Proof. We define an order relation on idempotents in \(R\) by \(e_1 \le e_2\) if \(e_1 \cdot e_2 = e_1 = e_2\cdot e_1\). If \(e_1,e_2\in R\) are idempotent, then there is another idempotent \(e\in R\) with \(e\cdot e_j = e_j = e_j\cdot e\) for \(j=1,2\) because \(R\) is a ring with local units. This makes the set of idempotents in \(R\) directed. By assumption, any element of \(R\) belongs to \(e R e\) for some idempotent \(e\in R\). This implies that \(R = \varinjlim e R e\). Here \(e R e\) are subrings in \(R\). They happen to be unital, but the inclusion homomorphism \(e_1 R e_1 \to e_2 R e_2\) for \(e_1 \le e_2\) is not unital. The statements about \(R e\) and \(e R\) are proven similarly. ◻

Proof. [en:loc95then95smooth951]\(\implies\)[en:loc95then95smooth952] is trivial. The conditions [en:loc95then95smooth953] and [en:loc95then95smooth954] are equivalent because \(m\in M\) belong to \(e\cdot M\subseteq M\) if and only if \(e\cdot m = m\).

We show [en:loc95then95smooth952]\(\implies\)[en:loc95then95smooth953]. Let \(m\in M\). Then we may write \(m=\sum_{i=1}^n r_i\cdot m_i\) for some \(r_i\in R\), \(m_i\in M\) by [en:loc95then95smooth952]. Let \(e\in R\) be a local unit for the finite set \(\{r_1,r_2,\dotsc,r_n\}\). Then \(e\cdot m= m\), verifying [en:loc95then95smooth953].

We show [en:loc95then95smooth953]\(\implies\)[en:loc95then95smooth951]. The hypothesis implies immediately that \(\mu_M\) is surjective. We prove that it is also injective. Let \(\sum_{i=1}^n r_i\otimes m_i\) be in the kernel of \(\mu_M\), that is, \(\sum_{i=1}^n r_i\cdot m_i=0\) in \(M\). Let \(e\in R\) be a local unit for \(\{r_i\}_{i=1}^n\). Then \[\sum_{i=1}^n r_i\otimes m_i = \sum_{i=1}^n e r_i\otimes m_i = e\otimes\biggl(\sum_{i=1}^n r_i\cdot m_i\biggr)=0\] in \(R\otimes_R M\). So \(\ker \mu_M = 0\) as needed.

The analogous result for right modules is proven in the same way. ◻

Proof. This is straightforward to check. ◻

Proof. Let \(M\) be a right \(R\)-module. First assume \(M\) to be fgp. Since \(M\) is finitely generated, there are elements \(m_1,\dotsc,m_k\) so that the map \[\pi\colon R^k \to M,\qquad (r_1,\dotsc,r_k)\mapsto \sum_{j=1}^k m_j r_j,\] is surjective. Since \(M\) is projective, \(\pi\) splits by a module homomorphism \(\sigma\colon M \to R^k\). The map \(\sigma\circ\pi\colon R^k \to R^k\) multiplies by the \(k\times k\)-matrix \(e\) with columns \(\sigma(m_j)\in R^k\), \(j=1,\dotsc,k\). Since \(\pi\circ\sigma = \mathrm{id}_M\), the map \(\sigma\pi\) is idempotent, and then \(e\) is idempotent as a matrix. The map \(\pi\) restricts to an isomorphism from the image \(e\cdot R^k\) of \(\sigma\pi\) onto \(M\).

Conversely, let \(M\cong e\cdot R^k\) for an idempotent \(e\in\mathbb{M}_k(R)\). Let \(e_1,\dotsc,e_k\in R^k\) be the columns of \(e\). These belong to \(M\) because \(e\cdot e_j = e_j\), and they form a finite generating set for \(M\). Let \(N\) be another right \(R\)-module and let \(q\colon N\to M\) be a surjective \(R\)-module map. Then there are \(\hat{e}_1,\dotsc,\hat{e}_k\in N\) with \(q(\hat{e}_j) = e_j\). The module map \(R^k \to N\), \((r_1,\dotsc,r_k) \mapsto \sum_{j=1}^k \hat{e}_j r_j\), restricts to a map \(\sigma\colon M =e\cdot R^k \to N\). The composite \(q\circ\sigma\colon M\to M\) maps \(x\mapsto e\cdot x\) for all \(x\in M\subseteq R^k\), so that it is the identity on \(M\). Thus \(\sigma\) is a section for \(q\). ◻

Proof. We first assume \(M\) to be fgp as a \(T\)-module. We use Lemma 2 to identify \(M\cong e \cdot T^k = e\cdot \mathbb{M}_{k,1}(T)\) for some idempotent \(e\in \mathbb{M}_k(T)\). Then any \(T\)-module map \(M\to N \otimes_S P\) extends canonically to \(T^k\) by composing with the projection \(T^k \to M\), \(x\mapsto e\cdot x\). Conversely, a \(T\)-module map \(f\colon T^k\to N \otimes_S P\) is such a composite if and only if \(f\circ e = f\), where we view \(e\) as an endomorphism of \(T^k\). The same applies to maps \(M\to P\). Here a right \(T\)-module map \(f\colon T^k\to P\) with \(f\circ e = f\) must be of the form \((t_j) \mapsto \sum_{j=1}^k t_j p_j\) with some \(p_j \in P\); namely, \(p_j\) is the image of the \(j\)th column of \(e\). Thus \(\mathop{\mathrm{Hom}}_{-,T}(M, P) \cong \mathbb{M}_{1,k}(P)\cdot e\), where we use the canonical right \(\mathbb{M}_k(T)\)-module structure on \(\mathbb{M}_{1,k}(P)\). Then \[N\otimes_S \mathop{\mathrm{Hom}}_{-,T}(M, P) \cong N\otimes_S \mathbb{M}_{1,k}(P)\cdot e \cong \mathbb{M}_{1,k}(N \otimes_S P)\cdot e.\] While a general \(T\)-module map \(T\to N \otimes_S P\) need not be of the form \(t\mapsto x\cdot t\) for some \(x\in N \otimes_S P\), this becomes so after multiplication with an element of \(T\). Therefore, any \(T\)-module map \(f\colon T^k\to N \otimes_S P\) with \(f\circ e = f\) must be of the form \((t_i)\mapsto \sum_{i=1}^k t_i x_i\) for unique \(x_1,\dotsc,x_k\in N \otimes_S P\). Then \[\mathop{\mathrm{Hom}}_{-,T}(M,N \otimes_S P) \cong \mathbb{M}_{1,k}(N \otimes_S P)\cdot e \cong N\otimes_S \mathop{\mathrm{Hom}}_{-,T}(M, P).\]

Next, we assume \(N\) to be fgp, that is, \(N\cong f \cdot S^k\) for some idempotent \(f\in\mathbb{M}_k(S)\). As above, we identify \(N\otimes_S P \cong f\cdot P^k\) and \[\begin{align} N\otimes_S \mathop{\mathrm{Hom}}_{-, T}(M,P) &\cong f\cdot \mathop{\mathrm{Hom}}_{-,T}(M,P)^k \\&\cong \mathop{\mathrm{Hom}}_{-,T}(M,f\cdot P^k) \cong \mathop{\mathrm{Hom}}_{-,T}(M,N \otimes_S P).\qedhere \end{align}\] ◻

Proof. We assume \(M\) to be proper and prove the first isomorphism. Lemma 1 implies \[\mathop{\mathrm{Hom}}_{-,T}(M,N \otimes_S P)R \cong \varinjlim \mathop{\mathrm{Hom}}_{-,T}(M,N \otimes_S P)\cdot e \cong \varinjlim \mathop{\mathrm{Hom}}_{-,T}(e\cdot M,N \otimes_S P),\] where \(e\) runs through a local unit in \(R\). Similarly, \(\mathop{\mathrm{Hom}}_{-, T}(M,P) R\) is the inductive limit of \(\mathop{\mathrm{Hom}}_{-,T}(e\cdot M,P)\). For fixed \(e\), Theorem 1 implies an isomorphism \[N\otimes_S \mathop{\mathrm{Hom}}_{-,T}(e\cdot M,P) \cong \mathop{\mathrm{Hom}}_{-,T}(e\cdot M,N \otimes_S P).\] These isomorphisms for all \(e\) are natural and thus induce the desired isomorphism on the inductive limits.

The second isomorphism is proven similarly. Writing both sides as inductive limits over idempotents \(f\) in \(U\), we see that it suffices to prove that the canonical map is an isomorphism \(f N\otimes_S \mathop{\mathrm{Hom}}_{-, T}(M,P)\xrightarrow\sim\mathop{\mathrm{Hom}}_{-,T}(M,f N \otimes_S P)\), and this follows from Theorem 1 because \(f N\) is fgp as an \(S\)-module. ◻

Proof. Let \(R,S,T\) be rings with local units and let \(M={_RM}_S\) and \(N={_SN}_T\) be proper smooth bimodules. We claim that \(M\otimes_SN\) is proper, that is, \(e\cdot(M\otimes_SN)\) is fgp as a right \(T\)-module for every idempotent \(e\in R\). Since \(e\cdot M\) is an fgp right \(S\)-module, \(e\cdot M \cong f\cdot S^k\) for some idempotent \(f\in \mathbb{M}_k(S)\). Then \[e\cdot (M\otimes_SN) \cong (e\cdot M)\otimes_S N \cong f\cdot S^k \otimes_S N \cong f\cdot N^k,\] where \(f\) acts on \(N^k\) by matrix-vector multiplication and the left \(S\)-module structure on \(N\). Since \(S\) is a ring with local units, there is an idempotent \(g\in S\) that is a local unit for each entry of \(f\). Therefore, \(f\cdot N^k\) is contained in \((g\cdot N)^k\). Even more, it is the direct summand of \((g\cdot N)^k\) given by left multiplication with \(f\). Since \(N\) is proper, \(g\cdot N\) is an fgp right \(T\)-module. Then \((g\cdot N)^k\) is fgp as well, and so is any direct summand of it. Thus \(e\cdot (M\otimes_SN)\) is an fgp right \(T\)-module as asserted. ◻

Proof. The subsets \(V U\), \(U W\) and \(\braket{U_1}{U_2}\) are slices by Antunes-Ko-Meyer:Groupoid_correspondences?*Lemma 7.4. In particular, they are Hausdorff and open. In addition, since \(\mathcal{G}^0\) and \(\mathcal{X}/\mathcal{G}\) are Hausdorff, the sets of pairs \((u_1,u_2)\), \((u,v)\) or \((v,w)\) for which \(\braket{u_1}{u_2}\), \(u\cdot v\) or \(v\cdot w\) are defined are closed subsets of the product spaces \(U_1 \times U_2\), \(U \times V\), and \(V\times W\), respectively. These products of compact spaces are again compact, and so are their closed subsets. Since the multiplication and the bracket map are continuous, it follows that \(V U\), \(U W\), and \(\braket{U_1}{U_2}\) are again compact.

Next we prove that \(W Z\subseteq \mathcal{Y}\) is open, Hausdorff and compact. The multiplication map \(\mu\colon \mathcal{G}\times_{\mathcal{G}^0} \mathcal{Y} \to \mathcal{Y}\) of any groupoid action is a local homeomorphism by Antunes-Ko-Meyer:Groupoid_correspondences?*Lemma 2.9. Therefore, \(W Z = \mu(W\times_{\mathcal{G}^0} Z)\) is open in \(\mathcal{Y}\). Since \(s|_W\) is a homeomorphism onto \(s(W)\), the projection \(W\times_{\mathcal{G}^0} Z \to \mathcal{Y}\), \((w,y)\mapsto y\), is a homeomorphism onto \(r_{\mathcal{Y}}^{-1}(s(W))\). So \(W\times_{\mathcal{G}^0} Z\) is a compact Hausdorff space. The restriction of \(\mu\) to \(W\times_{\mathcal{G}^0} Z\) is an injective local homeomorphism, hence a homeomorphism. So \(W Z\) is also compact and Hausdorff. ◻

Proof. Let \(U\in\mathcal{B}_{\mathcal{X}}\) and \(V\in\mathcal{B}_{\mathcal{Y}}\). We first prove that \(U V\subseteq \mathcal{X}\circ\mathcal{Y}\) is open, Hausdorff, and compact. The subset \(U\times_{\mathcal{G}^0} V\subseteq\mathcal{X}\times_{s,\mathcal{G}^0,r}\mathcal{Y}\) is open. It is Hausdorff and compact as a closed subset of the Hausdorff compact space \(U\times V\). We claim that the restriction of \(\pi\) to \(U\times_{s,\mathcal{G}^0,r}V\) is injective. Indeed, let \(\pi(u_1,v_1) = \pi(u_2,v_2)\) for some \(u_1,u_2\in U\), \(v_1,v_2\in V\) with \(s(u_j) = r(v_j)\) for \(j=1,2\). Then there is \(g\in\mathcal{G}\) with \(u_2 = u_1 g\), \(v_1 = g v_2\). Since the orbit space projection on \(U\) is injective, \(u_2 = u_1 g\) implies \(u_1 = u_2\) and \(g=1_{s(u_1)}\). Then \(v_1 = v_2\) follows. Since \(\pi\) is injective on the open set \(U\times_{\mathcal{G}^0} V\) and a local homeomorphism, it restricts to a homeomorphism from \(U\times_{\mathcal{G}^0} V\) onto an open subset \(U V\) of \(\mathcal{X}\circ\mathcal{Y}\). So \(U V\) is open, Hausdorff and compact.

If \(U\in\mathcal{B}_{\mathcal{X}}\), \(V\in\mathcal{B}_{\mathcal{Y}}\), then \(s(U) = \braket{U}{U}\in \mathcal{B}_{\mathcal{G}}\) and \(V' \mathrel{\vcentcolon=}\braket{U}{U} V\in \mathcal{B}_{\mathcal{Y}}\). We compute \(V' = r_{\mathcal{Y}}^{-1}(s(U)) \cap V\), so that \(U V = U V'\) and \(r(V') \subseteq s(U)\). So all elements of \(\mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}\) are of the form \(U V\) for \(U\in\mathcal{B}_{\mathcal{X}}\), \(V\in\mathcal{B}_{\mathcal{Y}}\), with \(s(U) \supseteq r(V)\) as claimed.

Let \(W\subseteq \mathcal{X}\circ \mathcal{Y}\) be an open subset and let \(\pi(x,y)\in W\). We claim that \(W\) contains a neighbourhood of \(\pi(x,y)\) of the form \(U V\). The subset \(\pi^{-1}(W) \subseteq\mathcal{X}\times_{s,\mathcal{G}^0,r}\mathcal{Y}\) is open. So there is an open subset \(\tilde{W}\subseteq \mathcal{X}\times\mathcal{Y}\) with \(\pi^{-1}(W) = \tilde{W}\cap \mathcal{X}\times_{s,\mathcal{G}^0,r}\mathcal{Y}\). Since \((x,y)\in \tilde{W}\) and \(\mathcal{B}_{\mathcal{X}}\) and \(\mathcal{B}_{\mathcal{Y}}\) are bases, there are \(U\in\mathcal{B}_{\mathcal{X}}\) and \(V\in\mathcal{B}_{\mathcal{Y}}\) with \((x,y)\in U\times V\subseteq\tilde{W}\). Then \(\pi(x,y)\in U V\subseteq W\).

It remains to show that the family of subsets \(\mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}\) is stable under intersections and set differences. Let \(U_j\in\mathcal{B}_{\mathcal{X}}\) and \(V_j\in\mathcal{B}_{\mathcal{Y}}\) for \(j=1,2\). We must show that \(U_1 V_1 \cap U_2 V_2\) and \(U_1 V_1 \setminus U_2 V_2\) are in \(\mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}\). First, we claim \[\label{eq:intersect95in95BXY} U_1 V_1 \cap U_2 V_2 = U_1 V_3,\qquad \text{with} \quad V_3 \mathrel{\vcentcolon=}V_1 \cap \braket{U_1}{U_2} V_2.\tag{1}\] Assume \(u_1 v_1 = u_2 v_2\) in \(\mathcal{X}\circ\mathcal{Y}\) for some \(u_j\in U_j\), \(v_j\in V_j\) for \(j=1,2\). This means that there is \(g\in \mathcal{G}\) with \(u_1 g = u_2\) and \(v_1 = g v_2\). This implies \(g = \braket{u_1}{u_2}\) and \(v_1 = g v_2 = \braket{u_1}{u_2} v_2 \in \braket{U_1}{U_2} V_2\). So \(u_1 v_1 \in U_1 V_3\) with \(V_3 = V_1 \cap \braket{U_1}{U_2} V_2\). Conversely, any element in \(U_1 V_3\) belongs to both \(U_1 V_1\) and \(U_1 \braket{U_1}{U_2} V_2 \subseteq U_2 V_2\). This proves the claim. So \(U_1 V_1 \cap U_2 V_2\in \mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}\).

Next, we claim that \[U_1 V_1 \setminus U_2 V_2 = U_1 V_1 \setminus (U_1 V_1 \cap U_2 V_2) = U_1 V_1 \setminus U_1 V_3 \overset{!}= U_1 (V_1 \setminus V_3) \in \mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}.\] Only the equality marked with \(!\) requires justification. Here we use that an element \(u_1 v_1 \in U_1 V_1\) is determined uniquely by \(v_1 \in V_1\) because \(s(u_1) = r(v_1)\) determines \(u_1\) in the slice \(U_1\). Therefore, \(u_1 v_1 \in U_1 V_1\) belongs to \(U_1 V_3\) if and only if \(v_1 \in V_3\), as needed. We have now verified all claims in the lemma. ◻

Proof. First assume that there are \(W_1\) and \(W_2\) satisfying 2 . Any element of \((u_j,v_j)\in U_j \times_{s,r} V_j\) satisfies \(s(u_j) = r(v_j) \in r(W_j)\), so that there is \(g\in W_j\) with \(r(g) = s(u_j) = r(v_j)\). Then \(\pi(u_j,v_j) = \pi(u_j g,g^{-1} v_j) \in (U_j W_j) (W_j^{-1} V_j)\). It is clear that any element of \((U_j W_j) (W_j^{-1} V_j)\) also belongs to \(U_j V_j\). Therefore, our assumptions imply that \(U_1 V_1 = (U_1 W_1) (W_1^{-1} V_1) = (U_2 W_2) (W_2^{-1} V_2) = U_2 V_2\).

Conversely, assume that \(U_1 V_1 = U_2 V_2\). Then \[\label{eq:UV95intersect} U_1 V_1 = U_1 V_1 \cap U_2 V_2 = U_1 (V_1 \cap \braket{U_1}{U_2} V_2)\tag{3}\] by 1 . Since \(U_1\) is a slice, the only way for two elements of the form \(\pi(u,v)\) and \(\pi(u',v')\) with \(u,u'\in U_1\), \(v,v'\in\mathcal{H}\) to have the same class in \(\mathcal{X}\circ \mathcal{Y}\) is if \(u= u'\) and \(v=v'\). Therefore, 3 implies that any element \(v_1\in V_1\) with \(r(v_1)\in s(U_1)\) is also contained in \(\braket{U_1}{U_2}V_2\). Since \(s(\braket{U_1}{U_2}) \subseteq s(U_2)\), we may rewrite this further as \(s(U_1) V_1 \subseteq \braket{U_1}{U_2} s(U_2) V_2\). The same argument with \(1,2\) exchanged gives \(s(U_2) V_2 \subseteq \braket{U_2}{U_1} s(U_1)V_1\). These two equations together imply \(s(U_2) V_2 = \braket{U_2}{U_1} s(U_1)V_1\) and \(s(U_1) V_1 = \braket{U_1}{U_2} s(U_2)V_2\). Then it follows that \(r(V_1) \cap s(U_1) \subseteq r(\braket{U_1}{U_2})\) because \(r(V_1) \cap s(U_1) = r(s(U_1) V_1)\). Exchanging \(1\) and \(2\), the same argument gives \(r(V_2) \cap s(U_2) \subseteq r(\braket{U_2}{U_1})\). Now let \[W_1\mathrel{\vcentcolon=}\braket{U_1}{U_2} \in \mathcal{B}_{\mathcal{G}},\qquad W_2\mathrel{\vcentcolon=}r(\braket{U_2}{U_1}) \in \mathcal{B}_{\mathcal{G}^0} \subseteq \mathcal{B}_{\mathcal{G}}.\] We have seen above that \(r(W_j) \supseteq r(V_j) \cap s(U_j)\) for \(j=1,2\). We claim that \((U_1 W_1, W_1^{-1} V_1) = (U_2 W_2, W_2^{-1} V_2)\) holds as well. To prove this, we use that \(U_1 \braket{U_1}{U_2} \subseteq U_2\) and \(s(U_1\braket{U_1}{U_2}) = r(\braket{U_2}{U_1})\) to conclude that \(U_1 W_1 = U_2 W_2\). The computations above show that \(W_1^{-1} V_1 = \braket{U_2}{U_1} V_1= \braket{U_2}{U_1} s(U_1)V_1 = s(U_2) V_2 = \braket{U_2}{U_1} s(U_1)V_1 = \braket{U_2}{U_1}\braket{U_1}{U_2} s(U_2) V_2 = W_2^{-1} V_2\). ◻

Proof. We prove that \(\ref{enum:steinalg1}\subseteq \ref{enum:steinalg2}\subseteq \ref{enum:steinalg3}\subseteq\ref{enum:steinalg1}\). The third inclusion is immediate, so it remains to prove the first two.

For the first inclusion, take a Hausdorff open subset \(U\subseteq X\) and a map \(\xi\in \mathrm{C_c}(U,R)\). Since subsets of Hausdorff spaces are Hausdorff, \(\mathop{\mathrm{supp}}(\tilde{\xi})=\mathop{\mathrm{supp}}(\xi)\) is Hausdorff. Since \(\xi\) is locally constant, \(\mathop{\mathrm{supp}}(\xi)\) is open in \(U\) and as \(U\) is open in \(X\), it is also open in \(X\). Thus \(\mathop{\mathrm{supp}}(\tilde{\xi})\) is a compact, Hausdorff open subset of \(X\). The restriction of \(\tilde{\xi}\) to its support remains continuous as a function on \(\mathop{\mathrm{supp}}(\tilde{\xi})\subseteq U\).

