Proof. Let $U\subseteq {\mathscr G}$ be a bisection. If $g_1,g_2\in U$ satisfy ${\varphi }(g_1)={\varphi }(g_2)$ , then, since ${\varphi }$ is iso-unital, we must have ${\mathop {\boldsymbol d}}(g_1)={\mathop {\boldsymbol d}}(g_2)$ ; and so $g_1=g_2$ as U is a bisection. Thus ${\varphi }|_U$ is injective. Clearly, if ${\varphi }$ is a local homeomorphism, it is open. Conversely, if ${\varphi }$ is open and $g\in {\mathscr G}$ , then, since ${\mathscr G}$ is étale, we can find an open bisection U with $g\in U$ . Then ${\varphi }|_U\colon U\to {\varphi }(U)$ is injective and open and, hence, a homeomorphism. Therefore, ${\varphi }$ is a local homeomorphism.

Proof. If ${\varphi }$ is injective, then ${\varphi }(s)=e\in E(T)$ implies that ${\varphi }(s^*s)=e^*e=e$ and so $s=s^*s\in E(S)$ . Conversely, if ${\varphi }^{-1}(E(T))=E(S)$ and ${\varphi }(s_1)={\varphi }(s_2)$ , then since ${\varphi }$ is idempotent bijective, $s_1^*s_1=s_2^*s_2$ . Also ${\varphi }(s_1s_2^*)={\varphi }(s_1s_1^*)\in E(T)$ and so $s_1s_2^*\in E(S)$ . Therefore, $s_1=s_1s_1^*s_1=s_1s_2^*s_2\leq s_2$ ; dually, $s_2\leq s_1$ and, hence, $s_1=s_2$ . Thus ${\varphi }$ is injective.

Similarly, if $\psi $ is injective, then $\psi (g)=x\in {\mathscr H}^{(0)}$ implies that $\psi (g)=\psi ({\mathop {\boldsymbol d}}(g))$ , and so $g={\mathop {\boldsymbol d}}(g)\in {\mathscr G}^{(0)}$ . On the other hand, if $\psi ^{-1}({\mathscr H}^{(0)})={\mathscr G}^{(0)}$ and $\psi (g_1)=\psi (g_2)$ , then ${\mathop {\boldsymbol d}}(g_1)={\mathop {\boldsymbol d}}(g_2)$ and ${\mathop {\boldsymbol r}}(g_1)={\mathop {\boldsymbol r}}(g_2)$ , as $\psi $ is iso-unital; moreover, $\psi (g_1g_2^{-1})=\psi (g_1g_1^{-1})\in {\mathscr H}^{(0)}$ , and so $g_1g_2^{-1} \in {\mathscr G}^{(0)}$ , whence $g_1=g_2$ .

Proof. We begin by showing that $\Gamma _c$ is a functor. Let ${\varphi }\colon {\mathscr G}\to {\mathscr H}$ be an open iso-unital functor. Let $U\in \Gamma _c({\mathscr G})$ . Then ${\varphi }(U)$ is compact open; we just need to show that it is a bisection. Let $g_1,g_2\in U$ satisfy ${\mathop {\boldsymbol d}}({\varphi }(g_1))={\mathop {\boldsymbol d}}({\varphi }(g_2))$ . Then ${\mathop {\boldsymbol d}}(g_1)={\mathop {\boldsymbol d}}(g_2)$ as ${\varphi }$ is iso-unital, and $g_1=g_2$ because U is a bisection. Thus ${\mathop {\boldsymbol d}}|_{{\varphi }(U)}$ is injective and the same argument shows that ${\mathop {\boldsymbol r}}|_{{\varphi }(U)}$ is injective. We conclude that ${\varphi }(U)\in \Gamma _c({\mathscr G})$ .

