d_4\) is surjective. Hence by exactness of 11 , \(\widetilde{y}_*\colon H_3(\widetilde{G};\mathbb{Z}/2)\to H_3(\widetilde{F};\mathbb{Z}/2)\) is an isomorphism. This shows [it:hom-G-i].

Now consider the integral Leray–Serre spectral sequence for the same fibration \(K(\mathbb{Z}/2,3)\to \widetilde{G}\to \widetilde{F}\). Since \(H_q(K(\mathbb{Z}/2,3);\mathbb{Z})=0\) for \(q=1,2,4\) and \(H_1(\widetilde{F};H_3(K(\mathbb{Z}/2,3);\mathbb{Z}))=0\), there are no nontrivial differentials out of \(H_5(\widetilde{F};\mathbb{Z})\). This proves [it:hom-G-ii]. ◻

Proof. We consider the James spectral sequences with \(E^2_{p,q}\) terms \(H_p(\widetilde{G};\Omega_q^{\mathop{\mathrm{Spin}}})\) and \(H_p(\widetilde{F};\Omega_q^{\mathop{\mathrm{Spin}}})\), converging to \(\Omega_4(\xi_{\widetilde{G}})\) and \(\Omega_4(\xi_{\widetilde{F}})\), respectively. The respective images of \(\Omega_4(\xi_{(\widetilde{G})^{(3)}})\) and \(\Omega_4(\xi_{\widetilde{F}^{(3)}})\) correspond to the \(F_{3,1}\) filtration steps in these spectral sequences.

Since \(\widetilde{y}_*\colon H_i(\widetilde{G};\mathbb{Z}/2)\to H_i(\widetilde{F};\mathbb{Z}/2)\) is an isomorphism for \(i\leq 3\), the terms \(E^2_{0,4},E^2_{2,2}\), and \(E^2_{3,1}\) agree in the spectral sequences for \(\widetilde{G}\) and \(\widetilde{F}\). In particular, \(E^2_{3,1}\cong E^2_{2,2}\cong (\mathbb{Z}/2)^2\). For \(\Omega_4(\xi_{\widetilde{F}})\) we have \(E^\infty_{3,1}\cong E^\infty_{2,2}\cong \mathbb{Z}/2\) as shown in the proof of 6 [it:wtF-ii]. Hence to show that the \(F_{3,1}\) filtration steps agree, it suffices to show that we also have \(E^\infty_{3,1}\cong E^\infty_{2,2}\cong \mathbb{Z}/2\) for \(\Omega_4(\xi_{\widetilde{G}})\).

Since \(p^*\mathfrak{o}=x_1x_2\) by 9, we have a fibration \[\widetilde{G}\to \widetilde{F}\xrightarrow{x_1x_2} K(\mathbb{Z}/2,4).\] By 2, \(\widetilde{F}\simeq Y=Y_1\times Y_2\times S^\infty\) with \(Y_i=K_2\). For each \(i=1,2\), one of \(x_1\) and \(x_2\) pulls back trivially to \(Y_i\). Hence the inclusion of \(Y_i\) into \(\widetilde{F}\simeq Y\) lifts to \(\widetilde{G}\). Thus the surjection \(\Omega_4(\xi_{Y_1})\oplus \Omega_4(\xi_{Y_2}) \cong \mathbb{Z}\oplus \mathbb{Z}\to \Omega_4(\xi_{\widetilde{F}})\) from  [it:wtF-ii] and [it:wtF-iii] of 6 factors through \(\widetilde{y}_*\). This implies, by the same logic used in the proof of 6 [it:wtF-ii] (but without the \(E_{4,0}^{\infty} \cong \mathbb{Z}/4 \oplus \mathbb{Z}/4\) term), that the \(F_{3,1}\) filtration step for \(\Omega_4(\xi_{\widetilde{G}})\) is a quotient of \((4\mathbb{Z}\oplus 4\mathbb{Z})/(16,-16)\).

Since \(\widetilde{y}_*\colon H_5(\widetilde{G};\mathbb{Z})\to H_5(\widetilde{F};\mathbb{Z})\) is surjective by 10 and the differential \(d_2\colon E^2_{5,0}\to E^2_{3,1}\) is nontrivial for \(\Omega_4(\xi_{\widetilde{F}})\) (see the proof of 6 [it:wtF-ii]), the same differential is nontrivial for \(\Omega_4(\xi_{\widetilde{G}})\). It follows that the \(E_{3,1}^\infty\) terms agree for \(\widetilde{G}\) and \(\widetilde{F}\) and are isomorphic to \(\mathbb{Z}/2\). Hence the images of \((4,0)\) and \((0,4)\) in the \(E_{3,1}^\infty\) term for \(\widetilde{G}\) agree and so \((4,-4)\) lies in \(F_{2,2}\). As \(E_{0,4}^\infty \to F_{2,2} \to E^{\infty}_{2,2} \to 0\) is exact and \(E^\infty_{2,2}\) is 2-torsion, it follows that the image of \((8,-8)\) maps trivially to \(E^\infty_{2,2}\) and so lies in \(E_{0,4}^\infty\). Since the images of \((8,0)\) and \((0,-8)\) generate \(E_{2,2}^\infty\), we have that \(E_{2,2}^\infty\cong \mathbb{Z}/2\) for \(\widetilde{G}\), as for \(\widetilde{F}\). As mentioned, the lemma follows, noting that, by the same logic as that used in 6 , the \(F_{3,1}\) filtration step for \(\Omega_4(\xi_{\widetilde{F}})\) is isomorphic to \((4\mathbb{Z}\oplus 4\mathbb{Z})/(4,-4)\). ◻

Proof. As in the proof of 7, the map \[\Omega_4(\xi_{(\widetilde{G})^{(3)}})\to\Omega_4(\xi_{\widetilde{G}}) \to \Omega_4(\xi_{G})\] factors through \(\mathbb{Z}\otimes_{\mathbb{Z}[\mathbb{Z}/2]}\Omega_4(\xi_{\widetilde{G}})\), where \(\mathbb{Z}/2\) acts by the deck transformation. By 11, the image of \(\Omega_4(\xi_{(\widetilde{G})^{(3)}})\) is isomorphic to \((4\mathbb{Z}\oplus 4\mathbb{Z})/(4,-4)\), and the deck transformation acts by \((z,z')\mapsto (-z',-z)\), as in the proof of 7. We compute in the quotient that \[(16,0)=(8,8)=(8,0)+(0,8)=(8,0)+(-8,0)=(0,0).\] Since \([K3]\) represents \((16,0)\), it follows that \([K3]=0\in \Omega_4(\xi_{G})\) as claimed. ◻