For the second inclusion, we take a map \(\xi\colon X\to R\) such that \(U\mathrel{\vcentcolon=}\mathop{\mathrm{supp}}(\xi)\subseteq X\) is compact Hausdorff open and \(\xi\vert_U\) is locally constant. Then \(\xi (U)\) is a finite subset \(\xi(U)=\{r_1,\dotsc,r_n\}\subseteq R\). The subsets \(U_i\mathrel{\vcentcolon=}\xi^{-1}(r_i)\subseteq \mathop{\mathrm{supp}}(\xi)\) for \(i=1,\dotsc, n\) are closed and open in \(U\) because \(\xi\) is locally constant. Since \(U\) is compact and Hausdorff, they are compact and Hausdorff as well. The subsets \(U_i\) are disjoint and \(X=\xi^{-1}(0)\sqcup U_1\sqcup\dotsb\sqcup U_n\). Then \(\xi=\sum_{i=1}^n r_i\mathbb{1}_{U_i}\) follows. ◻

Proof. We use Definition [enum:steinalg3] of the Steinberg module. Let \[\iota_i\colon A_R(X_i) \to A_R(X),\qquad \xi \mapsto \tilde{\xi},\] be the extension by zero map. It is a well-defined \(R\)-module homomorphism because \(X_i\subseteq X\) is open. These maps induce a map \(\iota\colon \bigoplus_{i\in I} A_R(X_i) \to A_R(X)\). Since the sets \(X_i\) are disjoint, the map \(\iota\) is clearly injective. It remains to prove that it is also surjective. Let \(U\subseteq X\) be compact, Hausdorff, and open. Then \(U_i \mathrel{\vcentcolon=}U\cap X_i\) is compact, Hausdorff, and open for \(i \in I\), and only finitely many of them are nonempty. So \(\mathbb{1}_{U} = \sum_{i\in F} \iota_i(\mathbb{1}_{U_i})\) for some finite subset \(F\subseteq I\). ◻

Proof. A finite disjoint union of compact, Hausdorff spaces is again compact and Hausdorff, and a finite union of open subsets is again open. Thus any finite disjoint union of subsets in \(\mathcal{B}\) is Hausdorff, compact and open in \(X\). Conversely, let \(U\subseteq X\) be compact, Hausdorff, and open. Since \(\mathcal{B}\) is a base, for any \(x\in U\) there is \(U_x\in\mathcal{B}\) with \(x\in U_x\) and \(U_x \subseteq U\). These sets cover \(U\). Since \(U\) is compact, there are finitely many \(U_i\in \mathcal{B}\) for \(i=1,\dotsc, n\) with \(U=\bigcup_{i=1}^n U_i\). As \(U\) is Hausdorff and each \(U_i\) is compact, \(U_i\) is relatively closed in \(U\). For \(i=1,\dotsc,n\), define \(W_i\mathrel{\vcentcolon=}U_i\setminus (\bigcup_{j=1}^{i-1} U_j)\). These sets are disjoint, closed and open in \(U\), and satisfy \(U=\bigsqcup_{i=1}^n W_i\). They belong to \(\mathcal{B}\) because we may get them by repeating the set difference operation. ◻

Proof. The map \(\pi\) is well-defined because \(\mathbb{1}_{B}\in A_R(X)\) for \(B\in\mathcal{B}\). Proposition 1 implies that the \(R\)-module generated by \(\mathbb{1}_{B}\) for \(B\in\mathcal{B}\) contains \(\mathbb{1}_{U}\) for any Hausdorff, compact, open subset \(U\subseteq X\). The latter functions generate \(A_R(X)\) by [enum:steinalg3]. Thus \(\pi\) is surjective.

Let \(T\) denote the \(R\)-submodule in \(\bigoplus_{B\in\mathcal{B}} R\) generated by \(\delta_{U\sqcup V} - \delta_U - \delta_V\) for disjoint \(U,V\in\mathcal{B}\) with \(U \sqcup V \in\mathcal{B}\). If \(U,V\in\mathcal{B}\) are disjoint, then \(\mathbb{1}_{U\sqcup V}=\mathbb{1}_{U}+\mathbb{1}_{V}\). This implies \(\pi|_T=0\). Let \(x = \sum_{j=1}^N a_j \delta_{U_j}\) be an element of \(\ker \pi\). We want to subtract an element of \(T\) so that the result is equal to such a sum where all \(U_j\) are disjoint. Recall that \(U_1,U_2\in\mathcal{B}\) implies \(U_1 \cap U_2\in\mathcal{B}\) and \(U_1 \setminus U_2 \in \mathcal{B}\) and our relations give \[\delta_{U_1} \equiv \delta_{U_1\cap U_2} + \delta_{U_1 \setminus U_2} \bmod T.\] Iterating this, an induction argument on \(\abs{I \cup J}\) shows that \(\mathcal{B}\) contains \[U_{I,J} \mathrel{\vcentcolon=}\bigcap_{i\in I} U_i \setminus \bigcup_{j\in J} U_j\] for any disjoint finite subsets \(I,J\subseteq \{1,\dotsc,N\}\) and that the sum of \(\delta_{U_{I,J}}\) for all disjoint \(I,J\) with \(i \in I\) and fixed \(I \cup J\) is equivalent to \(\delta_i\) modulo \(T\). By construction, if \(I_k,J_k\) are disjoint pairs with \(I_k \cup J_k = \{1,\dotsc,N\}\) for \(k=1,2\), then \(U_{I_1,J_1} \cap U_{I_2,J_2} = \emptyset\) unless \(I_1=I_2\) and \(J_1 = J_2\). The argument above shows that \(x \equiv \sum_{I,J} a_{I,J} \delta_{U_{I,J}} \bmod T\) for some \(a_{I,J}\in R\), where the sum runs over all pairs \(I,J\) of disjoint sets with \(I \cup J = \{1,\dotsc,N\}\). Now \(\pi(x)=0\) implies that \(a_{I,J}=0\) or \(U_{I,J}=\emptyset\) for all \(I,J\). Since \(\delta_\emptyset = -(\delta_\emptyset + \delta_\emptyset - \delta_{\emptyset \cup \emptyset})\in T\), this implies \(x\in T\) as desired. ◻

Proof. Let \(\mathcal{B}_X'\) be the set of compact open subsets \(U\in\mathcal{B}_X\) such that \(f|_U\) is injective. This subset of \(\mathcal{B}_X\) is an ample base for \(X\) as well because \(f\) is a local homeomorphism. If \(U\in \mathcal{B}_X'\), then \(f(U)\subseteq Y\) is compact, Hausdorff and open because \(f\) is a local homeomorphism, and \(f_*(\mathbb{1}_{U}) = \mathbb{1}_{f(U)}\). Thus \(f_*(\mathbb{1}_{U})\in A_R(Y)\). This implies that \(f_*\) maps \(A_R(X)\) to \(A_R(Y)\) because functions of the form \(\mathbb{1}_{U}\) for \(U\in \mathcal{B}_X'\) generate \(A_R(X)\) as an \(R\)-module by Theorem 3. If \(U\in\mathcal{B}_X\), then \(g^{-1}(U)\subseteq Y\) is open because \(g\) is continuous, and Hausdorff and compact because \(g\) is proper. It is clear that \(g^*(\mathbb{1}_{U}) = \mathbb{1}_{g^{-1}(U)}\). Then \(g^*\) maps \(A_R(X)\) to \(A_R(Y)\) by Theorem 3. ◻

Proof. We omit the proof of this well known result because we will prove a more general statement in Lemma 8. ◻

Proof. Let \(\xi_1,\dotsc,\xi_n\in A_R(\mathcal{X})\). Then \(\xi_j\) are \(R\)-linear combinations of characteristic functions of Hausdorff, compact, open subsets of \(\mathcal{X}\). Their images under the anchor maps in \(\mathcal{H}^0\) and \(\mathcal{G}^0\) are compact. Since the latter spaces are ample and Hausdorff, the unions of these compact subsets are again compact and admit compact open neighbourhoods \(U\subseteq \mathcal{H}^0\), \(W\subseteq \mathcal{G}^0\). Then \(\mathbb{1}_{U}* \xi_j = \xi_j = \xi_j * \mathbb{1}_{W}\) for \(j=1,\dotsc,n\). ◻

Proof. Since \(\mathbb{1}_{K}*\mathbb{1}_{V}=\mathbb{1}_{K V} = \mathbb{1}_{V\cap r^{-1}(K)}\) for any compact open slice \(V\), it follows that \(\mathbb{1}_{K}* A_R(\mathcal{X}) = A_R(r_{\mathcal{X}}^{-1}(K))\). The map \(\mathcal{X}/\mathcal{G}\to\mathcal{H}^0\) induced by \(r_{\mathcal{X}}\) is proper by assumption. So the right \(\mathcal{G}\)-space \(r_{\mathcal{X}}^{-1}(K)\) is cocompact, that is, \(\Bar{r}^{-1}_{\mathcal{X}}(K)=\Pi(r_{\mathcal{X}}^{-1}(K))\) is a compact subset of \(\mathcal{X}/\mathcal{G}\), where \(\Pi\colon \mathcal{X}\to\mathcal{X}/\mathcal{G}\) is the quotient map. Since \(K\) is also open, so is \(r_{\mathcal{X}}^{-1}(K) \subseteq \mathcal{X}\). Therefore, if \(x\in \Pi(r_{\mathcal{X}}^{-1}(K))\), then there is \(U_x\in\mathcal{B}_{\mathcal{X}}\) with \(x \in \Pi(U_x)\) and \(U_x \subseteq r_{\mathcal{X}}^{-1}(K)\). The sets \(\Pi(U_x)\) form an open cover of \(\Pi(r_{\mathcal{X}}^{-1}(K))\). Since the latter is compact, there are finitely many elements so that \(\bigcup_{i=1}^n U_{x_i}\cdot\mathcal{G}= r_{\mathcal{X}}^{-1}(K)\). The image \(\Pi(U_{x_i})\) is still compact open in \(\mathcal{X}/\mathcal{G}\). Now we replace these subsets by ones with disjoint images in \(\mathcal{X}/\mathcal{G}\), letting \[U_j \mathrel{\vcentcolon=}U_{x_j} \setminus \bigcup_{i=1}^{j-1} \Pi^{-1}(\Pi(U_{x_i})) = U_{x_j} \setminus \bigcup_{i=1}^{j-1} U_{x_i} \cdot \mathcal{G}\] for \(j=1,\dotsc,n\). Then \(\Pi(U_j) = \Pi(U_{x_j}) \setminus \bigcup_{i=1}^{j-1} \Pi(U_{x_i})\). So the subsets \(\Pi(U_j) \subseteq \mathcal{X}/\mathcal{G}\) are disjoint and their union is still \(\Pi(r_{\mathcal{X}}^{-1}(K))\). Thus \[r_{\mathcal{X}}^{-1}(K) = \bigsqcup U_j \cdot \mathcal{G}, \qquad A_R(r_{\mathcal{X}}^{-1}(K)) \cong \bigoplus A_R(U_j \cdot \mathcal{G}).\] We claim that \(U_j\in \mathcal{B}_{\mathcal{X}}\) for \(j=1,\dotsc,n\). We may construct \(U_j\) by repeated set differences. So it suffices to prove that \(U \setminus (V\cdot \mathcal{G}) \in\mathcal{B}_{\mathcal{X}}\) if \(U,V\in\mathcal{B}_{\mathcal{X}}\). Since \(U\) and \(V\) are compact and the right \(\mathcal{G}\)-action is proper, the set of \(g\in \mathcal{G}\) for which there are \(u\in U\), \(v\in V\) with \(u = v g\) is compact. By Proposition 1, we may write this set as a finite union \(\bigcup_{j=1}^\ell W_j\) of sets in \(\mathcal{B}_{\mathcal{G}}\). So \(U \setminus (V\cdot \mathcal{G}) = U\setminus \bigcup_{j=1}^\ell V\cdot W_j\). The assumptions in Lemma 5 imply \(V W_j\in \mathcal{B}_{\mathcal{X}}\) for all \(j\), and then repeated set difference shows that \(U \setminus (V\cdot \mathcal{G}) \in \mathcal{B}_{\mathcal{X}}\) as needed.

Since \(U_i\) is a slice, \(s|_{U_i}\) is a homeomorphism \(U_i \xrightarrow\sim s(U_i)\), and \(s(U_i)\) is a compact open subset of \(\mathcal{G}^0\). The map \(x\cdot g\mapsto s(x)\cdot g\) for \(x\in U_i\), \(g\in\mathcal{G}\) with \(s(x) = r(g)\) is a homeomorphism from \(U_i\cdot \mathcal{G}\subseteq \mathcal{X}\) onto \(\mathcal{G}_{s(U_i)}\) because the right \(\mathcal{G}\)-action on \(\mathcal{X}\) is free and proper. Thus \(A_R(U_i\cdot \mathcal{G}) \cong A_R(s(U_i)\cdot \mathcal{G}) = A_R(\mathcal{G}_{s(U_i)}) = \mathbb{1}_{s(U_i)} * A_R(\mathcal{G})\). Now the asserted direct sum decomposition follows. The assumptions in Lemma 5 also imply \(s(U_j) = \braket{U_j}{U_j} \in \mathcal{B}_{\mathcal{G}}\) as asserted. ◻

Proof. Proposition 1 implies that \(\mathbb{1}_{K} * A_R(\mathcal{X})\) for a compact open subset \(K\subseteq \mathcal{H}^0\) is fgp as a right \(A_R(\mathcal{G})\)-module. This implies the claim because \(\mathbb{1}_{K}\) for such \(K\) is a local unit in \(A_R(\mathcal{G})\) by Proposition 1. ◻

Proof. Recall that the compact slices form an ample base for the topology on \(\mathcal{X}\). Proposition 1 implies that \(A_R(f)\) maps \(A_R(\mathcal{X})\) to \(A_R(\mathcal{Y})\) and maps \(\mathbb{1}_{U}\) to \(\mathbb{1}_{f(U)}\) as claimed. If \(W\subseteq \mathcal{H}\), \(V\subseteq \mathcal{G}\) are compact slices, then \(A_R(f)\) maps \(\mathbb{1}_{U}* \mathbb{1}_{V} = \mathbb{1}_{U V}\) to \(\mathbb{1}_{f(U)V}= \mathbb{1}_{f(U)}*\mathbb{1}_{V}\) and \(\mathbb{1}_{W}* \mathbb{1}_{U} = \mathbb{1}_{W U}\) to \(\mathbb{1}_{W f(U)}= \mathbb{1}_{W}*\mathbb{1}_{f(U)}\). Therefore, \(A_R(f)\) is an \(A_R(\mathcal{H}) , A_R(\mathcal{G})\)-bimodule homomorphism. The functoriality of \(f\mapsto A_R(f)\) is trivial. ◻

Proof. The sum \([x,y] \mapsto \sum_{g\in \mathcal{G}_{s(x)}} \alpha(xg^{-1})\cdot\beta(gy)\) does not depend on the representation of \([x,y]\), since a different representative \((x\tilde{g}^{-1},\tilde{g} y)\in [x,y]\) only changes the order of the summands. It is easy to check that \(\tilde{\mu}_{\mathcal{X},\mathcal{Y}}(\mathbb{1}_{U}, \mathbb{1}_{V}) = \mathbb{1}_{U V}\).

In the following proof, we pick ample bases for \(\mathcal{X}\), \(\mathcal{G}\) and \(\mathcal{Y}\) that satisfy the assumptions in Lemma 5. For instance, taking the ample bases of all compact slices will do. The balanced tensor product commutes with direct sums and cokernels, and \(R\otimes_R R \cong R\). Since \(A_R(\mathcal{G})\) is spanned by \(\mathbb{1}_{W}\) for \(W\in\mathcal{B}_{\mathcal{G}}\), the balancing of the tensor product \(A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} A_R(\mathcal{Y})\) over \(A_R(\mathcal{G})\) has the same effect as dividing out \(x*\mathbb{1}_{W} \otimes y - x \otimes \mathbb{1}_{W}*y\) for all \(W\in\mathcal{B}_{\mathcal{G}}\) and all generators \(x\) and \(y\) for \(A_R(\mathcal{X})\) and \(A_R(\mathcal{Y})\). Therefore, Theorem 3 implies that \(A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} A_R(\mathcal{Y})\) is the quotient of the free \(R\)-module on the set of pairs \((U,V)\) with \(U\in\mathcal{B}_{\mathcal{X}}\), \(V\in\mathcal{B}_{\mathcal{Y}}\) by the subspace \(S\) generated by \[\delta_{(U_1 \sqcup U_2,V)} - \delta_{(U_1,V)} - \delta_{(U_2,V)},\quad \delta_{(U,V_1 \sqcup V_2)} - \delta_{(U,V_1)} - \delta_{(U,V_2)},\quad \delta_{U W,V} - \delta_{U, W V}\] for all \(U,U_1,U_2\in \mathcal{B}_{\mathcal{X}}\), \(V,V_1,V_2\in \mathcal{B}_{\mathcal{X}}\), and \(W\in\mathcal{B}_{\mathcal{G}}\) with \(U_1 \sqcup U_2 = U\) and \(V_1 \sqcup V_2 = V\).

We first implement the relation \(\delta_{U W,V} \equiv \delta_{U,W V}\). We claim that the quotient by this relation is the free module generated by the elements of the base \(\mathcal{B}_{\mathcal{X}\circ \mathcal{Y}}\) in Lemma 5. First, it is clear that \((U W) V = U (W V)\) holds in \(\mathcal{B}_{\mathcal{X}\circ \mathcal{Y}}\). Secondly, let \(U_j,V_j,W_j\) be as in Lemma 6. Let \(j\in \{1,2\}\). Since \(r(W_j) \supseteq r(V_j) \cap s(U_j)\), the compact open subset \(s(U_j) \setminus r(W_j)\) is disjoint from \(r(V_j)\). Since the latter is compact, there is a compact open set \(X_j\) with \(r(V_j) \subseteq X_j\) and \(X_j \cap s(U_j) \setminus r(W_j) = \emptyset\), so that \(X_j \cap s(U_j) \subseteq r(W_j)\). Then \(X_j V_j = V_j\), so that \((U_j,V_j) \equiv (U_j X_j, V_j) = (U_j W_j W_j^{-1}, V_j) \equiv (U_j W_j,W_j V_j)\). Thus \((U_1,V_1)\) and \((U_2,V_2)\) are identified when we divide out the relations \((U W,V) \equiv (U, W V)\).

Let \(U,U_1,U_2\in\mathcal{B}_{\mathcal{X}}\), \(V,V_1,V_2\in\mathcal{B}_{\mathcal{Y}}\) be such that \(U = U_1 \sqcup U_2\) and \(V = V_1 \sqcup V_2\). In \(A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} A_R(\mathcal{Y})\), we also divide out the relations \(\delta_{(U_1 \sqcup U_2,V)} - \delta_{(U_1,V)} - \delta_{(U_2,V)}\) and \(\delta_{(U,V_1 \sqcup V_2)} - \delta_{(U,V_1)} - \delta_{(U,V_2)}\). We claim that, after identifying the generators \(\delta_{(U_1,V_1)}\) and \(\delta_{(U_2,V_2)}\) when \(U_1 V_1 = U_2 V_2\), these relations are exactly the same relations as the relation \(\delta_X - \delta_{X_1} - \delta_{X_2}\) for all \(X,X_1,X_2 \in \mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}\) with \(X= X_1 \sqcup X_2\), which occur in the presentation of \(A_R(\mathcal{X}\circ\mathcal{Y})\) in Theorem 3.

First, the assumptions above imply \(U V = U_1 V \cup U_2 V = U V_1 \cup U V_2\). In addition, since \(U \supseteq U_1,U_2\) and \(V\supseteq V_1,V_2\) are slices, 1 implies that \(U_1 V \cap U_2 V = \emptyset\) and \(U V_1 \cap U V_2 = \emptyset\) in \(\mathcal{X}\circ\mathcal{Y}\). Therefore, \(\delta_{U V} - \delta_{U_1 V} - \delta_{U_2 V}\) and \(\delta_{U V} - \delta_{U V_1} - \delta_{U V_2}\) are among the generating relations in Theorem 3.

Conversely, let \(X_1\sqcup X_2 = X\) with disjoint \(X_1,X_2,X\in\mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}\). Write \(X = U V\) for \(U\in\mathcal{B}_{\mathcal{X}}\), \(V\in\mathcal{B}_{\mathcal{Y}}\) with \(s(U) \supseteq r(V)\). Equation 1 applied to the trivial statements \(X_j = X\cap X_j\) for \(j=1,2\) implies that \(X_j = U V_j\) for some \(V_1,V_2 \subseteq V\) with \(V_1,V_2\in\mathcal{B}_{\mathcal{Y}}\). In addition, it follows that \(X_1 \cap X_2 = U(V_1 \cap V_2)\) and \((X\setminus X_1) \setminus X_2 = U \bigl((V\setminus V_1) \setminus V_2\bigr)\). Since \(s(U) \supseteq r(V)\), these sets would be nonempty if \(V_1 \cap V_2\) or \((V\setminus V_1) \setminus V_2\) were nonempty, respectively. Thus, \(V_1 \cap V_2 = \emptyset\) and \(V_1 \sqcup V_2 =V\). So our identification maps the relation \(\delta_{(U,V)} - \delta_{(U,V_1)} - \delta_{(U,V_2)}\) from the presentation of \(A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} A_R(\mathcal{Y})\) to the relation \(\delta_X - \delta_{X_1} - \delta_{X_2}\) from the presentation of \(A_R(\mathcal{X}\circ\mathcal{Y})\). As a consequence, both \(A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} A_R(\mathcal{Y})\) and \(A_R(\mathcal{X}\circ\mathcal{Y})\) are the quotients of the free module on \(\mathcal{B}_{\mathcal{X}\circ\mathcal{Y}}\) by the same relations. This makes them isomorphic. The isomorphism is the map that maps \(\mathbb{1}_{U}\otimes \mathbb{1}_{V}\) to \(\mathbb{1}_{U V}\). The same formula holds for \(\tilde{\mu}_{\mathcal{X},\mathcal{Y}}\), and so the latter is an isomorphism as asserted.