Next we check that ${\varphi }(UV)={\varphi }(U){\varphi }(V)$ for $U,V\in \Gamma _c({\mathscr G})$ . If $g\in UV$ , then $g=g_1g_2$ for some $g_1\in U$ and $g_2\in V$ , and so ${\varphi }(g)={\varphi }(g_1){\varphi }(g_2)\in {\varphi }(U){\varphi }(V)$ . Thus ${\varphi }(UV)\subseteq {\varphi }(U){\varphi }(V)$ . Conversely, suppose that $h\in {\varphi }(U){\varphi }(V)$ . Then $h={\varphi }(g_1){\varphi }(g_2)$ for some $g_1\in U$ and $g_2\in V$ . Since ${\varphi }$ is iso-unital, this implies that ${\mathop {\boldsymbol d}}(g_1)={\mathop {\boldsymbol r}}(g_2)$ and so $h={\varphi }(g_1g_2)$ for some $g_1g_2\in UV$ . Thus ${\varphi }(UV)={\varphi }(U){\varphi }(V)$ as required. Observe that, since ${\varphi }|_{{\mathscr G}^{(0)}}\colon {\mathscr G}^{(0)}\to {\mathscr H}^{(0)}$ is a homeomorphism and $E(\Gamma _c({\mathscr G}))$ is the set of compact open subsets of ${\mathscr G}^{(0)}$ and $E(\Gamma _c({\mathscr H}))$ is the set of compact open subsets of ${\mathscr H}^{(0)}$ , it follows that $\Gamma _c({\varphi })$ is idempotent bijective. It is clear from the definition that $\Gamma _c$ is functorial.

Next we show that, if ${\varphi }\colon S\to T$ is an idempotent bijective homomorphism of Boolean inverse semigroups, then ${\mathcal G}({\varphi })$ is a well-defined, open iso-unital functor. To simplify our notation, we write ${\varphi }_*$ for ${\mathcal G}({\varphi })$ . To check that ${\varphi }_*$ is well defined, suppose that $u\leq s,s'$ satisfies $\chi (u^*u)=1$ . Then ${\varphi }(u)\leq {\varphi }(s),{\varphi }(s')$ and we have that $\chi \circ ({\varphi }_{E(S)})^{-1}({\varphi }(u)^*{\varphi }(u)) = \chi (u^*u)=1$ . Thus $[{\varphi }(s),\chi \circ ({\varphi }_{E(S)})^{-1}]=[{\varphi }(s'),\chi \circ ({\varphi }_{E(S)})^{-1}]$ . It is immediate from ${\varphi }$ being a homomorphism that ${\varphi }_*$ is a functor. It is bijective on unit spaces as $\chi \mapsto \chi \circ ({\varphi }_{E(S)})^{-1}$ is a bijection of Stone spaces. It remains to show that ${\varphi }_*$ is open and continuous. For continuity, we claim that ${\varphi }_*^{-1}(D(t)) = \bigcup _{{\varphi }(s)\leq t} D(s)$ . Indeed, if ${\varphi }(s)\leq t$ and $\chi (s^*s)=1$ , then ${\varphi }_*([s,\chi ]) = [{\varphi }(s),\chi \circ ({\varphi }_{E(S)})^{-1}] = [t,\chi \circ ({\varphi }|_{E(S)})^{-1}]\in D(t)$ . Conversely, if ${\varphi }_\ast ([s,\chi ])=[{\varphi }(s),\chi \circ ({\varphi }_{E(S)})^{-1}]\in D(t)$ , then there exists $u\leq {\varphi }(s),t$ with $\chi \circ ({\varphi }_{E(S)})^{-1}(u^*u)=1$ . Put $e=({\varphi }_{E(S)})^{-1}(u^*u)\leq ({\varphi }_{E(S)})^{-1}({\varphi }(s)^*{\varphi }(s))=s^*s$ and let $s_0=se$ . Then ${\varphi }(s_0) = {\varphi }(s){\varphi }(e)= {\varphi }(s)u^*u=u\leq t$ and $\chi (s_0^*s_0) = \chi (e) = \chi \circ ({\varphi }_{E(S)})^{-1}(u^*u)=1$ . Thus $[s,\chi ] = [s_0,\chi ]\in D(s_0)$ satisfies ${\varphi }(s_0)\leq t$ . To show that ${\varphi }_*$ is open, we claim that ${\varphi }_*(D(s)) = D({\varphi }(s))$ . From what we have just shown, ${\varphi }_*(D(s))\subseteq D({\varphi }(s))$ . Suppose that $[{\varphi }(s),\lambda ]\in D({\varphi }(s))$ . Let $\chi =\lambda \circ {\varphi }|_{E(S)}$ . Then $\chi (s^*s) = \lambda ({\varphi }(s^*s)) = 1$ and so $[s,\chi ]\in D(s)$ . Moreover, ${\varphi }_*([s,\chi ]) = [{\varphi }(s), \chi \circ ({\varphi }|_{E(S)})^{-1}] = [{\varphi }(s),\lambda ]$ , as required. It is clear that if ${\varphi },\psi $ are composable idempotent bijective homomorphisms, then ${\mathcal G}({\varphi }\psi ) = {\mathcal G}({\varphi })\circ {\mathcal G}(\psi )$ .