If \(X\in\mathcal{B}_{\mathcal{H}}\), \(Y\in\mathcal{B}_{\mathcal{K}}\), then \(\mathbb{1}_{X} * \mathbb{1}_{U} = \mathbb{1}_{X U}\), \(\mathbb{1}_{V} * \mathbb{1}_{Y} = \mathbb{1}_{V Y}\), \(\mathbb{1}_{X} * \mathbb{1}_{U V} = \mathbb{1}_{(X U) V}\), and \(\mathbb{1}_{U V} * \mathbb{1}_{Y} = \mathbb{1}_{U (V Y)}\). These computations on slices imply immediately that the map \(\mu_{\mathcal{X},\mathcal{Y}}\) is an \(A_R(\mathcal{H}) , A_R(\mathcal{K})\)-bimodule homomorphism. If \(f\colon \mathcal{X}\to\mathcal{X}'\) and \(g\colon \mathcal{Y}\to\mathcal{Y}'\) are \(2\)-arrows of groupoid correspondences, then \(f(U)\subseteq \mathcal{X}'\) and \(g(V) \subseteq \mathcal{Y}'\) are slices as well, and the induced map \(f\circ g\colon \mathcal{X}\circ \mathcal{Y} \to \mathcal{X}' \circ \mathcal{Y}'\) maps \(U V\) to \(f_*(U) g_*(V)\). This implies that \(\mu_{\mathcal{X},\mathcal{Y}}\) is natural. ◻

Proof. Being normal means that the identity groupoid correspondence on an ample groupoid \(\mathcal{G}\) is mapped to the identity bimodule on the Steinberg algebra \(A_R(\mathcal{G})\) and that the multiplicativity maps \(\mu_{\mathcal{G},\mathcal{Y}}\) and \(\mu_{\mathcal{X},\mathcal{G}}\) are the canonical bimodule maps. This is easy to check using that \(\mu_{\mathcal{X},\mathcal{Y}}\) maps \(\mathbb{1}_{U} \otimes \mathbb{1}_{V} \mapsto \mathbb{1}_{U V}\) for compact slices \(U,V\) in \(\mathcal{X},\mathcal{Y}\). It is also easy to check that \(\mu_{\mathcal{X},\mathcal{Y}}\) is natural with respect to maps of groupoid correspondences. For a pseudofunctor, it remains to check the following commuting diagram for three composable ample correspondences \(\mathcal{X}\colon \mathcal{H}\leftarrow \mathcal{G}\), \(\mathcal{Y}\colon\mathcal{G}\leftarrow \mathcal{K}\) and \(\mathcal{Z}\colon \mathcal{K}\leftarrow\mathcal{L}\): \[\begin{tikzcd}[column sep=large] {\bigl( A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} A_R(\mathcal{Y})\bigr)\otimes_{ A_R(\mathcal{K})} A_R(\mathcal{Z})} \arrow[r, "{\mu_{\mathcal{X},\mathcal{Y}}\otimes \mathrm{id}}", "{\cong}"'] \arrow[d, "\mathrm{assoc}", "{\cong}"'] & { A_R(\mathcal{X}\circ \mathcal{Y})\otimes_{A_R(\mathcal{K})} A_R(\mathcal{Z})} \arrow[d, "{\mu_{\mathcal{X}\circ \mathcal{Y}, \mathcal{Z}}}", "{\cong}"'] \\ { A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} \bigl(A_R(\mathcal{Y}) \otimes_{A_R(\mathcal{K})} A_R(\mathcal{Z})\bigr)} \arrow[d, "{\mathrm{id}\otimes\mu_{\mathcal{Y},\mathcal{Z}}}", "{\cong}"'] & { A_R({(\mathcal{X}\circ \mathcal{Y})\circ_{\mathcal{K}}\mathcal{Z}})} \arrow[d, "\mathfrak A(\mathrm{assoc})", "{\cong}"']\\ { A_R(\mathcal{X})\otimes_{ A_R(\mathcal{G})} A_R({\mathcal{Y}\circ_{\mathcal{K}}\mathcal{Z}})} \arrow[r, "{\mu_{\mathcal{X}, \mathcal{Y}\circ_{\mathcal{K}} \mathcal{Z}}}", "{\cong}"'] & { A_R({\mathcal{X}\circ (\mathcal{Y}\circ_{\mathcal{K}}\mathcal{Z})})} \end{tikzcd}\] This commutes because both ways around the diagram map \(\mathbb{1}_{U} \otimes \mathbb{1}_{V} \otimes \mathbb{1}_{W}\) for compact slices \(U\), \(V\) and \(W\) in \(\mathcal{X}\), \(\mathcal{Y}\) and \(\mathcal{Z}\) to \(\mathbb{1}_{U V W}\). The pseudofunctor maps proper correspondences to proper bimodules by Proposition 1. ◻

Proof. The equivalences in the bicategories of groupoid correspondences and of bimodules are exactly the Morita equivalences of groupoids and rings with local units, respectively. For the groupoid case, this is Meyer:Groupoid_models_relative?*Theorem 2.3. The case of rings is trivial if Morita equivalence is defined using equivalence bimodules. Since a pseudofunctor maps equivalences to equivalences, the Steinberg algebra pseudofunctor maps Morita equivalences of groupoids to Morita equivalences of rings with local units. ◻

Proof. It is trivial that the maps \(\tilde{\eta}_p\mathrel{\vcentcolon=} f\circ\tilde{\nu}_p\) for \(p\in P\) form a lax covariant representation if \((\tilde{\nu}_p)\) does and that \(\mathop{\mathrm{CovRep}}_{\mathrm{lax}}(-,\mathcal{F})\) is a functor. Assume now that \((\tilde{\nu}_p)_{p\in P}\) is a strong covariant representation. Let \(p\in P\). The bimodule homomorphism \(F_p\otimes_{F_1}D_2 \to D_2\) corresponding to \(\tilde{\eta}_p\) is the composite map \[F_p\otimes_{F_1}D_2 \cong F_p\otimes_{F_1}(D_1\otimes_{D_1}D_2) \cong (F_p\otimes_{F_1}D_1)\otimes_{D_1}D_2 \xrightarrow[\cong]{\nu_p} D_1\otimes_{D_1}D_2 \cong D_2.\] Since this is an isomorphism, \((\tilde{\eta}_p)_{p\in P}\) is a strong covariant representation. ◻

Proof. A covariant representation consists of maps \(\tilde{\nu}_p\colon F_p \to D\) for \(p\in P\) with some properties. The property \(\tilde{\nu}_p(x) \cdot \tilde{\nu}_q(y) = \tilde{\nu}_{p q}(\mu_{p,q}(x\otimes y))\) for all \(x\in F_p\), \(y\in F_q\), \(p,q\in P\) holds if and only if \(\bigoplus \tilde{\nu}_p\colon L \to D\) is a ring homomorphism. The map \(\nu_1\colon F_1 \otimes_{F_1} D \to D\) is an isomorphism if and only if the representation of \(F_1\) in \(D\) is nondegenerate, if and only if the representation of \(L\) in \(D\) is nondegenerate. In the nondegenerate case, the maps \(\nu_p\) in Definition 11 are isomorphisms for all \(p\) if and only if their restrictions \(e\nu_p\colon e F_p \otimes_{F_1} \mathrm{id}_D \to e D\) are isomorphisms for all \(e,p\). These maps are equivalent to the maps \(e\psi_p\otimes_{L} \mathrm{id}_D\).

We prove the last statement. If the map \(e\psi_p\otimes_{L} \mathrm{id}_D\) is an isomorphism, then so is the map \(f\psi_p\otimes_{L} \mathrm{id}_D\) for any idempotent \(f\) with \(f \le e\), that is, \(f e = f\). Therefore, if \(e\psi_p\otimes_{L} \mathrm{id}_D\) is an isomorphism for all \(e\) in a local unit, then it is an isomorphism for all idempotents \(e\in F_1\), and then \(\psi_p\otimes_L \mathrm{id}_D\) is an isomorphism. This property is hereditary for products in \(P\), so it suffices if this holds for \(p\) in a given set of generators. ◻

Proof. The Yoneda Lemma implies that all Cohn localisations are canonically isomorphic. The existence proof below uses the same idea as in the unital case. First, we use Lemma 2 to identify \(P_i \cong e_i R^{m_i}\), \(Q_i \cong f_i R^{n_i}\) using \(m_i,n_i\in\mathbb{N}\) and idempotent matrices \(e_i,f_i\). Any module map \(P_i \to Q_i\) such as \(u_i\) becomes the map of left multiplication by a matrix \(u_i' \in f_i \mathbb{M}_{n_i,m_i}(R) e_i \subseteq \mathbb{M}_{n_i,m_i}(R)\). To make the map \(u_i\) invertible, we need to adjoin quasi-inverses to the matrices \(u_i'\). More precisely, we let \(R'\) be the ring with adjoined elements \((u_i^\dagger)_{1\le j\le m_i, 1\le k\le n_i}\) for \(i\in I\) – that is, we adjoin the entries of the \(m_i,n_i\)-matrices \(u_i^\dagger\) to \(R\) – subject to the relations \(u_i^\dagger = e_i\cdot u_i^\dagger \cdot f_i\), \(u_i' \cdot u_i^\dagger = f_i\), \(u_i^\dagger \cdot u_i' = e_i\), rewritten in terms of the entries of these matrices, that is, as relations involving the entries in \(R\) or \(R'\) of the matrices \(u_i'\), \(e_i\), \(f_i\) and \(u_i^\dagger\). These relations express exactly that left multiplication by the matrix \(u_i^\dagger\) is a map \(f_i (R')^{n_i} \to e_i (R')^{m_i}\) that is inverse to left multiplication by \(u_i'\). This gives the desired inverse to \(u_i\). The resulting ring \(R'\) has the desired universal property. The universal property of the Cohn localisation of \(L\) at the maps \(e\psi_p\) is exactly the same as the universal property of the covariance ring of the diagram. ◻

Proof. The only choice was the factorisation of an element of \(F_p^*\) as \(T\cdot e\) for some \(T\in \mathop{\mathrm{Hom}}_{-,F_1}(F_p,F_1)\) and some \(e\in F_1\). The formulas above define a map \[\mathop{\mathrm{Hom}}_{-,F_1}(F_p,F_1) \otimes_{F_1} F_1 \to \mathcal{O},\qquad T\otimes e\mapsto \nu_1 (T\otimes_{F_1} \mathrm{id}_{\mathcal{O}}) \nu_p^{-1} (\tilde{\nu}_1(e)).\] This proves that \(\kappa_p^*\) is well-defined because the multiplication map is an isomorphism \(\mathop{\mathrm{Hom}}_{-,F_1}(F_p,F_1) \otimes_{F_1} F_1 \xrightarrow\sim \mathop{\mathrm{Hom}}_{-,F_1}(F_p,F_1) F_1\). ◻

Proof. We first prove the last claim and let \(U,V\subseteq\mathcal{X}\) be slices. If \(f=\mathbb{1}_{U^*}\), \(h = \mathbb{1}_{V}\), then \(\mathcal{I}_{\mathcal{X}}(\mathbb{1}_{U^*})(\mathbb{1}_{V})\) is the function on \(\mathcal{G}\) that maps \(\gamma\in \mathcal{G}\) to the number of \(x\in\mathcal{X}\) with \(s(x) = s(\gamma)\), \(x\gamma^{-1}\in U\) and \(x\in V\). Since \(s|_V\) is injective, this number is either \(0\) or \(1\). It is \(1\) if and only if \(x\in V\) as above exists. Recall that \(\braket{x}{y}\) for \(x,y\in \mathcal{X}\) is the unique \(g\in\mathcal{G}\) with \(r(g)= s(x)\) and \(x\cdot g = y\). So \(\gamma = \braket{x\gamma^{-1}}{x}\), and thus \(x\in\mathcal{X}\) with \(s(x) = s(\gamma)\), \(x\gamma^{-1}\in U\) and \(x\in V\) exist if and only if \(\gamma = \braket{x \gamma^{-1}}{x} \in \braket{U}{V}\). Therefore, \(\mathcal{I}_{\mathcal{X}}(\mathbb{1}_{U^*})(\mathbb{1}_{V}) = \mathbb{1}_{\braket{U}{V}}\) as claimed.

By Antunes-Ko-Meyer:Groupoid_correspondences?*Proposition 3.5, if \(U,V\subseteq \mathcal{X}\) and \(W\subseteq \mathcal{G}\), \(X\subseteq \mathcal{H}\) are slices, then \[\label{eq:braket95products} \braket{X U}{X V} = \braket{U}{V}, \qquad \braket{U}{V W} = \braket{U}{V} W,\qquad \braket{U W}{V} = W^{-1} \braket{U}{V}.\tag{7}\] The first condition is equivalent to \[\label{eq:braket95products952} \braket{X^{-1} U}{V} = \braket{U}{X V}.\tag{8}\] Characteristic functions of slices span the relevant Steinberg spaces as \(R\)-modules by Theorem 3, and the map \((f,h)\mapsto \mathcal{I}_{\mathcal{X}}(f)(h)\) is clearly \(R\)-bilinear. Therefore, 7 implies that the map \(\mathcal{I}_{\mathcal{X}}(f)\colon A_R(\mathcal{X}) \to A_R(\mathcal{G})\) is right \(A_R(\mathcal{G})\)-linear and that \(\mathcal{I}_{\mathcal{X}}\) is a right \(A_R(\mathcal{H})\)-linear map to \(\mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl( A_R(\mathcal{X}), A_R(\mathcal{G})\bigr)\). It is also left \(A_R(\mathcal{G})\)-linear by 8 . Since \(A_R(\mathcal{X}^*)\) is a nondegenerate right module over \(A_R(\mathcal{H})\), it follows that the range of \(\mathcal{I}_{\mathcal{X}}\) belongs to \(\mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl( A_R(\mathcal{X}), A_R(\mathcal{G})\bigr) A_R(\mathcal{H})\). So \(\mathcal{I}_{\mathcal{X}}\) is a well-defined \(A_R(\mathcal{G})\)-\(A_R(\mathcal{H})\)-bimodule map.

Let \(R,S\) be rings with local units and let \({}_R M_S\) be a smooth bimodule. Proposition 1 implies \(M \cong \varinjlim M\cdot e\) as left \(R\)-modules, where the inductive limit runs over the set of idempotents in \(S\). We use this for \(S=A_R(\mathcal{H})\). We may replace the set of all idempotents by the local unit consisting of the idempotents \(\mathbb{1}_{K}\in A_R(\mathcal{H})\) for compact open subsets \(K\subseteq \mathcal{H}^0\) because this is a cofinal subset. Since \(\mathcal{I}_{\mathcal{X}}\) is right \(A_R(\mathcal{H})\)-linear, it restricts to a map from \(A_R(\mathcal{X}^*)* \mathbb{1}_{K}\) to \(\mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl( A_R(\mathcal{X}), A_R(\mathcal{G})\bigr) A_R(\mathcal{H})* \mathbb{1}_{K}\). It suffices to prove that these maps are bijective for all \(K\). Here \[\begin{gather} \mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl( A_R(\mathcal{X}), A_R(\mathcal{G})\bigr) A_R(\mathcal{H})* \mathbb{1}_{K} =\mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl( A_R(\mathcal{X}), A_R(\mathcal{G})\bigr)* \mathbb{1}_{K} \\\cong \mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl(\mathbb{1}_{K}* A_R(\mathcal{X}), A_R(\mathcal{G})\bigr). \end{gather}\] Since \(\mathcal{X}\) is proper, Proposition 1 implies that \(\mathbb{1}_{K}* A_R(\mathcal{X})\) is an fgp right \(A_R(\mathcal{G})\)-module. Even more, the proof of the proposition shows that \(\mathbb{1}_{K}* A_R(\mathcal{X}) \cong \bigoplus_{i=1}^n \mathbb{1}_{s(W_i)}* A_R(\mathcal{G})\) for compact open slices \(W_i \subseteq \mathcal{X}\), \(i=1,\dotsc,n\), whose images in \(\mathcal{X}/\mathcal{G}\) cover \(r_*^{-1}(K) \subseteq \mathcal{X}/\mathcal{G}\). Thus \(A_R(\mathcal{X}^*)* \mathbb{1}_{K}\) is an fgp left \(A_R(\mathcal{G})\)-module and \(A_R(\mathcal{X}^*)* \mathbb{1}_{K} \cong \bigoplus_{i=1}^n A_R(\mathcal{G})* \mathbb{1}_{s(W_i)}\) as left \(A_R(\mathcal{G})\)-modules. By the proof of Theorem 1, evaluation at \(\mathbb{1}_{s(W_i)}\in \mathbb{1}_{s(W_i)}* A_R(\mathcal{G})\) is an isomorphism \[\mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl(\mathbb{1}_{s(W_i)}* A_R(\mathcal{G}), A_R(\mathcal{G})\bigr) \xrightarrow\sim A_R(\mathcal{G}) * \mathbb{1}_{s(W_i)}.\] The image of \(\mathbb{1}_{s(W_i)}* A_R(\mathcal{G})\) in \(A_R(\mathcal{X})\) is \(\mathbb{1}_{W_i}\). The restriction of \(\mathcal{I}_{\mathcal{X}}(\mathbb{1}_{U^*})\) to \(\mathbb{1}_{s(W_i)}* A_R(\mathcal{G})\) maps this to \(\mathbb{1}_{\braket{U}{W_i}}\). Therefore, we get an isomorphism \[\mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl(\mathbb{1}_{K}* A_R(\mathcal{X}), A_R(\mathcal{G})\bigr) \cong \bigoplus_{i=1}^n A_R(\mathcal{G})* \mathbb{1}_{s(W_i)} \cong A_R(\mathcal{X}^*) * \mathbb{1}_{K},\] and it maps \(\mathcal{I}_{\mathcal{X}}(\mathbb{1}_{U^*})\) for a slice \(U\subseteq \mathcal{X}\) to the sum of the characteristic functions of \(\braket{U}{W_i}W_i^* = (W_i\braket{W_i}{U})^*\). This sum is the characteristic function of \((U\cap r^{-1}(K))^*\). As a consequence, this isomorphism is inverse to the restriction of \(\mathcal{I}_{\mathcal{X}}\). This implies that the latter is an isomorphism. ◻

Proof of Theorem 7. Let \(\mathcal{Q}\) be the \(R\)-algebra with the presentation in the theorem. We are going to prove that \(\mathcal{Q}\) is isomorphic to the covariance ring of \(\mathfrak A*\mathfrak X\), which exists by Theorem 6. Theorem 3 says that each \(A_R(\mathcal{X}_p)\) is isomorphic to the \(R\)-module that is generated by elements \(\delta_U\) for \(U\in \mathcal{B}_p\) subject to the relations \(\delta_U = \delta_{U_1} + \delta_{U_2}\) if \(U,U_1,U_2\in\mathcal{B}_p\) satisfy \(U_1 \sqcup U_2 = U\). Then the \(P\)-graded ring \(L \mathrel{\vcentcolon=}\bigoplus_{p\in P} A_R(\mathcal{X}_p)\) defined in Section 4.1 is generated as an \(R\)-algebra by \(\delta_U\) for all \(U\in\mathcal{B}_p\), \(p\in P\), subject to the relation above and the relation \(\delta_U \delta_V = \delta_{U V}\) for all \(p,q\in P\), \(U\in\mathcal{B}_p\) and \(V\in\mathcal{B}_q\). Mapping each \(\delta_U\) above to the corresponding generator of \(\mathcal{Q}\) defines a homomorphism \(L \to \mathcal{Q}\). We claim that this satisfies the criterion in Lemma 10, so that it corresponds to a covariant representation of \(\mathfrak A*\mathfrak X\) and induces a homomorphism \(\mathcal{O}\to \mathcal{Q}\).

It suffices to check the condition in Lemma 10 for idempotents of the form \(\mathbb{1}_{K}\) for \(K\subseteq \mathcal{G}^0\) with \(K\in\mathcal{B}_1\) because any compact open subset of \(\mathcal{G}^0\) is a disjoint union of such subsets by Proposition 1, and the idempotent elements \(\mathbb{1}_{K}\) for \(K\subseteq \mathcal{G}^0\) form a local unit in \(A_R(\mathcal{G})\) by Proposition 1. Proposition 1 identifies \(\mathbb{1}_{K} * A_R(\mathcal{X}_p) \cong \bigoplus_{i=1}^n \delta_{s(U_i)} * A_R(\mathcal{G})\) with certain \(U_1,\dotsc,U_n \in \mathcal{B}_p\). So the criterion in Lemma 10 is that \[\psi_{K,p}\colon \bigoplus_{i=1}^n \delta_{s(U_i)} * \mathcal{Q} \xrightarrow\sim\delta_K * \mathcal{Q}, \qquad \sum_{i=1}^n x_i\mapsto \sum_{i=1}^n \delta_{U_i}* x_i\] is invertible for all \(p\in P\), \(K\subseteq \mathcal{G}^0\) with \(K\in \mathcal{B}_1\), and \(U_1,\dotsc,U_n\in\mathcal{B}_p\) as above. We claim that the map \[\psi^*_{K,p}\colon \delta_K * \mathcal{Q} \xrightarrow\sim\bigoplus_{i=1}^n \delta_{s(U_i)} * \mathcal{Q}, \qquad x \mapsto (\delta^*_{U_i}* x)_{i=1,\dotsc,n},\] is inverse to \(\psi_{K,p}\). Indeed, \(\psi_{K,p}\psi^*_{K,p}\) is the identity map on \(\delta_K * \mathcal{Q}\) because of the relation \(\delta_K = \sum_{j=1}^n \delta_{U_i} \delta^*_{U_i}\) in \(\mathcal{Q}\). The composite \(\psi_{K,p}^*\psi_{K,p}\) is multiplication by the matrix in \(\mathbb{M}_n(\mathcal{Q})\) with entries \(\delta_{U_i}^* \delta_{U_j} = \delta_{\braket{U_i}{U_j}}\). Since \(\Pi(U_i)\cap \Pi(U_j)=\emptyset\) for \(i\neq j\), this matrix is diagonal. Its \(i\)th entry is \(\delta_{\braket{U_i}{U_i}} = \delta_{s(U_i)}\), which describes the identity map on the summand \(\delta_{s(U_i)} * \mathcal{Q}\). So the criterion in Lemma 10 is satisfied. Thus the tautological formula \(\delta_U \mapsto \delta_U\) defines a covariant representation of \(\mathfrak A*\mathfrak X\) in \(\mathcal{Q}\) and induces a homomorphism \(\mathcal{O}\to\mathcal{Q}\).

Next, we construct a homomorphism \(\mathcal{Q} \to \mathcal{O}\). We must find images for all the generators \(\delta_U\) and \(\delta_U^*\) of \(\mathcal{Q}\). Of course, we map \(\delta_U\in \mathcal{Q}\) for \(U\in\mathcal{B}_p\), \(p\in P\) to the image of \(\mathbb{1}_{U}\in A_R(\mathcal{X}_p)\) in \(\mathcal{O}\). We interpret \(\delta_U^*\) as the characteristic function \(\mathbb{1}_{U^*} \in A_R(\mathcal{X}_p^*)\). Now we use the isomorphism \[\mathcal{I}_{\mathcal{X}_p}\colon A_R(\mathcal{X}_p^*) \to \mathop{\mathrm{Hom}}_{-, A_R(\mathcal{G})}\bigl( A_R(\mathcal{X}_p), A_R(\mathcal{G})\bigr) A_R(\mathcal{G}) = A_R(\mathcal{X}_p)^*\] in Proposition 1 and the canonical map \(A_R(\mathcal{X}_p)^* \to \mathcal{O}\) defined in Lemma 11 to map \(\mathbb{1}_{U^*}\) to an element of \(\mathcal{O}\), which we denote by \(\delta_U^*\). The relations \(\delta_{U_1} + \delta_{U_2} = \delta_{U}\) and \(\delta^*_{U_1} + \delta^*_{U_2} = \delta^*_U\) are satisfied because our maps on generators are parts of maps \(A_R(\mathcal{X}_p) \to \mathcal{O}\) and \(A_R(\mathcal{X}_p)^* \to \mathcal{O}\). The relation \(\delta_U \delta_V = \delta_{\mu_{p,q}(U V)}\) for all \(U\in\mathcal{B}_p\), \(V\in\mathcal{B}_q\), \(p,q\in P\) is already built into the covariance ring. Next we prove \(\delta_V^* \delta_U^* = \delta_{\mu_{p,q}(U V)}^*\) for \(U\in\mathcal{B}_p\), \(V\in\mathcal{B}_q\), \(p,q\in P\) in \(\mathcal{O}\). Recall that the canonical map \(\nu_t\colon F_t \otimes_{F_1} \mathcal{O} \to \mathcal{O}\) is an isomorphism for all \(t\in P\). Therefore, the map \(\nu_p (\nu_q\otimes \mathrm{id})\colon F_p \otimes_{F_1} F_q \otimes_{F_1} \mathcal{O}\to \mathcal{O}\) is an isomorphism. Since elements of the form \(\mathbb{1}_{W}\) for \(W\in \mathcal{B}_t\) generate \(F_t\) as an \(R\)-module, it follows that two elements \(x_1,x_2\in\mathcal{O}\) are equal if \(x_1 \cdot \mathbb{1}_{W} \cdot \mathbb{1}_{X} = x_2 \cdot \mathbb{1}_{W}\cdot \mathbb{1}_{X}\) for all \(W \in \mathcal{B}_p\), \(X\in \mathcal{B}_q\). Here \(\mathbb{1}_{W} \cdot \mathbb{1}_{X} = \mathbb{1}_{W X}\) with \(W X \in \mathcal{B}_{p q}\), \[\delta_V^* \cdot \delta_U^* \cdot \mathbb{1}_{W} \cdot \mathbb{1}_{X} = \delta_V^*\cdot \mathbb{1}_{\braket{U}{W}}\cdot \mathbb{1}_{X} = \delta_V^*\cdot \mathbb{1}_{\braket{U}{W} X} = \mathbb{1}_{\braket{V}{\braket{U}{W} X}}.\] and \(\delta_{U V}^*\cdot \mathbb{1}_{W X} = \mathbb{1}_{\braket{U V}{W X}}\). Thus our formula follows if \(\braket{V}{\braket{U}{W} X} = \braket{U V}{W X}\) as subsets of \(\mathcal{G}\). To prove this, we show that \(\braket{v}{\braket{u}{w} x} = \braket{u v}{w x}\) holds for all \(v,x\in \mathcal{X}_p\), \(u,w\in \mathcal{X}_p\) with \(s(u) = r(v)\) and \(s(w) = r(x)\); this equality includes the claim that one side is defined if and only if the other side is defined. Indeed, \(\braket{u v}{w x}\) is the unique element of \(\mathcal{G}\) with \(u v \braket{u v}{w x} = w x\). The equation \(u v \braket{u v}{w x} = w x\) says that there is \(h\in \mathcal{G}\) with \(u h = w\) and \(h^{-1} v \braket{u v}{w x} = x\). This means that \(\braket{u}{w}\) is defined and equal to \(h\) and \(v \braket{u v}{w x} = h x = \braket{u}{w} x\). The last equation says that \(\braket{v}{\braket{u}{w} x} = \braket{u v}{w x}\) as desired.

Next, we check the relation \(\delta_{U_1}^* \delta_{U_2} = \delta_{\braket{U_1}{U_2}}\) for \(U_1,U_2\in\mathcal{B}_p\). The element \(\delta_{U_2} \in \mathcal{O}\) is defined so that \(\delta_{U_2}\cdot x = \nu_p(\mathbb{1}_{U_2}\otimes x)\) for all \(x\in \mathcal{O}\). Thus \[\begin{gather} \delta_{U_1}^* \cdot \delta_{U_2} \cdot x = \nu_1(\mathcal{I}_{\mathcal{X}_p} (\mathbb{1}_{U_1})\otimes \mathrm{id}_{\mathcal{O}}) (\mathbb{1}_{U_2}\otimes x) \\= \nu_1(\mathcal{I}_{\mathcal{X}_p} (\mathbb{1}_{U_1})(\mathbb{1}_{U_2}) \otimes x) = \nu_1( \mathbb{1}_{\braket{U_1}{U_2}} \otimes x) = \mathbb{1}_{\braket{U_1}{U_2}} \cdot x. \end{gather}\] This implies that \(\delta_{U_1}^* \cdot \delta_{U_2}\) is the image \(\delta_{\braket{U_1}{U_2}}\) of \(\mathbb{1}_{\braket{U_1}{U_2}} \in A_R(\mathcal{G})\).

Finally, we check the relation \(\delta_K = \sum_{j=1}^n \delta_{U_i}\delta^*_{U_i}\) if \(p\in P\), \(K\subseteq \mathcal{G}^0\) with \(K\in \mathcal{B}_1\), and \(U_1,\dotsc,U_n\in\mathcal{B}_p\) are chosen as in Proposition 1. To check this, we map \(\mathcal{O}\) faithfully to \(\mathop{\mathrm{End}}_{-,\mathcal{O}}(\mathcal{O})\) and use the bimodule isomorphism \(\nu_p\) to identify the latter with the endomorphism ring of the right \(\mathcal{O}\)-module \(F_p \otimes_{F_1} \mathcal{O}\). So it suffices to prove that \(\delta_K\) and \(\sum_{j=1}^n \delta_{U_i}\delta^*_{U_i}\) induce the same map on \(F_p \otimes_{F_1} \mathcal{O}\). Here \(\delta_K\) acts by left multiplication with \(\mathbb{1}_{K}\in A_R(\mathcal{G})\) on the tensor factor \(F_p = A_R(\mathcal{X}_p)\). Thus \(\delta_K\) maps \(\mathbb{1}_{V}\otimes x\) for \(V\in \mathcal{B}_p\), \(x\in \mathcal{O}\) to \(\mathbb{1}_{K}*\mathbb{1}_{V}\otimes x = \mathbb{1}_{r_{\mathcal{X}_p}^{-1}(K)\cap V}\). Left multiplication by \(\delta^*_{U_i}\) applies \(\mathcal{I}_{\mathcal{X}_p}(\mathbb{1}_{U_i^*})\otimes \mathrm{id}_{\mathcal{O}}\) to \(\mathbb{1}_{V}\otimes x\), and then multiplies using \(\nu_1\). This gives \(\mathbb{1}_{\braket{U_i}{V}} \cdot x\). Then \(\delta_{U_i}\) maps this back to \(\mathbb{1}_{U_i} \otimes \mathbb{1}_{\braket{U_i}{V}} \cdot x\). Since the tensor product is balanced over \(A_R(\mathcal{G})\), this is equal to \(\mathbb{1}_{U_i} * \mathbb{1}_{\braket{U_i}{V}} \otimes x = \mathbb{1}_{U_i \cdot \braket{U_i}{V}} \otimes x\). Here \(U_i \braket{U_i}{V}\) is the set of all \(v\in V\) for which \(\Pi(v) \in \mathcal{X}_p/\mathcal{G}\) belongs to \(\Pi(U_i)\). As a consequence, the sum \(\sum_{i=1}^n \mathbb{1}_{U_i \cdot \braket{U_i}{V}}\) is the characteristic function of \[\bigsqcup_{i=1}^n V\cap \Pi^{-1}(\Pi(U_i)) = V \cap r^{-1}(K)\] by construction of \(U_1,\dotsc,U_n\). This finishes the proof that \(\delta_K\) and \(\sum_{j=1}^n \delta_{U_i}\delta^*_{U_i}\) give the same operator on \(A_R(\mathcal{X}_p)\otimes_{A_R(\mathcal{G})} \mathcal{O}\). This gives the required homomorphism \(\mathcal{Q} \to \mathcal{O}\).

The maps back and forth clearly send each generator \(\delta_U\) to itself. Since a covariant representation only specifies how the \(\delta_U\) act, a homomorphism \(\mathcal{O}\to \mathcal{O}\) is determined uniquely by its values on the generators \(\delta_U\). Thus the composite map \(\mathcal{O}\to \mathcal{Q} \to \mathcal{O}\) is the identity map. To finish the proof, we prove that the homomorphism \(\mathcal{O}\to \mathcal{Q}\) is surjective. It is enough to prove that its image contains all generators \(\delta_U\) and \(\delta_U^*\) for \(U\in\mathcal{B}_p\), \(p\in P\). The generators \(\delta_U\) belong to the image by construction. So do the generators \(\delta^*_{U_i}\) whenever \(U_i \in\mathcal{B}_p\) occurs in the situation of Proposition 1 for some \(K\subseteq \mathcal{G}^0\) with \(K\in\mathcal{B}_1\).

Let \(U\in\mathcal{B}_p\) for some \(p\in P\). Choose \(K\subseteq \mathcal{G}^0\) with \(r(U) \subseteq K\). If necessary, write \(K = \bigsqcup_{i=1}^\ell K_j\) with \(K_j\in \mathcal{B}_1\) by Proposition 1. Decompose \(r^{-1}(K) = \bigsqcup_{i=1}^n V_i \cdot \mathcal{G}\) as in Proposition 1. Then \(U = K U = \bigsqcup_{j=1}^\ell K_j U\). The sets \(K_j U\) all belong to \(\mathcal{B}_p\), and we may get any union of them from \(U\) by taking the ones not in the union away from \(U\). This implies that any union of these sets belongs to \(\mathcal{B}_p\). So the relation \(\delta_U^* = \sum_{j=1}^\ell \delta_{K_j U}^*\) holds in \(\mathcal{Q}\). Therefore, it suffices to show that \(\delta_{K_j U}^*\) belongs to the image of the homomorphism \(\mathcal{O}\to \mathcal{Q}\). All this shows is that we may assume without loss of generality that there is \(K \in\mathcal{B}_1\) with \(r(U) \subseteq K\). Let \(V_1,\dotsc,V_n\in\mathcal{B}_p\) be such that \(r^{-1}(K) = \bigsqcup_{i=1}^n V_i \cdot \mathcal{G}\) as in Proposition 1. Then the relation \(\delta_K = \sum_{j=1}^n \delta_{V_j} \delta_{V_j}^*\) is imposed in \(\mathcal{Q}\).

The relations defining \(\mathcal{Q}\) imply \(\delta_K^* = \delta_K\) because both multiplication by \(\delta_K^*\) and by \(\delta_K\) are inverse to multiplication by \(\delta_K\) as a map \(\delta_K \mathcal{Q} \to \delta_K \mathcal{Q}\), and the inverse of any map is unique. Therefore, \[\delta_U^* = \delta_{K U}^* = \delta_U^* \delta_K^* = \delta_U^* \delta_K = \delta_U^* \sum_{j=1}^n \delta_{V_j} \delta_{V_j}^* = \sum_{j=1}^n \delta_{\braket{U}{V_j}} \delta_{V_j}^*.\] Since \(\delta_{V_j}^*\) and \(\delta_{\braket{U}{V_j}}\) belong to the image of the homomorphism \(\mathcal{O}\to \mathcal{Q}\), so does \(\delta_U^*\). Thus the latter homomorphism is surjective and our two homomorphisms are inverse to each other. ◻

Proof. It is manifest that \(\iota\) preserves all the relations in the presentation in the theorem, and so it defines a homomorphism \(\mathcal{O} \to \mathcal{O}^\mathrm{op}\). ◻

Proof. Let \(D\) be a ring with local units and let \(E\) be a smooth \(F_1,D\)-bimodule; we do not assume \(E\) to be proper. Roughly speaking, we are going to construct a canonical bijection from covariant representations \(\tilde{\nu}_p\colon F_p\to \mathop{\mathrm{End}}_{-,D}(E)\) to cones over \(\mathcal{F}\) with summit \(D\) based on the \(F_1,D\)-bimodule \(E\). More precisely, since covariant representations are implicitly assumed to be nondegenerate, we replace \(\mathop{\mathrm{End}}_{-,D}(E)\) by the target ring \(F_1 \mathop{\mathrm{End}}_{-,D}(E) F_1\); a local unit in \(F_1\) also provides one in the latter ring.

On the one hand, a cone over \(\mathcal{F}\) with summit \(D\) based on the \(F_1,D\)-bimodule \(E\) means a family of \(F_1,D\)-bimodule isomorphisms \(\nu_p\colon F_p \otimes_{F_1} E \xrightarrow\sim E\) such that \(\nu_1\) is the given left \(F_1\)-module structure on \(E\) and the diagram 9 commutes. As in Definition 11, \(\nu_p\) corresponds to a map \(\tilde{\nu}_p\colon F_p \to \mathop{\mathrm{End}}_{-,D}(E)F_1\) given by \(\tilde{\nu}_p(x)(v) = \nu_p(x\otimes v)\). Here \(\tilde{\nu}_1\) is the given \(F_1\)-module structure and 9 is equivalent to \(\tilde{\nu}_p(x)\tilde{\nu}_q(y) = \tilde{\nu}_{p q}(x y)\) for all \(p,q\in P\), \(x\in F_p\), \(y\in F_q\). In particular, each \(\tilde{\nu}_p\) is \(F_1\)-linear on the left and right. Since \(F_p\) is a smooth \(F_1\)-bimodule, it follows that \(\tilde{\nu}_p\) is a map to \(F_1 \mathop{\mathrm{End}}_{-,D}(E) F_1\). Conversely, a family of maps \(\tilde{\nu}_p\colon F_p \to F_1\mathop{\mathrm{End}}_{-,D}(E)F_1\) comes from a cone over \(\mathcal{F}\) with underlying \(F_1\)-bimodule \(E\) if and only if \(\tilde{\nu}_p(x)\tilde{\nu}_q(y) = \tilde{\nu}_{p q}(x y)\) for all \(p,q\in P\), \(x\in F_p\), \(y\in F_q\), \(\tilde{\nu}_1\) is the given left \(F_1\)-module structure on \(E\), and the following maps for \(p\in P\) are bijective: \[\label{eq:bijection95for95cone} \nu_p\colon F_p \otimes_{F_1} E \to E,\qquad x\otimes v \mapsto \tilde{\nu}_p(x)(v).\tag{10}\] On the other hand, a covariant representation of our diagram in \(F_1\mathop{\mathrm{End}}_{-,D}(E)F_1\) is a family of maps \((\tilde{\nu}_p)_{p\in P}\) such that \(\tilde{\nu}_p(x)\tilde{\nu}_q(y) = \tilde{\nu}_{p q}(x y)\) for all \(p,q\in P\), \(x\in F_p\), \(y\in F_q\), \(\tilde{\nu}_1\) is the given left \(F_1\)-module structure on \(E\), and the following maps for \(p\in P\) are bijective: \[\label{eq:bijection95for95covrep} \nu_p\colon F_p \otimes_{F_1} F_1\mathop{\mathrm{End}}_{-,D}(E)F_1 \to F_1\mathop{\mathrm{End}}_{-,D}(E)F_1,\qquad x\otimes T \mapsto \tilde{\nu}_p(x)\cdot T.\tag{11}\] The only difference is that 11 replaces 10 . So it remains to prove that these two conditions are equivalent. To begin with, let us assume that 11 is bijective. Then the maps \(\tilde{\nu}_p\) induce a nondegenerate representation of the covariance ring \(\mathcal{O}_\mathcal{F}\) in \(F_1 \mathop{\mathrm{End}}_{-,D}(E)F_1\) by the universal property of the covariance ring. Since \(E\) is a nondegenerate left \(F_1\)-module, the ring \(F_1 \mathop{\mathrm{End}}_{-,D}(E)F_1\) acts nondegenerately on \(E\). Hence \(\mathcal{O}_\mathcal{F}\otimes_{\mathcal{O}_\mathcal{F}} E \cong E\). The universal covariant representation in \(\mathcal{O}_\mathcal{F}\) contains \(F_1,\mathcal{O}_\mathcal{F}\)-bimodule isomorphisms \(F_p \otimes_{F_1} \mathcal{O}_\mathcal{F} \cong \mathcal{O}_\mathcal{F}\). Therefore, \[F_p \otimes_{F_1} E \cong F_p \otimes_{F_1} \mathcal{O}_\mathcal{F}\otimes_{\mathcal{O}_\mathcal{F}} E \cong \mathcal{O}_\mathcal{F}\otimes_{\mathcal{O}_\mathcal{F}} E \cong E.\] Thus the maps in 10 are bijective if the maps in 11 are bijective.

Conversely, assume that the maps in 10 are bijective. Theorem 2 implies \[\begin{gather} F_p \otimes_{F_1} F_1 \mathop{\mathrm{End}}_{-,D}(E) F_1 = F_1 F_p \otimes_{F_1} \mathop{\mathrm{Hom}}_{-,D}(E,E) F_1 \\\cong F_1 \mathop{\mathrm{Hom}}_{-,D}(E, F_p \otimes_{F_1} E) F_1 \cong F_1 \mathop{\mathrm{Hom}}_{-,D}(E, E) F_1 = F_1 \mathop{\mathrm{End}}_{-,D}(E) F_1 \end{gather}\] because \(F_p\) is a proper \(F_1\)-bimodule and 10 is bijective.

Along the way, we have replaced covariant representations in \(F_1 \mathop{\mathrm{End}}_{-,D}(E) F_1\) based on the given action of \(F_1\) by homomorphisms \(\mathcal{O}_\mathcal{F}\to F_1 \mathop{\mathrm{End}}_{-,D}(E) F_1\) extending the given homomorphism on \(F_1\). The latter are also in bijection with homomorphisms \(\mathcal{O}_\mathcal{F}\to \mathop{\mathrm{End}}_{-,D}(E)\) because \(F_1 \to \mathcal{O}_\mathcal{F}\) is nondegenerate. So we get a bijection between cones over \(F\) based on the \(F_1,D\)-correspondence \(E\) and \(\mathcal{O}_\mathcal{F},D\)-correspondences that restrict to \(E\) when we restrict the left \(\mathcal{O}_\mathcal{F}\)-action to \(F_1\).

These bijections for different \(E\) are natural in two ways. First, an \(F_1,D\)-correspondence map \(\varphi\colon E\to E'\) intertwines the representations of \(\mathcal{O}_\mathcal{F}\) if and only if it intertwines the covariant representations of the diagram, if and only if it is a morphism of cones. Therefore, our bijections combine to an isomorphism of categories between the categories of cones over \(F\) with summit \(D\) and of \(\mathcal{O}_\mathcal{F},D\)-correspondences. Secondly, given a \(D,D'\)-correspondence \(X\), the bijections for \(E\) and \(E\otimes_D X\) are related as expected. Namely, if \((\nu_p)\) is a cone based on \(E\), then \((\nu_p\otimes_D X)\) is a cone based on \(E\otimes_D X\), and the bijection maps the latter cone to the representation of \(\mathcal{O}_\mathcal{F}\) that is induced by the covariant representation given by \(\tilde{\nu}_p(\xi) \otimes_D \mathrm{id}_X\) for \(p\in P\), \(\xi\in F_p\). Therefore, the isomorphisms of categories above are natural in \(D\), giving a pseudonatural transformation of pseudofunctors from \(\mathfrak{Rings}\) (or \(\mathfrak{Rings}_\mathrm{prop}\)) to \(\operatorname{\mathfrak{Cat}}\). This finishes the proof that the covariance ring \(\mathcal{O}_\mathcal{F}\) represents the cones pseudofunctor from \(\mathfrak{Rings}\) or \(\mathfrak{Rings}_\mathrm{prop}\) to \(\operatorname{\mathfrak{Cat}}\). So it is a bicategorical limit as asserted. ◻

Proof. Since \(F_{p_1}\) is a proper \(F_1\)-bimodule by assumption, Theorem 2 applies here and gives the asserted bimodule isomorphism. The bimodule \(F_{p_2} \otimes_{F_1} F_{p_1}^*\) is smooth on the left because \(F_{p_2}\) is, and smooth on the right by construction. ◻

Proof. This follows because \(T(d\cdot x) = T(d)\cdot x\) for all \(d,x\in D\) and \(T\in\mathop{\mathrm{End}}_{-,D}(D)\). ◻

Proof. The formula for \(\varphi_{p_1,p_2,q}\) defines an \(F_1\)-bimodule homomorphism \[\mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2}, F_{p_1}) \to \mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2q}, F_{p_1q})\] because each \(\mu_{p_2,q}\) is an \(F_1\)-bimodule homomorphism. As a bimodule map, \(\varphi_{p_1,p_2,q}\) maps \(\mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2}, F_{p_1}) F_1\) to \(\mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2q}, F_{p_1q}) F_1\). Equation 12 follows because \(\mu_{p_i,1}\) for \(i=1,2\) is natural for bimodule maps. Equation 13 follows because the following diagram commutes for all \(T\in \mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2},F_{p_1})F_1\): \[\begin{tikzcd}[ampersand replacement=\&,column sep=2.4em] {F_{p_1q}\otimes_{F_1} F_t} \& {F_{p_1}\otimes_{F_1} F_q \otimes_{F_1} F_t} \& {F_{p_2}\otimes_{F_1} F_q \otimes_{F_1} F_t} \& {F_{p_2q}\otimes_{F_1} F_t} \\ {F_{p_1qt}} \& {F_{p_1}\otimes_{F_1} F_{qt}} \& {F_{p_2}\otimes_{F_1} F_{qt}} \& {F_{p_2qt}} \arrow["{{\mu_{p_1q,t}}}", "{\cong}"', from=1-1, to=2-1] \arrow["{{\mathrm{id}\otimes \mu_{q,t}}}", "{\cong}"', from=1-2, to=2-2] \arrow["{{\mu_{p_1,q}\otimes \mathrm{id}}}"',"{\cong}", from=1-2, to=1-1] \arrow["{{\mu_{p_2,q}\otimes \mathrm{id}}}","{\cong}"', from=1-3, to=1-4] \arrow["{{\mathrm{id}\otimes \mu_{q,t}}}","{\cong}"', from=1-3, to=2-3] \arrow["{T \otimes\mathrm{id}\otimes\mathrm{id}}"', from=1-3, to=1-2] \arrow["{{\mu_{p_2q,t}}}","{\cong}"', from=1-4, to=2-4] \arrow["{{\mu_{p_1,qt}}}"',"{\cong}", from=2-2, to=2-1] \arrow["{{\mu_{p_2,qt}}}","{\cong}"', from=2-3, to=2-4] \arrow["{T \otimes\mathrm{id}}"', from=2-3, to=2-2] \end{tikzcd}\] Here the left and right squares commute by Definition 9 and the middle square commutes because the tensor product is a bifunctor. ◻

Proof. The definition of \(w_{g,h}\) uses representatives \((p_1,p_2)\in R_g\), \((q_1,q_2)\in R_h\) for \(g,h \in G\) and \(t_1,t_2\in P\) such that \(p_2t_1=q_1t_2\). First, we check that the value of \(w_{g,h}\) does not depend on the choice of \(t_1, t_2\). So take \(t_1,t_2\) as above and let \(u_1,u_2\in P\) also satisfy \(p_2 u_1=q_1u_2\). Condition [enum:O1] provides \(x_1,x_2\in P\) with \(t_1x_1= u_1x_2\). Then \(q_1 t_2x_1=p_2 t_1 x_1=p_2 u_1 x_2=q_1 u_2 x_2\). Condition [enum:O2] gives \(n\in P\) with \(t_2 x_1n = u_2 x_2 n\). Let \(b_1\mathrel{\vcentcolon=}x_1 n\) and \(b_2\mathrel{\vcentcolon=}x_2 n\). Then \(t_1 b_1= u_1 b_2\) and \(t_2b_1= u_2b_2\). To show that \(w_{g,h}\) does not depend on \(t_1,t_2\), we must prove that \[\varphi_{p_1 t_1,q_2 t_2,b_1} \bigl(\varphi_{p_1,p_2,t_1}(x)\circ\varphi_{q_1,q_2,t_2}(y)\bigr) = \varphi_{p_1 u_1,q_2 u_2,b_2} \bigl(\varphi_{p_1,p_2,u_1}(x)\circ\varphi_{q_1,q_2,u_2}(y)\bigr).\] This works because the diagram in Figure 1 commutes.