It now remains to show that $\eta $ and $\varepsilon $ from Theorem 3.3 are natural isomorphisms. Let ${\varphi }\colon {\mathscr G}\to {\mathscr H}$ be an open iso-unital functor. Let $g\in {\mathscr G}$ and U be a compact open bisection containing g. Note that ${\varphi }(U)$ is a compact open bisection containing ${\varphi }(g)$ and ${\mathcal G}(\Gamma _c({\varphi }))(\eta _{{\mathscr G}}(g)) = {\mathcal G}(\Gamma _c({\varphi }))[U,\chi _{{\mathop {\boldsymbol d}}(g)}] = [{\varphi }(U),\chi _{{\mathop {\boldsymbol d}}({\varphi }(g))}] = \eta _{{\mathscr H}}({\varphi }(g))$ . Next we check that $\varepsilon $ is a natural transformation. If ${\varphi }\colon S\to T$ is an idempotent bijective homomorphism of inverse semigroups, then the computation in the previous paragraph shows that ${\mathcal G}({\varphi })(\varepsilon _S(s)) = {\mathcal G}({\varphi })(D(s)) = D({\varphi }(s))=\varepsilon _T({\varphi }(s))$ . This completes the proof.

Proof. Let ${\varphi }\colon {\mathscr G}\to {\mathscr H}$ be an open iso-unital functor. Suppose first that ${\varphi }$ is injective. Then, for $U\in \Gamma _c({\mathscr G})$ , we have that ${\varphi }(U)\in E(\Gamma _c({\mathscr H}))$ implies ${\varphi }(U)\subseteq {\mathscr H}^{(0)}$ and, hence, $U\subseteq {\mathscr G}^{(0)}$ by Lemma 3.2. Thus $U\in E(\Gamma _c({\mathscr G}))$ and we conclude that $\Gamma _c({\varphi })$ is injective by another application of Lemma 3.2.

Next suppose that ${\varphi }$ is a surjective and let $V\in \Gamma _c({\mathscr H})$ . Note that V is compact Hausdorff, being homeomorphic to the compact Hausdorff space ${\mathop {\boldsymbol d}}(V)$ . We claim that, for each $h\in V$ , there is a compact open neighborhood $V_h\subseteq V$ of h such that there is a continuous section $s\colon V_h\to {\mathscr G}$ of ${\varphi }$ . Choose $g\in {\mathscr G}$ with ${\varphi }(g)=h$ . Then, since ${\varphi ^{-1}}(V)$ is open and the compact open bisections form a basis for the topology of ${\mathscr G}$ , as ${\mathscr G}$ is ample, we can find a compact open bisection $U_g$ with $g\in U_g$ and ${\varphi }(U_g)\subseteq V$ . Then $V_h={\varphi }(U_g)$ is a compact open set containing h (since ${\varphi }$ is open). Moreover, since ${\varphi }$ is iso-unital and $U_g$ is a bisection, ${\varphi }|_{U_g}$ is injective by Proposition 3.1. Then $s=({\varphi }|_{U_g})^{-1}$ is our desired section, as ${\varphi }$ is open.