Next, we check that \(w_{g,h}\) is independent of the choices of the representatives of \(\bigl[x,(p_1,p_2)\bigr]\) and \(\bigl[y,(q_1,q_2)\bigr]\). Let \(\bigl[x,(p_1,p_2)\bigr] = \bigl[x', (p_1',p_2')\bigr]\) and \(\bigl[y,(q_1,q_2)\bigr] = \bigl[y',(q_1',q_2')\bigr]\). Then there are \(n,n'\in P\) with \[(p_1 n,p_2 n) = (p_1' n', p_2' n')\quad\text{and}\quad \varphi_{p_1, p_2,n}(x) =\varphi_{p_1', p_2', n'}(x').\] Similarly, there are \(m,m'\in P\) with \((q_1 m,q_2 m) = (q_1' m', q_2' m')\) and \(\varphi_{q_1,q_2,m}(y) = \varphi_{q_1', q_2', m'}(y')\). Next, there are \(t_1,t_2\in P\) with \(p_2 n t_1=q_1 m t_2\). Then \(p_2' n' t_1=q_1' m' t_2\). Using Lemma 14, this implies \[\begin{align} \varphi_{p_1,p_2, nt_1}(x) &\circ\varphi_{q_1,q_2, mt_2}(y) = \varphi_{p_1 n,p_2 n, t_1}\bigl(\varphi_{p_1,p_2,n}(x)\bigr) \circ \varphi_{q_1 m ,q_2 m ,t_2}\bigl(\varphi_{q_1,q_2,m}(y)\bigr)\\ &= \varphi_{p_1' n',p_2' n', t_1}\bigl(\varphi_{p_1', p_2', n'}(x')\bigr) \circ \varphi_{q_1' m' ,q_2' m' ,t_2}\bigl(\varphi_{q_1', q_2', m'}(y')\bigr)\\ &= \varphi_{p_1',p_2', n' t_1}(x')\circ \varphi_{q_1',q_2', m' t_2}(y'). \end{align}\] Hence \(w_{g,h}\) is well-defined. Next we prove that \(w_{g,h}\) is additive in both variables. Since \(w_{g,h}\) is independent of the choice of representatives, it suffices to check this when both summands are in the same \(\mathop{\mathrm{Hom}}_{-, F_1}(F_{p_2}, F_{p_1})F_1\). Since \(\varphi_{p_1,p_2,t}\) is additive by Lemma 14 and composition distributes over addition, \(w_{g,h}\) is additive in each argument. It is also easy to see that \(w_{g,h}(a x b, y c) = a w_{g,h}(x, b y) c\) for \(a,b,c\in F_1\), \(x\in \mathcal{O}_{g}\), \(y\in \mathcal{O}_{h}\). This implies that \(w_{g,h}\) descends to an \(F_1\)-bimodule map \(\mathcal{O}_{g}\otimes_{F_1}\mathcal{O}_{h} \to \mathcal{O}_{g h}\). ◻

Proof. Let \(\tilde{x}\in\mathcal{O}_{g}\), \(\tilde{y}\in\mathcal{O}_{h}\) and \(\tilde{z}\in\mathcal{O}_{k}\). We may choose representatives of the form \(\tilde{x}=\bigl[x,(p_1,p_2)\bigr]\), \(\tilde{y}=\bigl[y,(p_2,p_3)\bigr]\) and \(\tilde{z}=\bigl[z,(p_3,p_4)\bigr]\) because such triples of pairs \((p_1,p_2),(p_2,p_3),(p_3,p_4)\) are cofinal in \(\mathcal{C}_P^{g}\times\mathcal{C}_P^{h}\times\mathcal{C}_P^{k}\). Then \[\label{eq:w95nice} w_{gh,k}\bigl(w_{g,h}(\tilde{x} \otimes \tilde{y}) \otimes\tilde{z}\bigr) = w_{g,hk}\bigl(\tilde{x} \otimes w_{h,k}(\tilde{y} \otimes \tilde{z})\bigr)\tag{14}\] because composition of maps is associative. Equation 14 with \(g=h = k= e\) says that \(\mathcal{O}_{e}\) is an associative ring. Similarly, 14 implies that \(w_{e,g}\) and \(w_{g,e}\) make \(\mathcal{O}_{g}\) an \(\mathcal{O}_{e}\)-bimodule and that the map \(w_{g,h}\) is \(\mathcal{O}_{e}\)-balanced and so descends to an \(\mathcal{O}_{e}\)-bimodule map \(w_{g,h}\colon\mathcal{O}_{g}\otimes_{\mathcal{O}_{e}} \mathcal{O}_{h} \to\mathcal{O}_{gh}\). Finally, 14 says that the diagram in the statement commutes. ◻

Proof. As an inductive limit of nondegenerate \(F_1\)-bimodules, \(\mathcal{O}_{g}\) is also a nondegenerate \(F_1\)-bimodule. The \(\mathcal{O}_{e}\)-bimodule structure on \(\mathcal{O}_{g}\) extends the \(F_1\)-bimodule structure with respect to the canonical map \(F_1 \to \mathcal{O}_{e}\). Therefore, the idempotents in \(F_1\) form a local unit in \(\mathcal{O}_{e}\) and \(\mathcal{O}_{g}\) is nondegenerate as an \(\mathcal{O}_{e}\)-bimodule. ◻

Proof. We use Lemma 12 to replace \(\mathop{\mathrm{Hom}}_{-,F_1}(F_{p_1}, F_{p_2})F_1\) in the definition of \(\mathcal{O}_{g}\) by \(F_{p_2} \otimes_{F_1} F_{p_1}^*\). If \(p,p_1,p_2\in P\), then there is a chain of isomorphisms \[{!}{ \begin{tikzcd}[ampersand replacement=\&,column sep= small, row sep=tiny] {F_p\otimes_{F_1} \mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2},F_{p_1})F_1} \& {F_p\otimes_{F_1} F_{p_1}\otimes_{F_1} F_{p_2}^*F_1} \& {F_{pp_1}\otimes_{F_1} F_{p_2}^*F_1} \& {\mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2},F_{pp_1})F_1,} \\ {x\otimes y\psi(-)} \& {x\otimes y\otimes\psi} \& {\mu_{p,p_1}(x\otimes y)\otimes\psi} \& {\mu_{p,p_1}(x\otimes y)\psi(-)}, \arrow["\cong"', from=1-2, to=1-1] \arrow["\cong", from=1-2, to=1-3] \arrow["\cong", from=1-3, to=1-4] \arrow[maps to, from=2-2, to=2-3] \arrow[maps to, from=2-3, to=2-4] \arrow[maps to, from=2-2, to=2-1] \end{tikzcd} }\] The composite isomorphisms from left to right for \((p_1,p_2) \in R_g\) are compatible with the maps \(\varphi_{p_1,p_2,t}\) in Definition 18. That is, they provide an isomorphism of diagrams over \(\mathcal{C}_P^{g}\). The inductive system on the left is \(F_p\otimes_{F_1} H_{{\mathcal{F},g}}\), and its inductive limit is \(F_p \otimes_{F_1} \mathcal{O}_{g}\) because the tensor product commutes with colimits. We claim that the functor \(\mathcal{C}_P^{g}\to \mathcal{C}_P^{pg}\) that maps \((p_1,p_2) \in R_g\) to \((p p_1, p_2)\) is cofinal. This implies that the inductive system on the right has the same colimit as \(H_{{\mathcal{F},pg}}\). So the isomorphism of inductive systems above implies the desired isomorphism \(F_p \otimes_{F_1} \mathcal{O}_{g} \cong \mathcal{O}_{p g}\). It remains to prove the asserted cofinality.

Let \(q_1,q_2\in P\) be such that \(q_1 q_2^{-1} = p g = p p_1 p_2^{-1}\). The latter means that there are \(t_1,t_2\in P\) with \(p p_1 t_1 =q_1 t_2\) and \(p_2 t_1 = q_2 t_2\). So \((q_1,q_2,t_2)\) is an arrow in \(\mathcal{C}_P^{pg}\) from \((q_1,q_2)\) to \((p p_1 t_1, p_2 t_1) = (q_1 t_2, q_2 t_2)\), which is in the image of our functor. Let \(u_1,u_2\in P\) be another choice with \(p p_1 u_1 =q_1 u_2\) and \(p_2 u_1 = q_2 u_2\), so that \((q_1,q_2,u_2)\) is another arrow in \(\mathcal{C}_P^{pg}\) from \((q_1,q_2)\) to an object \((p p_1 u_1, p_2 u_1) = (q_1 u_2, q_2 u_2)\) in the image of our functor. Then [enum:O1] gives \(v,w\in P\) with \(u_2 v = t_2 w\). Then \(p_2 (u_1 v) = q_2 u_2 v = q_2 t_2 w = p_2 (t_1 w)\). Then [enum:O2] gives \(x\in P\) with \(u_1 v x= t_1 w x\). Since \(u_2 v x = t_2 w x\) as well, the arrows \((p p_1 t_1,p_2 t_1,w x)\colon (p p_1 t_1,p_2 t_1) \to (p p_1 t_1 w x, p_2 t_1 w x)\) and \((p p_1 u_1,p_2 u_1,v x)\colon (p p_1 u_1,p_2 u_1) \to (p p_1 u_1 v x, p_2 u_1 v x)\) in the image of \(\mathcal{C}_P^{g}\) equalise the two arrows above. This finishes the proof of cofinality. ◻

Proof. We use Proposition 1 twice: \[\mathcal{O}_{p}\otimes_{\mathcal{O}_{e}}\mathcal{O}_{g} \cong F_p\otimes_{F_1} \mathcal{O}_{e}\otimes_{\mathcal{O}_{e}}\mathcal{O}_{g} \cong F_p\otimes_{F_1} \mathcal{O}_{g}\cong\mathcal{O}_{pg}.\qedhere\] ◻

Proof. The isomorphisms \(\kappa_{p,g}\) give the isomorphism \(\kappa_p\) because the tensor product commutes with direct sums and \(g\mapsto p g\) is a bijection on the group \(G\). We check that these isomorphisms form a covariant representation. The only nontrivial point is that \(\tilde{\kappa}_p(x)\cdot \tilde{\kappa}_q(y) = \tilde{\kappa}_{pq}(\mu_{p,q}(x\otimes y))\) for all \(p,q\in P\), \(x\in F_p\) and \(y\in F_q\). The left hand side here is represented by the composite map \[F_1 \to F_q \to F_p \otimes_{F_1} F_q \cong F_{p q}, \qquad z\mapsto \mu_{p,q}(x\otimes \mu_{q,1}(y\otimes z)).\] The coherence conditions of the diagram in Definition 9 imply that this is \(\mu_{pq, 1}(\mu_{p,q}(x\otimes y)\otimes z)\). This is a representative for \(\tilde{\kappa}_{pq}(\mu_{p,q}(x\otimes y))\) as needed. ◻

Proof. Let \((\tilde{\nu}_p)_{p\in P}\) be a covariant representation of \(\mathcal{F}\) in the ring \(D\). Then \(\tilde{\nu}_p\) for \(p\in P\) induces an \(F_1,D\)-bimodule isomorphism \(\nu_p\colon F_p\otimes_{F_1}D\xrightarrow\sim D\). In particular, \(D\) is a smooth \(F_1,D\)-bimodule. Let \(g\in G\) and let \((p_1,p_2)\in P\times P\) satisfy \(p_1 p_2^{-1} = g\). Define a map \[\begin{align} \tilde{\psi}_{p_1,p_2}\colon \mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2},F_{p_1}) &\to\mathop{\mathrm{End}}_{-,D}(D),\\ T&\mapsto\nu_{p_1}\circ(T\otimes\mathrm{id}_D)\circ\nu_{p_2}^{-1}. \end{align}\] As an \(F_1\)-bimodule map, this maps \[\mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2},F_{p_1})F_1\to \mathop{\mathrm{End}}_{-,D}(D)F_1 \subseteq \mathop{\mathrm{End}}_{-,D}(D)D = D,\] where the last step is Lemma 13. So \(\tilde{\psi}_{p_1,p_2}\) restricts to an \(F_1\)-bimodule map \[\psi_{p_1,p_2}\colon \mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2},F_{p_1})F_1 \to D.\] The covariance condition in Definition 11 is equivalent to \[\nu_{p_1 p_2}\circ(\mu_{p_1,p_2}\otimes\mathrm{id}_D) = \nu_{p_1}\circ(\mathrm{id}_{F_{p_1}}\otimes\nu_{p_2})\] for all \(p_1,p_2\in P\). By a routine computation, this implies that the maps \(\tilde{\psi}_{p_1,p_2}\) form a cone under the filtered diagram \(H_{{\mathcal{F},g}}\). This is inherited by the maps \(\psi_{p_1,p_2}\). Hence they induce an \(F_1\)-bimodule map \(\varphi_g\colon \mathcal{O}_{g}\to D\) on the inductive limit. Let \(\varphi\colon \mathcal{O}_\mathcal{F}\to D\) be the map that restricts to \(\varphi_g\) on the summand \(\mathcal{O}_{g}\). An easy computation shows that \(\varphi\) is a ring homomorphism. Since \(D\) is nondegenerate as an \(F_1\)-bimodule and the idempotents in \(F_1\) form a local unit in \(\mathcal{O}_{\mathcal{F}}\), \(D\) is nondegenerate as an \(\mathcal{O}_{\mathcal{F}}\)-module as well. That is, the homomorphism \(\varphi\) is nondegenerate.

It remains to prove that the map that sends the covariant representation \((\tilde{\nu}_p)_{p\in P}\) to the nondegenerate homomorphism \(\varphi\) is inverse to \(\beta_D\). In one direction, we must compute \(\beta_D(\varphi)\). Lemma 13 identifies \(\mathop{\mathrm{End}}_{-,D}(D)D\) with the subring isomorphic to \(D\) consisting of the left multiplication operators. This works also for \(F_1\), so that \(F_1^* = \mathop{\mathrm{End}}_{-,F_1}(F_1) F_1 \cong F_1\). Then \(F_p \otimes_{F_1} F_1^* \cong F_p\). A simple inspection shows that the map \(\psi_{p,1}\) becomes the given map \(\tilde{\nu}_p\) after the resulting identification of \(\mathop{\mathrm{Hom}}_{-,F_1}(F_1,F_p)F_1\) with \(F_p\). This says that \(\beta_D\) maps \(\varphi\) back to \((\tilde{\nu}_p)_{p\in P}\). Conversely, let us start with a nondegenerate homomorphism \(f\colon \mathcal{O}_\mathcal{F}\to D\). This is mapped by \(\beta_D\) to the covariant representation \(\tilde{\nu}_p = f\circ \tilde{\kappa}_p\). Let \(\varphi\colon \mathcal{O}_\mathcal{F}\to D\) be the nondegenerate homomorphism associated to \((\tilde{\nu}_p)_{p\in P}\). We claim that \(f=\varphi\). It suffices to check this on an element of \(\mathcal{O}_\mathcal{F}\) that is the image of \(\xi\in \mathop{\mathrm{Hom}}_{-,F_1}(F_{p_2},F_{p_1})F_1\) for some \(p_1,p_2\in P\). Since \(D\) is a ring with local units, \(f(\xi)=\varphi(\xi)\) follows if \(f(\xi)\cdot d=\varphi(\xi)\cdot d\) holds for all \(d\in D\). We know that \(\nu_{p_2}\colon F_{p_2} \otimes_{F_1} D \to D\) is an isomorphism. Therefore, it suffices to prove \(f(\xi)\cdot \nu_{p_2}(x\otimes d) = \varphi(\xi)\cdot \nu_{p_2}(x\otimes d)\) for all \(x\in F_{p_2}\), \(d\in D\). We compute \[\begin{gather} \varphi(\xi)(\nu_{p_2}(x\otimes d)) = \nu_{p_1}(\xi(x)\otimes d) = \tilde{\nu}_{p_1}(\xi(x))\cdot d = f(\kappa_{p_1}(\xi(x)))\cdot d \\= f(\kappa_{p_1 p_2^{-1}}(\xi)) f(\kappa_{p_2}(x)) \cdot d = f(\xi)(\nu_{p_2}(x\otimes d)). \end{gather}\] Thus the two constructions are inverse to each other and so \(\beta_D\) is bijective. ◻

Proof. See Albandik:Thesis?*Lemma 3.6. ◻

Proof. The first map is a local homeomorphism by Albandik:Thesis?*Lemma 3.9, and this implies that the second map is so as well. ◻

Proof. It is shown in Albandik:Thesis?*Proposition 3.10 that \(\mathcal{H}\) is a locally compact, étale, topological groupoid. It is ample because \(\mathcal{G}\) is ample and \(\mathcal{H}^0=\mathcal{G}^0\). ◻

Proof. The isomorphisms in the beginning of the statement follow from Theorem 4, Proposition 1, and Theorem 2, respectively. Here we use that \(A_R(\mathcal{X}_{p_1})\) is a proper bimodule because \(\mathcal{X}_{p_1}\) is proper. ◻

Proof. Let \(x_k,y_k\in\mathcal{X}_{p_k}\) for \(k=1,2\) satisfy \(s(x_1) = s(x_2)\) and \(s(y_1) = s(y_2)\), so that \(x_1 x_2^*, y_1 y_2^* \in \mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\). By definition, \(x_1 x_2^* = y_1 y_2^*\) if and only if there is \(g\in \mathcal{G}\) with \(r(g) = s(x_1) = s(x_2)\), \(x_1 g = y_1\) and \(x_2 g = y_2\). This implies \([x_1]=[y_1]\) in \(\mathcal{X}_{p_1}/\mathcal{G}\) and \([x_2]=[y_2]\) in \(\mathcal{X}_{p_2}/\mathcal{G}\). So the maps \(\Pi_1\) and \(\Pi_2\) are well-defined. They are local homeomorphisms because the orbit space projections \(\mathcal{X}_{p_1} \times_{s,s} \mathcal{X}_{p_2} \to \mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\) and \(\mathcal{X}_{p_j} \to \mathcal{X}_{p_j}/\mathcal{G}\) and the coordinate projections \(\mathcal{X}_{p_1} \times_{s,s} \mathcal{X}_{p_2} \to \mathcal{X}_{p_j}\) are all local homeomorphisms. Now assume \(x_k,y_k \in U_k\) for \(k=1,2\). Fix \(j\in \{1,2\}\) and assume that \(\Pi_j(x_1 x_2^*) = \Pi_j(y_1 y_2^*)\). This means \(\Pi(x_j) = \Pi(y_j)\) and implies \(x_j = y_j\) because \(U_j\) is a slice. Then \(s(x_{3-j}) = s(x_j) = s(y_j) = s(y_{3-j})\). This implies \(x_{3-j} = y_{3-j}\) because \(U_{3-j}\) is a slice. Thus \(\Pi_j\) is injective on \(U_1U_2^*\). ◻

Proof. It is clear that any compact open subset of a slice in \(\mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\) is again a slice. Lemma 5. implies that the subsets \(U_1 U_2^*\) with compact open slices \(U_j\subseteq \mathcal{X}_{p_j}\) for \(j=1,2\) form an ample base. Since \(U_1 U_2^*\) is a slice by Lemma 20, the compact open slices in \(\mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\) form an ample base as well. ◻

Proof. Let \(u_1\in\mathcal{X}_{p_1}\) and \(u_2,w\in \mathcal{X}_{p_2}\). Then \(u_1 \braket{u_2}{w}\) is defined if and only if \(\Pi(u_2) = \Pi(w)\) in \(\mathcal{X}_{p_2}/\mathcal{G}\) and \(s(u_1) = s(u_2)\) because \(r(\braket{u_2}{w}) = s(u_2)\). The element \(u_1 \braket{u_2}{w}\) only depends on \(u_1 u_2^* \in \mathcal{X}_{p_1}\circ\mathcal{X}_{p_2}^*\) because \(\braket{u_2 g}{w} = g^{-1} \braket{u_2}{w}\) for all \(g\in\mathcal{G}\) with \(r(g) = s(u_2)\). Thus the map in the statement is well-defined, and its image \(U(W)\) consists of all elements that may be written as \(u_1 \braket{u_2}{w}\) with \(u_1 u_2^*\in U\) and \(w\in W\). We claim that the map is also injective. Let \(w' = u_1 \braket{u_2}{w}\) for some \(u_1 u_2^*\in U\) and \(w\in W\) with \(\Pi(u_2) = \Pi(w)\). Then \(\braket{u_2}{w} = \braket{u_1}{w'}\) and \(\Pi(w')= \Pi(u_1) = \Pi_1(u_1 u_2^*)\). The latter determines \(u_1 u_2^* \in U\) because \(U\) is a slice, and then \(w = u_2 \braket{u_2}{w} = u_2 \braket{u_1}{w'}\) is also determined. Thus the map in the statement is injective. In addition, \(s(u_1 \braket{u_2}{w}) = s(w)\) or \(\Pi(u_1 \braket{u_2}{w}) = \Pi(u_1) = \Pi_1(u_1 u_2^*)\) determine \(w\) or \(u_1 u_2^*\), respectively, because \(W\) and \(U\) are slices. Since \(\Pi|_W\) and \(\Pi_2|_U\) are injective as well, the condition \(\Pi_2(u_1 u_2^*) = \Pi(w)\) then determines both \(w\) and \(u_1 u_2^*\). As a consequence, \(s|_{U(W)}\) and \(\Pi|_{U(W)}\) are injective.

The domain \(U \times_{\Pi_2,\mathcal{X}_{p_2}/\mathcal{G},\Pi} W\) of the map in ?? is a Hausdorff, open subset of \((\mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*) \times_{\Pi_2,\mathcal{X}_{p_2}/\mathcal{G},\Pi} \mathcal{X}_{p_2}\) because \(U\) and \(W\) are open and Hausdorff. It is compact because \(U\times W\) is compact and the equality in the Hausdorff space \(\mathcal{X}_{p_2}/\mathcal{G}\) defines a closed subset. Thus \(U \times_{\Pi_2,\mathcal{X}_{p_2}/\mathcal{G},\Pi} W\) is also compact. The map \(\mathcal{X}_{p_2} \times_{\Pi,\mathcal{X}_{p_2}/\mathcal{G},\Pi} \mathcal{X}_{p_2} \to \mathcal{G}\), \((u_2,w)\mapsto \braket{u_2}{w}\), is a local homeomorphism by Antunes-Ko-Meyer:Groupoid_correspondences?*Proposition 3.5 because the right action on \(\mathcal{X}_{p_2}\) is basic. The multiplication \(\mathcal{X}_{p_1}\times_{s,\mathcal{G}^0,r} \mathcal{G}\to \mathcal{X}_{p_1}\) is a local homeomorphism as well by Antunes-Ko-Meyer:Groupoid_correspondences?*Lemma 2.9. Therefore, the map \(\mathcal{X}_{p_1} \times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2} \times_{\Pi,\mathcal{X}_{p_2}/\mathcal{G},\Pi} \mathcal{X}_{p_2} \to \mathcal{X}_{p_1}\), \((u_1,u_2,w)\mapsto u_1 \braket{u_2}{w}\), is a local homeomorphism. The induced map on the orbit space of the \(\mathcal{G}\)-action \((u_1,u_2,w) \cdot g = (u_1 g,u_2 g,w)\) is still a local homeomorphism because orbit space projections are surjective local homeomorphisms by Antunes-Ko-Meyer:Groupoid_correspondences?*Lemma 2.10. Since its restriction to \(U \times_{\Pi_2,\mathcal{X}_{p_2}/\mathcal{G},\Pi} W\) is also injective, this restriction is a homeomorphism onto its image \(U(W)\), and the latter is a compact open subset in \(\mathcal{X}_{p_1}\). Since \(\Pi\) and \(s\) are injective on \(U(W)\), it is a slice.

Now we prove that \(\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U})(\mathbb{1}_{W}) = \mathbb{1}_{U(W)}\) for compact slices \(U\subseteq \mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\) and \(W\subseteq \mathcal{X}_{p_2}\). First, we assume that \(U = U_1 U_2^*\) for compact slices \(U_j \subseteq \mathcal{X}_{p_j}\) for \(j=1,2\). The map \(\mathcal{I}_{p_1,p_2}\) is defined as the composite of three isomorphisms \[\begin{align} A_R(\mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*) &\xrightarrow\sim A_R(\mathcal{X}_{p_1}) \otimes_{A_R(\mathcal{G})} A_R(\mathcal{X}_{p_2}^*) \\&\xrightarrow\sim A_R(\mathcal{X}_{p_1}) \otimes_{A_R(\mathcal{G})} \mathop{\mathrm{Hom}}_{-,A_R(\mathcal{G})} \bigl( A_R(\mathcal{X}_{p_2}), A_R(\mathcal{G})\bigr) A_R(\mathcal{G}) \\&\xrightarrow\sim\mathop{\mathrm{Hom}}_{-,A_R(\mathcal{G})} \bigl( A_R(\mathcal{X}_{p_2}), A_R(\mathcal{X}_{p_1})\bigr) A_R(\mathcal{G}). \end{align}\] The formula in Theorem 4 shows that the first isomorphism maps \(\mathbb{1}_{U_1 U_2^*}\) to \(\mathbb{1}_{U_1} \otimes \mathbb{1}_{U_2^*}\). The formula in Proposition 1 shows that the second isomorphism maps this on to \(\mathbb{1}_{U_1} \otimes \mathcal{I}_{\mathcal{X}_{p_2}}(\mathbb{1}_{U_2^*})\), where \(\mathcal{I}_{\mathcal{X}_{p_2}}(\mathbb{1}_{U_2^*})\) is the map \(A_R(\mathcal{X}_{p_2}) \to A_R(\mathcal{G})\) that is uniquely determined by mapping \(\mathbb{1}_{W}\) to \(\mathbb{1}_{\braket{U_2}{W}}\). The third isomorphism gives the \(R\)-module map that sends \(\mathbb{1}_{W}\) to \(\mathbb{1}_{U_1} \cdot \mathbb{1}_{\braket{U_2}{W}}\). This is equal to \(\mathbb{1}_{U_1\braket{U_2}{W}}\) by Antunes-Ko-Meyer:Groupoid_correspondences?*Lemma 7.4. If \(U = U_1 U_2^*\), then \(U(W) = U_1 \braket{U_2}{W}\). Thus \(\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U})(\mathbb{1}_{W}) = \mathbb{1}_{U(W)}\) is true if \(U = U_1 U_2^*\) with compact slices \(U_j\subseteq \mathcal{X}_{p_j}\) for \(j=1,2\).

Any compact slice \(U\subseteq \mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\) is compact and Hausdorff. The slices of the form \(U_1 U_2^*\) already form an ample base by Lemma 5. Therefore, Proposition 1 allows us to write \(U\) as a disjoint union of sets of the form \(U_{1,k} U_{2,k}^*\) with compact slices \(U_{j,k}\subseteq \mathcal{X}_{p_j}\) for \(j=1,2\), \(k=1,\dotsc,\ell\). Since the map in ?? is a homeomorphism onto \(U(W)\), the sets \((U_{1,k} U_{2,k}^*)(W) = U_{1,k} \braket{U_{2,k}}{W}\) for \(k=1,\dotsc,\ell\) are all compact, open and disjoint. Since \(\mathcal{I}_{p_1,p_2}\) is additive, it follows that \[\mathcal{I}_{p_1,p_2}(\mathbb{1}_{U})(\mathbb{1}_{W}) = \sum_{k=1}^\ell \mathcal{I}_{p_1,p_2} (\mathbb{1}_{U_{1,k} U_{2,k}^*})(\mathbb{1}_{W}) = \sum_{k=1}^\ell \mathbb{1}_{U_{1,k} \braket{U_{2,k}}{W}} = \mathbb{1}_{U(W)}.\qedhere\] ◻

Proof. Let \(\Pi_{p_1 p_2^*}\colon \mathcal{X}_{p_1}\times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2}^* \to \mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\) be the canonical quotient map and define \(\Pi_{p_1 p_2^*, q}\colon \mathcal{X}_{p_1}\times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2}^* \times_{s,\mathcal{G}^0,r} \mathcal{X}_q \to \mathcal{X}_{p_1 q}\circ \mathcal{X}_{p_2 q}^*\), \((u_1,u_2,z) \mapsto (u_1 z)(u_2 z)^*\). The map \(\Pi_{p_1 p_2^*}\) is an orbit space projection for a basic groupoid action, so that it is a local homeomorphism. We claim that the second map is a local homeomorphism as well. This follows because for any slices \(U_1\subseteq \mathcal{X}_{p_1}\), \(U_2 \subseteq \mathcal{X}_{p_2}\), \(X\subseteq \mathcal{X}_q\), the restriction of \(\Pi_{p_1 p_2^*,q}\) to \(U_1 \times_{s,s} U_2 \times_{s,r} X\) is a homeomorphism onto the slice \((U_1 X) (U_2 X)^*\).

The set \(\Pi_{p_1 p_2^*}^{-1}(U)\) is open because \(\Pi_{p_1 p_2^*}\) is continuous. Then \(\Pi_{p_1 p_2^*}^{-1}(U) \times_{s,\mathcal{G}^0,r} \mathcal{X}_q\) is an open subset of \(\mathcal{X}_{p_1}\times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2}^* \times_{s,\mathcal{G}^0,r} \mathcal{X}_q\). The set \(\tilde{\alpha}_{p_1,p_2}^q(U)\) is its image under the local homeomorphism \(\Pi_{p_1 p_2^*,q}\), so it is also open.

Next, we show that \(\tilde{\alpha}_{p_1,p_2}^q(U)\) is a slice. We prove that \(\Pi_1\) is injective on it. The proof for \(\Pi_2\) is the same. So let \(x_1^k\in \mathcal{X}_{p_1}\), \(x_2^k\in \mathcal{X}_{p_2}\) and \(z^k\in \mathcal{X}_q\) satisfy \(s(x^k) = s(y^k) = r(z^k)\) and \(x_1^k (x_2^k)^* \in U\) for \(k=1,2\). We assume that \(x_1^1 z^1 = x_1^2 z^2\) in \(\mathcal{X}_{p_1 q}\). This means that there are \(g,h\in\mathcal{G}\) with \(r(g) = s(x_1^1)\) and \(r(h) = s(z^1)\) such that \(x_1^2 = x_1^1 g\) and \(z^2 = g^{-1} z^1 h\). This implies \(\Pi_1(x_1^1 (x_2^1)^*)\) = \(\Pi_1(x_1^2 (x_2^2)^*)\). Since \(U\) is a slice and \(x_1^1 (x_2^1)^*, x_1^2 (x_2^2)^* \in U\), it follows that \(x_2^2 = x_2^1 g\). So \[(x_1^2 z^2) (x_2^2 z^2)^* = (x_1^1 g\, g^{-1} z^1 h) (x_2^1 g\, g^{-1} z^1 h)^* = (x_1^1 z^1) (x_2^1 z^1)^*.\]

The proof above also shows that there is a well-defined map \(f\colon \tilde{\alpha}_{p_1,p_2}^q(U) \to U\), \((x_1 z)(x_2 z)^* \mapsto x_1 x_2^*\). The fibre of this map at \(x_1 x_2^* \in U\) is homeomorphic to the set of \(z\in\mathcal{X}_q\) with \(r(z) = s(x_1) = s(x_2)\). This set is compact because \(r_*\colon \mathcal{X}_q/\mathcal{G}\to \mathcal{G}^0\) is proper. Therefore, \(f\) is proper and so \(\tilde{\alpha}_{p_1,p_2}^q(U)\) is compact. This finishes the proof that \(\tilde{\alpha}_{p_1,p_2}^q(U)\) is a compact slice.

Next, we claim that \(U(W) X = \tilde{\alpha}_{p_1,p_2}^q(U)(W X)\) holds if \(W\subseteq \mathcal{X}_{p_2}\) and \(X\subseteq \mathcal{X}_q\) are slices. The first set consists of all \(u_1 \braket{u_2}{w} x\) for \(u_1 u_2^* \in U\), \(w\in W\) and \(x\in X\) with \(\Pi(u_2) = \Pi(w)\), \(s(w) = r(x)\). The second set consists of all \(u_1 z \braket{u_2 z}{w x}\) for \(u_1 u_2^* \in U\), \(z\in\mathcal{X}_q\), \(w\in W\) and \(x\in X\) with \(s(u_1) = s(u_2) = r(z)\), \(s(w) = r(x)\), and \(\Pi(u_2 z) = \Pi(w x)\). The last equation says that there are \(g,h\in\mathcal{G}\) with \(u_2 g = w\) and \(g^{-1} z h = x\). In particular, \(\Pi(u_2) = \Pi(w)\), and \(g= \braket{u_2}{w}\) is uniquely determined. Then \(\braket{u_2 z}{w x} = h\) and so \(u_1 z \braket{u_2 z}{w x} = u_1 z h = u_1 \braket{u_2}{w} x\). This computation shows that indeed \(U(W) X = \tilde{\alpha}_{p_1,p_2}^q(U)(W X)\).

Proposition 1 computes \(\mathcal{I}_{p_1,p_2} (\mathbb{1}_{\tilde{\alpha}_{p_1,p_2}^q(U)})\): it maps \(\mathbb{1}_{V}\) for a compact slice \(V\subseteq \mathcal{X}_{p_2 q}\) to \(\mathbb{1}_{\tilde{\alpha}_{p_1,p_2}^q(U)(V)}\). Similarly, \(\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U})\) maps \(\mathbb{1}_{W}\) for \(W\subseteq \mathcal{X}_{p_2}\) to \(\mathbb{1}_{U(W)}\). The slices of the form \(W X\) for slices \(W\subseteq \mathcal{X}_{p_2}\) and \(X\subseteq \mathcal{X}_q\) form an ample base for \(\mathcal{X}_{p_2 q}\) by Lemma 5. Theorem 4 implies that \(\mathbb{1}_{W X} = \mu_{\mathcal{X}_{p_2},\mathcal{X}_q}(\mathbb{1}_{W} \otimes \mathbb{1}_{X})\) for the canonical isomorphism \(\mu_{\mathcal{X}_{p_2},\mathcal{X}_q}\colon A_R(\mathcal{X}_{p_2}) \otimes_{A_R(\mathcal{G})} A_R(\mathcal{X}_q) \xrightarrow\sim A_R(\mathcal{X}_{p_2 q})\). Thus the operator \(\varphi_{p_1,p_2,q}(\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U}))\colon A_R(\mathcal{X}_{p_2 q}) \to A_R(\mathcal{X}_{p_1 q})\) maps \(\mathbb{1}_{W X}\) to \[\begin{align} \varphi_{p_1,p_2,q}(\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U})) (\mathbb{1}_{W X}) &= \mu_{\mathcal{X}_{p_1},\mathcal{X}_q} (\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U})(\mathbb{1}_{W}) \otimes \mathbb{1}_{X}) \\&= \mu_{\mathcal{X}_{p_1},\mathcal{X}_q} (\mathbb{1}_{U(W)}) \otimes \mathbb{1}_{X} = \mathbb{1}_{U(W) X}; \end{align}\] here \(U(W) \subseteq \mathcal{X}_{p_1}\) is a compact slice by Proposition 1, so that \(U(W) X \subseteq \mathcal{X}_{p_1 q}\) is a compact slice by Lemma 5. The functions of the form \(\mathbb{1}_{W X}\) for slices \(W\) and \(X\) as above span \(A_R(\mathcal{X}_{p_2 q})\) by Theorem 3 and Lemma 5. We already know that \(U(W) X = \tilde{\alpha}_{p_1,p_2}^q(U)(W X)\). So the computations above show that \(\Phi_{p_1,p_2,q}(\mathbb{1}_{U}) = \mathbb{1}_{\tilde{\alpha}_{p_1,p_2}^q(U)}\). ◻

Proof. Let \((u_1 u_2^*, v_1 v_2^*)\) be in the domain of the map in 15 . Then \(s(u_1) = s(u_2)\), \(s(v_1) = s(v_2)\) and \(\Pi(u_2) = \Pi_2(u_1 u_2^*) = \Pi_1(v_1 v_2^*) = \Pi(v_1)\). So \(u_1 \braket{u_2}{v_1} v_2^*\) is defined. The properties of the bracket map say that \(u_1 \braket{u_2}{v_1} v_2^* = u_1 g \braket{u_2 g}{v_1 h} (v_2 h)^*\) for all \(g,h\in\mathcal{G}\) with \(r(g) = s(u_1) = s(u_2)\), \(r(h) = s(v_1) = s(v_2)\), so that \(u_1\braket{u_2}{v_1} v_2^* \in \mathcal{X}_{p_1} \circ \mathcal{X}_{p_3}^*\) depends only on \(u_1 u_2^* \in \mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\) and \(v_1 v_2^* \in \mathcal{X}_{p_2} \circ \mathcal{X}_{p_3}^*\). Thus the map in 15 is well-defined. Of course, we let \(U\circ V\) be its image, so that this map becomes surjective.

We compute \(\Pi_1(u_1 \braket{u_2}{v_1} v_2^*) = \Pi(u_1) = \Pi_1(u_1 u_2^*)\) and \(\Pi_2(u_1 \braket{u_2}{v_1} v_2^*) = \Pi(v_2) = \Pi_2(v_1 v_2^*)\). Since \(U\) and \(V\) are slices, these determine \(u_1 u_2^*\in U\) or \(v_1 v_2^*\in V\), respectively. Thus the map in 15 is injective. Since \(\Pi(u_2) = \Pi(v_1)\), and \(U\) and \(V\) are slices, either \(u_1 u_2^*\in U\) or \(v_1 v_2^*\in V\) determines the other. As a consequence, \(\Pi_1\) and \(\Pi_2\) restrict to injective maps on \(U\circ V\). An argument as in the proof of Proposition 1 shows that the same formula as in 15 defines a local homeomorphism \((\mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*) \times_{\Pi_2,\mathcal{X}_{p_2}/\mathcal{G},\Pi_1} (\mathcal{X}_{p_2} \circ \mathcal{X}_{p_3}^*) \to \mathcal{X}_{p_1} \circ \mathcal{X}_{p_3}^*\). Since the restriction of this local homeomorphism in 15 is bijective, it follows that this map is a homeomorphism onto an open subset. Therefore, \(U\circ V\) is open in \(\mathcal{X}_{p_1} \circ \mathcal{X}_{p_3}^*\) and compact Hausdorff. Since \(\Pi_1\) and \(\Pi_2\) restrict to injective maps on \(U\circ V\), it is a compact slice.

Let \(X\subseteq \mathcal{X}_{p_3}\) be a slice. By Proposition 1, \(\mathcal{I}_{p_2,p_3} (\mathbb{1}_{V})\) maps \(\mathbb{1}_{X}\) to \(\mathbb{1}_{V(X)}\), the slice consisting of all \(v_1\braket{v_2}{x}\) with \(v_1 v_2^* \in V\), \(x\in X\) and \(\Pi(v_2) = \Pi(x)\). Then \(\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U})\) maps this on to \(\mathbb{1}_{U(V(X))}\), the slice consisting of all \(u_1 \braket{u_2}{v_1\braket{v_2}{x}} = u_1\braket{u_2}{v_1} \braket{v_2}{x}\) for all \(u_1 u_2^*\in U\), \(v_1 v_2^* \in V\) and \(x\in X\) with \(\Pi(v_2) = \Pi(x)\) and \(\Pi(u_2) = \Pi(v_1)\). This is the same as \(\mathbb{1}_{(U\circ V)(X)}\).

Recall that the product in \(\mathcal{O}_\mathcal{F}\) of \[\begin{align} \mathcal{I}_{p_2,p_3} (\mathbb{1}_{V}) &\in \mathop{\mathrm{Hom}}\bigl(A_R(\mathcal{X}_{p_3}),A_R(\mathcal{X}_{p_2})\bigr)A_R(\mathcal{G}),\\ \mathcal{I}_{p_1,p_2} (\mathbb{1}_{U}) &\in \mathop{\mathrm{Hom}}\bigl(A_R(\mathcal{X}_{p_2}),A_R(\mathcal{X}_{p_1})\bigr)A_R(\mathcal{G}) \end{align}\] is represented by the composite map in \(\mathop{\mathrm{Hom}}\bigl(A_R(\mathcal{X}_{p_3}),A_R(\mathcal{X}_{p_1})\bigr)A_R(\mathcal{G})\). This implies that \(f*h\) as in the statement of the lemma exists. Since \(\mathcal{I}_{p_1,p_3}(\mathbb{1}_{U\circ V})(\mathbb{1}_{X}) = \mathbb{1}_{U(V(X))} = \mathcal{I}_{p_1,p_2} (\mathbb{1}_{U}) \circ \mathcal{I}_{p_2,p_3} (\mathbb{1}_{V})\) for all compact slices \(X\), it follows that \[\mathcal{I}_{p_1,p_2} (\mathbb{1}_{U}) \circ \mathcal{I}_{p_2,p_3} (\mathbb{1}_{V}) = \mathcal{I}_{p_1,p_3}(\mathbb{1}_{U \circ V}).\qedhere\] ◻

Proof. The map in 16 is \(\mathcal{G}\)-equivariant for the relevant \(\mathcal{G}\)-actions on the spaces \(\mathcal{X}_{p_1} \times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2} \times_{s,\mathcal{G}^0,\pi_1^\infty} \Omega\) and \(\mathcal{X}_{p_1} \times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2}\), and so it induces a map \(\pi_{p_1,p_2}\) on the orbit spaces as in the lemma. The two actions on \(\mathcal{X}_{p_1} \times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2} \times_{s,\mathcal{G}^0,\pi_1^\infty} \Omega\) and \(\mathcal{X}_{p_1} \times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2}\) are basic because the right actions on \(\mathcal{X}_{p_1}\) and \(\mathcal{X}_{p_2}\) are basic. Therefore, their orbit space projections are surjective local homeomorphisms by Antunes-Ko-Meyer:Groupoid_correspondences?*Lemma 2.10.

A continuous map \(f\colon X\to Y\) between two spaces with an ample base is proper if and only if any point in \(Y\) has a Hausdorff, compact open neighbourhood \(U\) such that \(f^{-1}(U)\) is Hausdorff, compact. In the case at hand, such \(U\) are provided by compact slices of the form \(U_1 U_2^*\) with compact slices \(U_j \subseteq\mathcal{X}_{p_j}\) for \(j=1,2\). The preimage \(\pi_{p_1,p_2}^{-1}(U_1 U_2^*)\) is the set of all \((u_1, \omega) (u_2, \omega)^*\) with \(u_1\in U_1\), \(u_2\in U_2\), and \(\omega \in \Omega\) such that \(s(u_1) = s(u_2) = r(\omega)\). Since \(\Pi|_{U_1}\) is injective, the relation \((u_1, \omega) (u_2, \omega)^* = (u_1 g, g^{-1}\omega) (u_2 g, g^{-1}\omega)^*\) does not identify any of the elements \((u_1, \omega) (u_2, \omega)^*\) above. So \(\pi_{p_1,p_2}^{-1}(U_1 U_2^*)\) is homeomorphic to \(U_1 \times_{s,\mathcal{G}^0,s} U_2 \times_{r,\mathcal{G}^0,\pi_1^\infty} \Omega\), and the latter is Hausdorff and compact because the spaces \(U_1\), \(U_2\) and \((\pi_1^\infty)^{-1}(s(U_2)\cap s(U_1))\) are and \(\mathcal{G}^0\) is Hausdorff. ◻

Proof. We prove [en:lambda95pi95properties950] and [en:lambda95pi95properties951] together. We first apply \(\pi_{p_1,p_2}^*\) and then \((\lambda_{p_1,p_2})_*\).

It is clear that \(\pi_{p_1,p_2}^*\) in 17 maps \(\mathbb{1}_{U}\) to \(\mathbb{1}_{\pi_{p_1,p_2}^{-1}(U)}\). This forces \(\pi_{p_1,p_2}^{-1}(U)\) to be a compact, open, Hausdorff subset of \(\mathcal{X}_{p_1} \Omega\circ_{\mathcal{G}\Omega} (\mathcal{X}_{p_2} \Omega)^*\). Of course, the reason for this is that \(\pi_{p_1,p_2}\) is a proper continuous map.

By definition, \(\pi_{p_1,p_2}^{-1}(U)\) consists of all \((u_1, \omega)(u_2, \omega)^*\) with \(u_1 u_2^*\in U\) and \(\omega\in \Omega\) satisfying \(s(u_1) = s(u_2) = r(\omega)\), and where \((u_1, \omega) (u_2, \omega)^* = (u_1 g, g^{-1}\omega) (u_2 g, g^{-1}\omega)^*\) for all \(g\in\mathcal{G}\) with \(s(u_1) = s(u_2) = r(\omega) = r(g)\). We claim that \(\pi_{p_1,p_2}^{-1}(U)\) is a slice in the sense that the projections to \(\mathcal{X}_{p_1} \circ_{\mathcal{G}} \Omega\) and \(\mathcal{X}_{p_2} \circ_{\mathcal{G}} \Omega\) that map \((u_1, \omega)(u_2, \omega)^*\) to \(u_1, \omega\) and \(u_2, \omega\), respectively, are injective on \(\pi_{p_1,p_2}^{-1}(U)\).

It suffices to prove this for one of the two projections. Assume \((u_1, \omega) = (u_1', \omega')\) for some \((u_1, \omega)(u_2, \omega)^*\) and \((u_1', \omega')(u_2', \omega')^*\) in \(\pi_{p_1,p_2}^{-1}(U)\). That is, there is \(g\in\mathcal{G}\) with \(s(u_1) = r(g)\), \(u_1' = u_1 g\) and \(\omega' = g^{-1} \omega\). Then \(\Pi_1(u_1 u_2^*) = \Pi(u_1) = \Pi(u_1') = \Pi_1(u_1' (u_2')^*)\). Since \(U\) is a slice, it follows that \(u_1 u_2^* = u_1' (u_2')^*\), that is, \(u_1 h = u_1'\) and \(u_2 h = u_2'\) for some \(h\in\mathcal{G}\) with \(s(u_1) = s(u_2) = r(h)\). Here \(g=h\) because the right action on \(\mathcal{X}_{p_1}\) is free. Then \((u_1, \omega)(u_2, \omega)^* = (u_1', \omega')(u_2', \omega')^*\) in \(\mathcal{X}_{p_1} \Omega\circ_{\mathcal{G}\Omega} (\mathcal{X}_{p_2} \Omega)^*\).

Recall that there are canonical homeomorphisms \(\mathcal{X}_{p_j} \circ \Omega \xrightarrow\sim\Omega\) which, roughly speaking, identify the class of \((u, \omega) \in \mathcal{X}_{p_j}\times_{s,\mathcal{G}^0,\pi_1^\infty} \Omega\) for the equivalence relation \((u_1,\omega) \sim (u_1 g,g^{-1}\omega)\) with the infinite word \(u \omega \in \Omega\); more precisely, \(u\omega \in \varprojlim \mathcal{X}_{p_j q}/\mathcal{G}\), which is isomorphic to \(\Omega\) because the relevant subcategory is cofinal in the category \(PO\) defining \(\Omega\). We write an element of \(\mathcal{X}_{p_j} \Omega\) with a comma to distinguish it from the corresponding element of \(\mathcal{X}_{p_j} \circ \Omega \cong \Omega\). When we map \(\mathcal{X}_{p_1} \Omega \circ (\mathcal{X}_{p_2}\Omega)^*\) to \(\mathcal{H}\) using  \(\lambda_{p_1,p_2}\), then the maps that send \((u_1, \omega)(u_2, \omega)^*\) to \(u_1 \omega\in\Omega\) and \(u_2 \omega\in\Omega\) become the range and source maps \(r_{\mathcal{H}},s_{\mathcal{H}}\colon \mathcal{H}\rightrightarrows \Omega\). So the slice property shown above for \(\pi_{p_1,p_2}^{-1}(U)\) says that the two maps \[r_{\mathcal{H}}\circ \lambda_{p_1,p_2}, s_{\mathcal{H}}\circ \lambda_{p_1,p_2}\colon \mathcal{X}_{p_1} \Omega\circ_{\mathcal{G}\Omega} (\mathcal{X}_{p_2} \Omega)^* \to \Omega\] are injective on the subset \(\pi_{p_1,p_2}^{-1}(U)\). This implies that the restriction of \(\lambda_{p_1,p_2}\) to \(\pi_{p_1,p_2}^{-1}(U)\) is injective. Then it follows that \((\lambda_{p_1,p_2})_* (\mathbb{1}_{\pi_{p_1,p_2}^{-1}(U)}) = \mathbb{1}_{\lambda_{p_1,p_2} (\pi_{p_1,p_2}^{-1}(U))}\), which is the formula asserted in [en:lambda95pi95properties951]. In particular, this subset must be compact, open and Hausdorff. The argument above also shows that \(r_{\mathcal{H}}\) and \(s_{\mathcal{H}}\) restrict to injective maps on it. That is, \(\lambda_{p_1,p_2}(\pi_{p_1,p_2}^{-1}(U))\) is a slice in \(\mathcal{H}\).

We have now proven [en:lambda95pi95properties950]–[en:lambda95pi95properties951]. Next, we prove [en:lambda95pi95properties952]. Lemma 21 shows that \(\Phi_{p_1,p_2,q}(\mathbb{1}_{U})\) for a compact slice \(U\subseteq \mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\) is the characteristic function of the compact slice \(\tilde{\alpha}_{p_1,p_2}^q(U) \subseteq \mathcal{X}_{p_1 q} \circ \mathcal{X}_{p_2 q}^*\) defined in Lemma 21. So \((\pi_{p_1 q,p_2 q})^* \circ \Phi_{p_1,p_2,q}\) maps \(\mathbb{1}_{U}\) to the characteristic function of \(\pi_{p_1 q,p_2,q}^{-1}(\tilde{\alpha}_{p_1,p_2}^q(U))\).

Since the diagram \(\mathfrak X\Omega\) is tight, Definition 21 provides well-defined maps \[\alpha_{p_1,p_2}^q\colon \mathcal{X}_{p_1} \Omega\circ_{\mathcal{G}\Omega} (\mathcal{X}_{p_2} \Omega)^* \to \mathcal{X}_{p_1 q} \Omega\circ_{\mathcal{G}\Omega} (\mathcal{X}_{p_2 q} \Omega)^*, \qquad x y^* \mapsto (x z) (y z)^*.\] By construction, \(\lambda_{p_1 q,p_2 q} \circ \alpha_{p_1,p_2}^q = \lambda_{p_1,p_2}\). We claim that \[\label{eq:claim95pi95alpha95order} \pi_{p_1 q,p_2,q}^{-1}(\tilde{\alpha}_{p_1,p_2}^q(U)) = \alpha_{p_1,p_2}^q(\pi_{p_1,p_2}^{-1}(U)).\tag{19}\] Together with \(\lambda_{p_1 q,p_2 q} \circ \alpha_{p_1,p_2}^q = \lambda_{p_1,p_2}\), this implies [en:lambda95pi95properties952].

The set \(\tilde{\alpha}_{p_1,p_2}^q(U)\) consists of all \((u_1 z) (u_2 z)^* \in \mathcal{X}_{p_1 q} \circ \mathcal{X}_{p_2 q}^*\) with \(u_1 u_2^*\in U\), \(z\in \mathcal{X}_q\) and \(s(u_1) = s(u_2) = r(z)\). Thus the preimage \(\pi_{p_1 q,p_2,q}^{-1}(\tilde{\alpha}_{p_1,p_2}^q(U))\) consists of all \((u_1 z, \omega) (u_2 z, \omega)^* \in \mathcal{X}_{p_1 q}\Omega \circ_{\mathcal{G}\Omega} (\mathcal{X}_{p_2 q} \Omega)^*\) with \(u_1 u_2^* \in U\), \(z\in \mathcal{X}_q\), \(\omega \in \Omega\), \(s(u_1) = s(u_2) = r(z)\), and \(s(z) = \pi_1^\infty(\omega)\).

The set \(\pi_{p_1 q,p_2,q}^{-1}(U)\) consists of \((u_1, \omega') (u_2, \omega')^*\) with \(u_1 (u_2)^*\in U\) and \(\omega'\in\Omega\) such that \(s(u_1') = s(u_2') = \pi_1^\infty(\omega')\). Since the range map induces a homeomorphism \(\mathcal{X}_q\Omega/\mathcal{G}\Omega \cong \Omega\), there is \((z,\omega) \in \mathcal{X}_q \Omega\) with \(s(z) = \pi_1^\infty(\omega)\) and \(r_{\mathcal{X}_q\Omega}(z,\omega) = \omega'\), and its \(\mathcal{G}\)-orbit is unique. This implies \(r(z) = s(u_1) = s(u_2)\). By definition, \(\alpha_{p_1,p_2}^q((u_1, \omega') (u_2, \omega')^*) = (u_1 z, \omega) (u_2 z, \omega)^*\). Here \(u_1,u_2,z,\omega\) satisfy the conditions above to define an element of \(\pi_{p_1 q,p_2,q}^{-1}(\tilde{\alpha}_{p_1,p_2}^q(U))\). Conversely, given any such \(u_1,u_2,z,\omega\), we may take \(\omega' = r_{\mathcal{X}_q\Omega}(z, \omega)\) to witness that we have an element of \(\pi_{p_1 q,p_2 q}^{-1}(\tilde{\alpha}_{p_1,p_2}^q(U))\). This proves 19 and finishes the proof of [en:lambda95pi95properties952].

Finally, we prove [en:lambda95pi95properties953]. It suffices to prove this for \(f = \mathbb{1}_{U}\), \(h = \mathbb{1}_{V}\) for slices \(U\subseteq \mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\) and \(V\subseteq \mathcal{X}_{p_2} \circ \mathcal{X}_{p_3}^*\). Lemma 22 shows that \(\mathbb{1}_{U} * \mathbb{1}_{V} = \mathbb{1}_{U \circ V}\) for a certain slice \(U\circ V\subseteq\mathcal{X}_{p_2} \circ \mathcal{X}_{p_3}^*\). We have seen above that \(\tilde{\alpha}^\infty_{p_1,p_2}\) maps \(\mathbb{1}_{U}\) to the characteristic function of the slice \(\tilde{\alpha}^\infty_{p_1,p_2}(U) = \lambda_{p_1,p_2}( \pi_{p_1,p_2}^{-1}(U))\), and similarly for the other relevant characteristic functions. Since these sets are slices, the convolution of the characteristic functions of \(\tilde{\alpha}_{p_1,p_2}^\infty(U)\) and \(\tilde{\alpha}_{p_2,p_3}^\infty(V)\) is the characteristic function of the product slice \(\tilde{\alpha}_{p_1,p_2}^\infty(U) \cdot \tilde{\alpha}_{p_2,p_3}^\infty(V)\). So what we have to prove is that this product slice is equal to \(\tilde{\alpha}_{p_1,p_3}^\infty(U\circ V)\).

The multiplication in \(\mathcal{H}\) is described in Definition 23 by the equation \(x_1 x_2^* \cdot x_2 x_3^* = x_1 x_3^*\). Here \(x_j \in \mathcal{X}_{p_j}\Omega\) for \(j=1,2,3\). Since \(x_1 x_2^* = x_1 g (x_2 g)^*\) if \(g\in\mathcal{G}\Omega\) is composable with \(x_2\), this implies \(x_1 g (x_2 g)^* \cdot x_2 x_3^* = x_1 x_3^*\). The product \(x_1 x_2^* \cdot x_2' x_3^*\) is only defined if \(x_2' g = x_2\) for some \(g\in \mathcal{G}\Omega\). Here \(g\) is unique, namely, \(g = \braket{x_2'}{x_2}\). So \(x_1 x_2^* \cdot x_2' x_3^* = x_1 \braket{x_2}{x_2'} x_3^*\) if \(\braket{x_2}{x_2'}\) is defined, and \(x_1 x_2^* \cdot x_2' x_3^*\) is not defined otherwise.

Elements of \(\tilde{\alpha}_{p_1,p_2}^\infty(U)\) are represented by \((u_1, \omega) (u_2, \omega)^*\) with \(u_1 u_2^*\in U\), \(\omega\in\Omega\) such that \(s(u_1) = s(u_2) = \pi_1^\infty(\omega)\). The equivalence relation on these representatives is generated by \((u_1, \omega) (u_2, \omega)^* \sim (u_1 g, g^{-1}\omega) (u_2 g, g^{-1}\omega)^*\) for \(g\in\mathcal{G}\) with \(r(g) = s(u_1) = s(u_2) = r(\omega)\). Similarly, elements of \(\tilde{\alpha}_{p_2,p_3}^\infty(V)\) are represented by \((v_1, \omega') (v_2, \omega')^*\) with \(v_1 v_2^*\in V\) and \(\omega'\in\Omega\) such that \(s(v_1) = s(v_2) = \pi_1^\infty(\omega')\), with a similar equivalence relation. Now \((u_1, \omega) (u_2, \omega)^* \cdot (v_1, \omega') (v_2, \omega')^*\) is only defined if \((u_2, \omega) = (v_1, \omega')\) in \(\Omega\). Equivalently, there is \(g\in\mathcal{G}\) with \(r(g) = s(u_2)\), \(v_1 = u_2 g\), and \(\omega' = g^{-1} \omega\). Here \(g = \braket{u_2}{v_1}\) and so \(\omega = \braket{u_2}{v_1} \omega'\). Since \((u_1, \omega) (u_2, \omega)^* = (u_1 g, g^{-1} \omega) (u_2 g, g^{-1}\omega)^*\), the product is equal to \((u_1 \braket{u_2}{v_1}, \omega') (v_2,\omega')^*\). Now Lemma 22 shows that the product slice is indeed equal to \(\tilde{\alpha}_{p_1,p_3}^\infty(U\circ V)\). This finishes the proof of [en:lambda95pi95properties953]. ◻

Proof. By Lemma 24.[en:lambda95pi95properties950] these subsets are compact open slices in \(\mathcal{H}\). First, we prove that they generate the topology on \(\mathcal{H}\). As a projective limit, the space \(\Omega\) comes with canonical maps \(\pi_p^\infty\colon \Omega \to \mathcal{X}_p/\mathcal{G}\). The topology of \(\Omega\) is generated by subsets of the form \((\pi_p^\infty)^{-1}(U)\) for compact open subset \(U\subseteq \mathcal{X}_p/\mathcal{G}\). The topology on \(\mathcal{X}_p\Omega = \mathcal{X}_p \times_{s,\mathcal{G}^0,\pi_1^\infty} \Omega\) is the canonical one on the fibre product, so sets of the form \(U \times_{s,\pi_1^\infty} (\pi_q^\infty)^{-1}(V)\) for compact open subsets \(U\subseteq \mathcal{X}_p\) and \(V\subseteq \mathcal{X}_q/\mathcal{G}\) generate its topology. This induces a canonical topology on the spaces \(\mathcal{X}_{p_1} \Omega \circ (\mathcal{X}_{p_2} \Omega)^*\). A base for it is formed, for instance, by sets of the form \[\label{eq:base95set95in95Bipp95Omega} \setgiven{(u_1, \omega)(u_2, \omega)^*}{ u_1 u_2^*\in U,\; \omega\in (\pi_q^\infty)^{-1}(\Pi(V)),\;s(u_1) = s(u_2) = \pi_1^\infty(\omega)}\tag{20}\] for slices \(U\subseteq \mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\), \(q\in P\) and compact slices \(V\subseteq \mathcal{X}_p\). Any element of \((\pi_q^\infty)^{-1}(\Pi(V))\) may be written as \(v \omega'\) for \(v\in V\) and \(\omega'\in\Omega\) with \(s(v) = r(\omega')\). Here \(v\omega' \in\mathcal{X}_q \circ_{\mathcal{G}} \Omega \cong \Omega\). We claim that the \(\lambda_{p_1,p_2}\)-images of the base sets in 20 in \(\mathcal{H}\) form a base of the topology. Let \(W\subseteq \mathcal{H}_g\) be an open neighbourhood of some \(h\in \mathcal{H}\). There are \(p_1,p_2\in P\) and \(x\in \mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\) such that \(\lambda_{p_1,p_2}(x)=h\). Then \(\lambda_{p_1,p_2}^{-1}(W)\subseteq\mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\) is an open neighbourhood of \(x\). Since the sets in 20 form a base, one of them is contained in \(\lambda_{p_1,p_2}^{-1}(W)\). Then its \(\lambda_{p_1,p_2}\)-image is a neighbourhood of \(x\) contained in \(W\). The subset in 20 has the same \(\lambda_{p_1,p_2}\)-image in \(\mathcal{H}\) as the following subset of \(\mathcal{X}_{p_1 q} \Omega \circ (\mathcal{X}_{p_2 q} \Omega)^*\): \[\setgiven{(u_1v, \omega')(u_2 v, \omega')^*}{u_1 u_2^*\in U,\; v\in V,\; \omega'\in \Omega,\;s(u_1) = s(u_2) = r(v),\; s(v) = \pi_1^\infty(\omega')}\] Here the elements \((u_1 v) (u_2 v)^* \in \mathcal{X}_{p_1 q}\circ \mathcal{X}_{p_2 q}^*\) for \(u_1 u_2^*\in U\) and \(v\in V\) form a compact open subset of the slice \(\tilde{\alpha}_{p_1,p_2}^q(U)\). Therefore, the set above is a compact open slice in \(\mathcal{X}_{p_1 q}\circ \mathcal{X}_{p_2 q}^*\). As a consequence, already the sets of the form \(\tilde{\alpha}_{p_1,p_2}^\infty(U)\) for compact slices \(U\subseteq \mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\) and \(p_1,p_2\in P\) form a base for the topology on \(\mathcal{H}\).

Next, we claim that any compact open subset of a set of the form \(\tilde{\alpha}_{p_1,p_2}^\infty(U)\) may also be written in the same form, but usually with different \(p_1,p_2\). By Lemma 24.[en:lambda95pi95properties950], the map \(\lambda_{p_1,p_2}\) maps \(\pi_{p_1,p_2}^{-1}(U)\) for a slice \(U\) homeomorphically onto \(\tilde{\alpha}_{p_1,p_2}^\infty(U)\). Therefore, any compact open subset of \(\tilde{\alpha}_{p_1,p_2}^\infty(U)\) is the image of a compact open subset \(W\subseteq \pi_{p_1,p_2}^{-1}(U) \subseteq \mathcal{X}_{p_1}\Omega \circ (\mathcal{X}_{p_2}\Omega)^*\). Since \(\Omega\) is defined as a projective limit, we may cover \(W\) by cylinder sets of the form \(X \times_{s,\mathcal{G}^0,\pi_1^\infty} \pi_q^{-1}(\Pi(Y))\) with slices \(X\subseteq \mathcal{X}_{p_1}\) and \(Y \subseteq \mathcal{X}_q\) for some \(q\in P\). Since \(W\) is compact, finitely many such sets suffice. Since \(P\) is an Ore monoid, there is one \(q\in P\) that dominates all of them, so that we only need sets as above with this fixed \(q\). Then we replace subsets of \(\mathcal{X}_{p_1}\Omega \circ (\mathcal{X}_{p_2} \Omega)^*\) by subsets of \(\tilde{\alpha}_{p_1,p_2}^q(U) \subseteq \mathcal{X}_{p_1 q}\Omega \circ (\mathcal{X}_{p_2 q} \Omega)^*\) as above. This shows that \(W = \tilde{\alpha}_{p_1 q,p_2 q}^\infty(W')\) for a compact open subset \(W'\subseteq \tilde{\alpha}_{p_1,p_2}^q(U)\). Now \(W'\) is a compact slice in \(\mathcal{X}_{p_1 q} \circ \mathcal{X}_{p_2 q}^*\) because it is a compact open subset of a compact slice. Thus \(W = \tilde{\alpha}_{p_1 q,p_2 q}^\infty(W')\) for a compact slice in \(\mathcal{X}_{p_1 q} \circ \mathcal{X}_{p_2 q}^*\). This finishes the proof that compact open subsets of sets in \(\mathcal{B}_{\mathcal{H}}\) again belong to \(\mathcal{B}_{\mathcal{H}}\). ◻

Proof. The second statement about equality follows easily from the first statement about inclusions because \(X = Y\) if and only if both \(X\supseteq Y\) and \(Y\supseteq X\). Here we use that \(P\) is an Ore monoid because we need to find an upper bound for two elements in \(P\). Thus it suffices to prove the statement about inclusions. Since \(\tilde{\alpha}_{p_1,p_2}^\infty(U) = \tilde{\alpha}_{p_1 a,p_2 a}^\infty(\tilde{\alpha}_{p_1,p_2}^a(U))\) by Lemma 24.[en:lambda95pi95properties952], an inclusion \(\tilde{\alpha}_{p_1,p_2}^a(U) \supseteq \tilde{\alpha}_{q_1,q_2}^b(V)\) as slices in \(\mathcal{X}_{p_1 a}\circ \mathcal{X}_{p_2 a}^* = \mathcal{X}_{q_1 b}\circ \mathcal{X}_{q_2 b}^*\) for some \(a,b\in P\) with \(p_1 a = q_1 b\), \(p_2 a = q_2 b\) implies that \(\tilde{\alpha}_{p_1,p_2}^\infty(U) \supseteq \tilde{\alpha}_{q_1,q_2}^\infty(V)\). The main point is the converse implication.

We may assume without loss of generality that already \(p_1=q_1\) and \(p_2= q_2\). If \(p_1 p_2^{-1} \neq q_1 q_2^{-1}\) in \(G\), then all sets we consider are disjoint anyway. If \(p_1 p_2^{-1} = q_1 q_2^{-1}\) in \(G\), then we may choose \(a\) and \(b\) to move both sets to the same \(\mathcal{X}_{t_1} \circ \mathcal{X}_{t_2}^*\) for some \(t_1,t_2\in P\). So we assume \(p_1 = p_2\) from now on. In addition, we assume that there is no \(q\in P\) with \(\tilde{\alpha}_{p_1,p_2}^q(U) \supseteq \tilde{\alpha}_{p_1,p_2}^q(V)\). Equivalently, the sets \(\tilde{\alpha}_{p_1,p_2}^q(V) \setminus \tilde{\alpha}_{p_1,p_2}^q(U)\) are nonempty for all \(q\). These sets are compact and open in \(\mathcal{X}_{p_1 q}\circ \mathcal{X}_{p_2 q}^*\) because both \(\tilde{\alpha}_{p_1,p_2}^q(U)\) and \(\tilde{\alpha}_{p_1,p_2}^q(V)\) are compact open subsets.

We claim that the spaces \(\tilde{\alpha}_{p_1,p_2}^q(V)\) form a projective system using the maps \[\label{eq:projective95system95in95base} \tilde{\alpha}_{p_1,p_2}^{q t}(V) \to \tilde{\alpha}_{p_1,p_2}^q(V),\qquad (v_1 w x) (v_2 w x)^* \mapsto (v_1 w) (v_2 w)^*\tag{21}\] for all \(v_1 v_2^*\in V\), \(w\in \mathcal{X}_q\), \(x\in\mathcal{X}_t\). Recall that the notation \(v_1 v_2^* \in V\) means that \((v_1,v_2)\in \mathcal{X}_{p_1} \times_{s,\mathcal{G}^0,s} \mathcal{X}_{p_2}\) and that its image in \(\mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\) belongs to \(V\). The words \((v_1 w x) (v_2 w x)^*\) and \((v_1 w) (v_2 w)^*\) use the lifts \(v_j\in\mathcal{X}_{p_j}\) and not just the image \(v_1 v_2^* \in \mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*\). We must show that the maps in 21 are well-defined. The equivalence relation that defines \(\mathcal{X}_{p_1 q t} \circ \mathcal{X}_{p_2 q t}^*\) is \[(v_1 w x) (v_2 w x)^* \sim (v_1 g_1\,g_1^{-1} w h_1\,h_1^{-1} x k) (v_2 g_2\,g_2^{-1} w h_2\,h_2^{-1} x k)^*\] for \(g_1,g_2,h_1,h_2,k\in\mathcal{G}\) with \(r(g_1) = r(g_2) = s(v_1)= s(v_2)\), \(r(h_1) = r(h_2) = s(w)\), \(r(k) = s(x)\). In addition, \((v_1 g_1)(v_2 g_2)^* \in V\), \(g_1^{-1} w h_1 = g_2^{-1} w h_2\), and \(h_1^{-1} x k = h_2^{-1} x k\) are needed in order for the word \((v_1 g_1\,g_1^{-1} w h_1\,h_1^{-1} x k) (v_2 g_2\,g_2^{-1} w h_2\,h_2^{-1} x k)^*\) to belong to \(\tilde{\alpha}_{p_1,p_2}^q(V)\). Since \(V\) is a slice and \(\Pi_1(v_1 v_2^*) = \Pi(v_1) = \Pi(v_1 g_1) = \Pi_1(v_1 g_1 (v_2 g_2)^*)\), this forces \(v_1 g_1 (v_2 g_2)^* = v_1 v_2^*\). This is equivalent to \(g_1 = g_2\) because the right \(\mathcal{G}\)-actions on \(\mathcal{X}_{p_1}\) and \(\mathcal{X}_{p_2}\) are free. Then \(h_1 = h_2\) follows because the right \(\mathcal{G}\)-action on \(\mathcal{X}_q\) is free. And the latter is what we need for \((v_1 g_1\,g_1^{-1} w h_1) (v_2 g_2\,g_2^{-1} w h_2)^* = (v_1 w)(v_2 w)^*\) in \(\mathcal{X}_{p_1 q} \circ \mathcal{X}_{p_2 q}^*\). This finishes the proof that the map in 21 is well-defined. It is clear that these maps form a contravariant functor on the filtered category \(PO\) used in Definition 24.

The projective limit of the projective system defined in 21 is canonically identified with the set \(\tilde{\alpha}_{p_1,p_2}^\infty(V) \subseteq \lambda_{p_1,p_2}(\mathcal{X}_{p_1}\Omega (\mathcal{X}_{p_2}\Omega)^*)\) because \(\Omega = \varprojlim \mathcal{X}_q/\mathcal{G}\). Roughly speaking, in the limit the finite words \(w\in \mathcal{X}_q\) approximate an element of \(\Omega\). A crucial point here is that \((v_1 w) (v_2 w)^* \sim (v_1 w h) (v_2 w h)^*\) for all \(h\in\mathcal{G}\) with \(r(h) = s(w)\).

The same remarks apply to \(U\) instead of \(V\). Even more, we claim that the spaces \(\tilde{\alpha}_{p_1,p_2}^q(V) \setminus \tilde{\alpha}_{p_1,p_2}^q(U) \subseteq \tilde{\alpha}_{p_1,p_2}^q(V) \subseteq \mathcal{X}_{p_1 q}\circ \mathcal{X}_{p_2 q}^*\) also form a projective system by the same maps. To see this, we show that if \((v_1 w) (v_2 w)^* \in \mathcal{X}_{p_1 q}\circ \mathcal{X}_{p_2 q}^*\) has another representative that belongs to \(\tilde{\alpha}_{p_1,p_2}^q(U)\), then \((v_1 w x) (v_2 w x)^*\) has another representative that belongs to \(\tilde{\alpha}_{p_1,p_2}^{q t}(U)\). Another representative for \((v_1 w) (v_2 w)^*\) is of the form \((v_1 g_1\, g_1^{-1} w h) (v_2 g_2\, g_2^{-1} w h)\) for some \(g_1,g_2,h\in \mathcal{G}\) with \(r(g_1) = r(g_2) = s(v_1) = s(v_2)\), \(r(h) = s(w)\), and \((v_1 g_1)(v_2 g_2)^* \in U\). Then \((v_1 g_1\, g_1^{-1} w h\, h^{-1} x) (v_2 g_2\, g_2^{-1} w h\, h^{-1} x)\) is another representative for \((v_1 w x) (v_2 w x)^*\) that belongs to \(\tilde{\alpha}_{p_1,p_2}^{q t}(U)\). This proves that the spaces \(\tilde{\alpha}_{p_1,p_2}^q(V) \setminus \tilde{\alpha}_{p_1,p_2}^q(U)\) also form a projective system. By assumption, all these spaces are compact and nonempty. Then their projective limit is compact and nonempty as well. Their projective limit is \(\tilde{\alpha}_{p_1,p_2}^\infty(V) \setminus \tilde{\alpha}_{p_1,p_2}^\infty(U)\), and so \(\tilde{\alpha}_{p_1,p_2}^\infty(U)\) does not contain \(\tilde{\alpha}_{p_1,p_2}^\infty(V)\). This finishes the proof. ◻

Proof. If one of the sets \(\tilde{\alpha}_{p_1,p_2}^\infty(U_1)\), \(\tilde{\alpha}_{q_1,q_2}^\infty(U_2)\) or \(\tilde{\alpha}_{t_1,t_2}^\infty(U_3)\) is empty, then the statement follows from Lemma 25. So we may assume that they are all nonempty. Then all slices must belong to the same homogeneous component \(\mathcal{H}_g\) because otherwise their union cannot be of the form \(\tilde{\alpha}_{t_1,t_2}^\infty(U_3)\). As a consequence, there are \(a,b,c\) with \(p_j a = q_j b = t_j c\) for \(j=1,2\). We may replace \(U_1\), \(U_2\) and \(U_3\) by \(\tilde{\alpha}_{p_1,p_2}^a(U_1)\), \(\tilde{\alpha}_{q_1,q_2}^b(U_2)\), and \(\tilde{\alpha}_{t_1,t_2}^c(U_3)\) to arrange, without loss of generality, that already \(p_j = q_j = t_j\) for \(j=1,2\). Next, since \(\tilde{\alpha}_{p_1,p_2}^\infty(U_j) \subseteq \tilde{\alpha}_{p_1,p_2}^\infty(U_3)\) for \(j=1,2\), Lemma 25 shows that there is \(a\in P\) with \(\tilde{\alpha}_{p_1,p_2}^a(U_j) \subseteq \tilde{\alpha}_{p_1,p_2}^a(U_3)\) for \(j=1,2\). Replacing \(U_j\) by \(\tilde{\alpha}_{p_1,p_2}^a(U_j)\) for \(j=1,2,3\), we may therefore arrange without loss of generality that already \(U_1\cup U_2 \subseteq U_3\). This makes \(U_1\cup U_2\) a slice. Our assumption \(\tilde{\alpha}_{p_1,p_2}^\infty(U_1) \cup \tilde{\alpha}_{p_1,p_2}^\infty(U_2) = \tilde{\alpha}_{p_1,p_2}^\infty(U_3)\) is equivalent to \(\tilde{\alpha}_{p_1,p_2}^\infty(U_1 \cup U_2) = \tilde{\alpha}_{p_1,p_2}^\infty(U_3)\) because \(\tilde{\alpha}_{p_1,p_2}^\infty\) commutes with unions. Therefore, Lemma 25 shows that there is \(a\in P\) with \(\tilde{\alpha}_{p_1,p_2}^a(U_1 \cup U_2) = \tilde{\alpha}_{p_1,p_2}^a(U_3)\). Therefore, we may assume without loss of generality that already \(U_1 \cup U_2 = U_3\). Finally, \(U_1 \cap U_2\) and \(\emptyset\) are slices in \(\mathcal{X}_{p_1}\circ\mathcal{X}_{p_2}^*\), and Lemma 25 also applies to them. Since \[\tilde{\alpha}_{p_1,p_2}^\infty(U_1 \cap U_2) \subseteq \tilde{\alpha}_{p_1,p_2}^\infty(U_1) \cap \tilde{\alpha}_{p_1,p_2}^\infty(U_2) = \emptyset = \tilde{\alpha}_{p_1,p_2}^\infty(\emptyset)\] Lemma 25 provides \(a\in P\) with \(\tilde{\alpha}_{p_1,p_2}^a(U_1 \cap U_2) \subseteq \tilde{\alpha}_{p_1,p_2}^a(\emptyset) = \emptyset\). So the reduction steps above eventually lead to a true statement. ◻

Proof. We fix \(g\in G\) to use the inductive limit description of \(\mathcal{O}_{g}\subseteq \mathcal{O}_\mathcal{F}\) and the corresponding inductive limit desccription of \(\mathcal{H}_g\). Theorem 3 describes \(A_R(X)\) for any space \(X\) through generators and relations, given any ample base \(\mathcal{B}\). Namely, let \[\mathcal{B}^{(2)} \mathrel{\vcentcolon=} \setgiven{(V_1,V_2)\in\mathcal{B}^2}{V_1\cap V_2 = \emptyset,\; V_1 \sqcup V_2\in\mathcal{B}};\] then \(A_R(X)\) is isomorphic to the cokernel of the map \[\bigoplus_{(V_1,V_2)\in\mathcal{B}^{(2)}} R \to \bigoplus_\mathcal{B}R,\qquad \delta_{(V_1,V_2)} \mapsto \delta_{V_1} + \delta_{V_2} - \delta_{V_1\sqcup V_2}.\] By Proposition 1, the sets \(\tilde{\alpha}^\infty_{p_1,p_2}(U)\) for \(p_1,p_2\in P\) with \(p_1 p_2^{-1} =g\) and slices \(U\subseteq \mathcal{X}_{p_1} \circ \mathcal{X}_{p_2}^*\) form an ample base \(\mathcal{B}_{\mathcal{H}_g}\) for \(\mathcal{H}_g\).

We describe \(A_R(\mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*)\) as in Theorem 3 using the ample base \(\mathcal{B}_{p_1,p_2}\) of all slices. These ample bases with the maps \(\tilde{\alpha}^q_{p_1,p_2}\) form an inductive system by Lemma 24. Lemma 25 says that our chosen ample base for \(\mathcal{H}_g\) is the inductive limit \(\varinjlim \mathcal{B}_{p_1,p_2}\). Lemma 26 says that same for \(\mathcal{B}^{(2)}\). Therefore, \(A_R(\mathcal{H}_g)\) is the cokernel of the induced map \[\bigoplus_{\varinjlim \mathcal{B}_{p_1,p_2}^{(2)}} R \to \bigoplus_{\varinjlim \mathcal{B}_{p_1,p_2}} R,\qquad \delta_{(V_1,V_2)} \mapsto \delta_{V_1} + \delta_{V_2} - \delta_{V_1\sqcup V_2}.\] Now the construction of free \(R\)-modules and the construction of cokernels both commute with arbitrary colimits and, in particular, with this inductive limit. So \(A_R(\mathcal{H}_g)\) is also isomorphic to the inductive limit of the system of cokernels, which is \(\mathcal{O}_{g} = \varinjlim A_R(\mathcal{X}_{p_1}\circ \mathcal{X}_{p_2}^*)\). The way we defined the maps, it is clear that this isomorphism is the restriction of \(\varrho\) defined above. ◻

Proof. It suffices to prove this when \(f= \mathbb{1}_{U}\) and \(g=\mathbb{1}_{V}\) for compact open slices \(U,V\subseteq \mathcal{X}_p\). Then \(\kappa_p(f) = \delta_U\), \(\kappa_p(g)^* = \delta_V^*\), and \[\kappa_p(f)^* \kappa_p(g) = \delta_U^* \delta_V = \delta_{\braket{U}{V}} = \kappa_1(\mathbb{1}_{\braket{U}{V}}) = \kappa_1(\braket{f}{g}).\qedhere\] ◻

Proof. The natural isomorphism in Theorem 9 is of the form \(\tilde{\nu}_p = \varphi \circ \kappa_p\) for all \(p \in P\) by the Yoneda Lemma. If \(\varphi\) is a \(^*\)-homomorphism, then \(\tilde{\nu}_p(f)^* \tilde{\nu}_p(g) = \tilde{\nu}_e(\braket{f}{g})\) for \(f, g \in A(\mathcal{X}_p)\), \(p\in P\) follows immediately from Lemma 27. Conversely, assume \(\tilde{\nu}_p(f)^* \tilde{\nu}_p(g) = \tilde{\nu}_e(\braket{f}{g})\) for all \(f, g \in A(\mathcal{X}_p)\), \(p \in P\). Since \(\varphi\) is a ring homomorphism, Lemma 27 implies \[\begin{align} \varphi(\kappa_p(f)^*) \cdot \tilde{\nu}_p(g) &= \varphi(\kappa_p(f)^* \kappa_p(g)) = \varphi(\kappa_e(\braket{f}{g})) = \tilde{\nu}_e(\braket{f}{g}) = \tilde{\nu}_p(f)^* \tilde{\nu}_p(g). \end{align}\] Therefore, \((\varphi(\kappa_p(f)^*) - \tilde{\nu}_p(f)^*) \cdot \tilde{\nu}_p(g) = 0\) for all \(g \in A(\mathcal{X}_p)\). Since \(\tilde{\nu}_p(A(\mathcal{X}_p))D\) is dense in \(D\) for any convariant representation, this implies \(\varphi(\kappa_p(f)^*) = \tilde{\nu}_p(f)^* = \varphi(\kappa_p(f))^*\). This implies that \(\varphi\) is compatible with the involution on all the generators \(\delta_U\) and \(\delta_U^*\). Thus, \(\varphi\) is a \(^*\)-homomorphism. ◻

Proof. By definition, \(\mathrm C^**\mathfrak X\) is the pseudofunctor from \(P\) to the \(\mathrm C^*\)-correspondence bicategory that we get by composing \(\mathfrak X\) with the pseudofunctor \(\mathrm C^*\) from the groupoid to the \(\mathrm C^*\)-correspondence bicategory. Such a pseudofunctor is identified with a product system in Albandik-Meyer:Colimits?. It consists of proper \(\mathrm C^*\)-correspondences because \(\mathfrak X\) consists of proper groupoid correspondences. Therefore, the absolute Cuntz–Pimsner algebra \(\mathcal{Q}\) of this product system is defined. Being absolute means that we impose the Cuntz–Pimsner relation on all elements of \(\mathrm C^*(\mathcal{G})\), which all act by compact operators on \(\mathrm C^*(\mathcal{X}_p)\). As a consequence, a nondegenerate \(^*\)-homomorphism from \(\mathcal{Q}\) to a \(\mathrm C^*\)-algebra \(D\) is a family of Cuntz–Pimsner covariant Toeplitz representations \(S_p\colon \mathrm C^*(\mathcal{X}_p) \to D\) for all \(p\in P\) that also satisfy \(S_p(f)S_q(g) = S_{p q}(\mu_{p,q}(f\otimes g))\) for all \(f\in \mathrm C^*(\mathcal{X}_p)\), \(g\in \mathrm C^*(\mathcal{X}_q)\). By Albandik-Meyer:Product?*Proposition 2.5, the Cuntz–Pimsner covariance condition is equivalent to \(\mathrm C^*(\mathcal{X}_p) \cdot D\) being dense in \(D\). It suffices to check the conditions for Toeplitz representations on the dense subspaces \(A(\mathcal{X}_p) \subseteq \mathrm C^*(\mathcal{X}_p)\) and \(A(\mathcal{G}) \subseteq \mathrm C^*(\mathcal{G})\). Therefore, \((S_p)\) is the same as a covariant representation of \(\mathfrak A*\mathfrak X\) in \(A(\mathcal{G}) D A(\mathcal{G})\) that satisfies the extra condition \(\tilde{\nu}_p(f)^* \tilde{\nu}_p(g) = \tilde{\nu}_e(\braket{f}{g})\) for all \(f, g \in A(\mathcal{X}_p)\), \(p \in P\), which is characteristic for Toeplitz representations. Now Proposition 1 provides a natural bijection between nondegenerate \(^*\)-homomorphisms \(\mathcal{Q} \to D\) and \(\mathcal{O}\to A(\mathcal{G}) D A(\mathcal{G})\) or, equivalently, \(\mathrm C^*(\mathcal{O})\to D\). This implies \(\mathrm C^*(\mathcal{O}) \cong \mathcal{Q}\). ◻

Proof. This follows from Theorem 12 and the isomorphism \(\mathrm C^*(\mathcal{O}) \cong \mathrm C^*(\mathcal{H})\) that we deduced above if \(P\) is an Ore monoid. ◻

Proof. First, we consider the relation \(\delta_x \delta_y = \delta_{x y}\) if \(s(x)=r(y)\) and \(\delta_x\delta_y=0\) if \(s(x)\neq r(y)\). If \(x,y\in\mathcal{X}_1=\mathcal{G}= \mathcal{G}^0\), this simply says that the elements \(\delta_x\) for \(x\in\mathcal{G}\) are orthogonal idempotents. We have seen in the proof of Theorem 7 that \(\delta_x = \delta_x^*\) for all \(x\in \mathcal{G}^0\). So these elements are orthogonal projections. In addition, \(\delta_{r(x)} \delta_x \delta_{s(x)} = \delta_x\) and \(\delta_{s(x)} \delta^*_x \delta_{r(x)} = \delta_x\) holds for all \(x\in \mathcal{X}_p\), \(p\in\mathcal{G}\).

The relation \(\delta_{x_1}^* \delta_{x_2} = \delta_{\braket{x_1}{x_2}}\) simplifies because \(\braket{x_1}{x_2}\) for \(x_1,x_2\in \mathcal{X}_p\) is only defined if \(x_1 = x_2\), and then it is \(s(x_1)=s(x_2)\). So this relation says \(\delta_{x_1}^* \delta_{x_2} =0\) for \(x_1\neq x_2\) and \(\delta_x^* \delta_x = \delta_{s(x)}\) for all \(x\in \bigsqcup\mathcal{X}_p\). Together with the relations \(\delta_{r(x)} \delta_x \delta_{s(x)} = \delta_x\) and \(\delta_{s(x)} \delta^*_x \delta_{r(x)} = \delta_x\), this implies that \(\delta_x\) is a partial isometry. The last relation in Corollary 3 simplifies to \[\label{eq:CP-relation95graph} \delta_x = \sum_{\setgiven{y\in\mathcal{X}_p}{r(y)=x}} \delta_y\delta_y^*\tag{23}\] because the right \(\mathcal{G}\)-orbits are just singletons. With these points clarified, it becomes easy to check that the relations in Corollary 3 and those in Brown-Yetter:Conduche?*Definition 2.7 are equivalent. ◻

Proof. The map \(\mu_{p,q}\) is \(\mathcal{G}\)-equivariant for the left and right \(\mathcal{G}\)-actions on \(\mathcal{X}_p\) and satisfies \(\mu_{p,q}(x_1 \bullet g,x_2) = x_1 \vartheta_p(g) \vartheta_g(x_2) = \mu_{p,q}(x_1, g \bullet x_2)\). Since \(\vartheta_1 = \mathrm{id}_{\mathcal{G}}\), the correspondence \(\mathcal{X}_1\) is the identity correspondence on \(\mathcal{G}\) and the maps \(\mu_{1,q}\) and \(\mu_{p,1}\) are the canonical multiplication maps. The assumption \(\vartheta_p \circ \vartheta_q = \vartheta_{p q}\) is equivalent to the associativity of the multiplication maps. ◻

Proof. Stammeier’s generators \(u_g\) and \(s_p\) produce our generators as \(\delta_{g,p} = u_g s_p\) and \(\delta^*_{g,p} = s_p^* u_{g^{-1}}\) for all \(g\in\mathcal{G}\), \(p\in P\). The first relation in Corollary 7 is equivalent to \(u_g u_h = u_{g h}\), \(s_p s_q = s_{p q}\), \(s_p u_g = u_{\vartheta_p(g)} s_p\), \(s_p^* s_q^* = s^*_{q p}\), and \(u_g s_p^* = s_p^* u_{\vartheta_p(g)}\) for all \(g,h\in\mathcal{G}\), \(p,q\in P\). In the presence of the first relation, the second relation in Corollary 7 is equivalent to \(s_p^* s_p = u_1 = 1\) and \(s_p^* u_g s_p = 0\) if \(g \notin \vartheta_p(G)\) for all \(p\in P\), \(g\in \mathcal{G}\). The third relation says that \(1 = \sum_{j=1}^n u_{g_j} s_p s_p^* u_{g_j}^{-1}\) for all \(p\in P\), where \(g_1,\dotsc,g_n\) are a system of representatives for the cosets in \(\mathcal{G}/\vartheta_p(\mathcal{G})\). Stammeier also has a relation about \(s_p^* u_g s_q\) for different \(p,q\in P\). Here he uses that \(P\) has more structure to simplify this expression. We use the relation \(1 = \sum_{j=1}^n u_{h_j} s_{p q} s_{p q}^* u_{h_j}^*\) for the element \(p q \in P\) and a set of coset representatives \((h_j)\) for \(G/\vartheta_p\vartheta_q(G)\) to rewrite \[s_p^* u_g s_q = \sum_{j=1}^n s_p^* u_g u_{h_j} s_{p q} s_{p q}^* u_{h_j^{-1}} s_q = \sum_{j=1}^n s_p^* u_{g h_j} s_p s_q s_p^* s_q^* u_{h_j^{-1}} s_q.\] Our relations imply \(s_p^* u_{g h_j} s_p=0\) unless \(g h_j \in \vartheta_p(\mathcal{G})\) and \(s_q^* u_{h_j^{-1}} s_q=0\) unless \(h_j^{-1} \in \vartheta_q(\mathcal{G})\). Thus all summands vanish unless \(g = g h_j h_j^{-1} \in \vartheta_p(\mathcal{G}) \vartheta_q(\mathcal{G})\). So we get \(s_p^* u_g s_q=0\) in this case, as in (CNP2) in Stammeier:Irreversible?*Definition 3.1. Assume now that \(g = \vartheta_p(g_1) \vartheta_q(g_2)\). Let \(t\) be a common lower bound for \(p,q\), so that \(p = t p'\) and \(q = t q'\). Then the relations above imply \[s_p^* u_g s_q = u_{g_1} s_p^* s_q u_{g_2} = u_{g_1} s_{p'}^* s_t^* s_t s_{q'} u_{g_2} = u_{g_1} s_{p'}^* s_{q'} u_{g_2},\] which is (CNP2) in Stammeier:Irreversible?*Definition 3.1 when \(t\) is a greatest common lower bound. So our relations imply (CNP2) in full generality if \(P\) is as in Stammeier:Irreversible?. ◻