Since V is compact, we can cover V by finitely many compact open subsets $V_1,\ldots , V_n$ such that there is a continuous section $s_i\colon V_i\to {\mathscr G}$ of ${\varphi }$ . Since V is compact Hausdorff, its compact open subsets form a Boolean algebra. The finitely many compact open subsets $V_1,\ldots , V_n$ of V generate a finite Boolean algebra with maximum element $V=\bigcup _{i=1}^nV_i$ . Let $W_1,\ldots , W_k$ be the atoms of this Boolean algebra. Then the $W_i$ are pairwise disjoint. Also since $W_i=V\cap W_i=(V_1\cap W_i)\cup \cdots \cup (V_n\cap W_i)$ , we see, using that $W_i$ is an atom, that $W_i\subseteq V_j$ for some j. Hence, there is a continuous section $t_i=s_j|_{W_i}\colon W_i\to {\mathscr G}$ of ${\varphi }$ . Since the maximum of a finite Boolean algebra is the join of its atoms, we deduce that $V=W_1\cup \cdots \cup W_k$ ; moreover this is a disjoint union. Hence, we can define a continuous section $s\colon V\to {\mathscr G}$ by putting $s|_{W_i}=t_i$ for $i=1,\ldots , k$ . Note that, since ${\varphi }$ is a local homeomorphism by Proposition 3.1, every continuous section of ${\varphi }$ defined on an open subset of ${\mathscr H}$ is an open mapping. Put $U=s(V)$ and note that U is compact open and ${\varphi }(U)=V$ . Moreover, ${\varphi }|_U$ is injective as ${\varphi }\circ s=1_V$ . We claim that $U\in \Gamma _c({\mathscr G})$ . If $g,g'\in U$ satisfy ${\mathop {\boldsymbol d}}(g)={\mathop {\boldsymbol d}}(g')$ , then ${\mathop {\boldsymbol d}}({\varphi }(g))={\mathop {\boldsymbol d}}({\varphi }(g'))$ and so ${\varphi }(g)={\varphi }(g')$ since V is a bisection. Therefore, $g=g'$ as ${\varphi }|_U$ is injective. Similarly, ${\mathop {\boldsymbol r}}|_U$ is injective and so $U\in \Gamma _c({\mathscr G})$ . Thus $\Gamma _c({\varphi })$ is surjective.

Suppose now that ${\varphi }\colon S\to T$ is an idempotent bijective homomorphism of Boolean inverse semigroups. First assume that ${\varphi }$ is injective and that $[{\varphi }(s),\chi \circ ({\varphi }|_{E(S)})^{-1}] {=}\,{\mathcal G}({\varphi })([s,\chi ])\in {\mathcal G}(T)^{(0)}$ . Then there is an idempotent $f\in E(T)$ such that $f\leq {\varphi }(s)$ and $\chi \circ ({\varphi }|_{E(S)})^{-1} (f)=1$ . Putting $e=({\varphi }|_{E(S)})^{-1}(f)$ , we have $\chi (e)=1$ and ${\varphi }(se)={\varphi }(s)f=f$ . Since ${\varphi ^{-1}}(E(T))=E(S)$ by Lemma 3.2, $se\in E(S)$ and $se\leq s$ . Moreover, $\chi (es^*se) =1$ and so $[s,\chi ]=[se,\chi ]\in {\mathcal G}^{(0)}$ . Therefore, ${\mathcal G}({\varphi })$ is injective by Lemma 3.2.

Next assume that ${\varphi }$ is surjective and let $[t,\chi ]\in {\mathcal G}(T)$ . Then $t={\varphi }(s)$ for some $s\in S$ and $\chi \circ {\varphi }|_{E(S)}(s^*s)=\chi (t^*t)=1$ . Hence, $[s,\chi \circ {\varphi }|_{E(S)}]\in {\mathcal G}(S)$ and ${\mathcal G}({\varphi })([s,\chi \circ {\varphi }|_{E(S)}]) = [{\varphi }(s),\chi ]=[t,\chi ]$ . Therefore, ${\mathcal G}({\varphi })$ is surjective.

Proof. We need only prove the equivalence of the first and second items as the equivalence of the first and third items follows by duality. Suppose first that j as in (2) exists. Then, if $e\in E(T)$ , we have $j(e) = j(ee)=j(e)j(e)$ and so j preserves idempotents. Also, if $t_1\leq t_2$ , we may write $t_1=t_2e$ for some $e\in E(T)$ . Therefore, $j(t_1)=j(t_2e) = j(t_2)j(e)\leq j(t_2)$ , as j preserves idempotents, and so j is order-preserving. Thus (2) implies (1).

Suppose that there is a map j satisfying the conditions specified in (1). Let $t\in T$ and $e\in E(T)$ . First we claim that $j(t)^*j(t)j(e)=j(te)^*j(te)$ . Indeed, applying ${\varphi }$ to both sides yields $t^*te$ . Consequently, $j(t)^*j(t)j(e)= j(te)^*j(te)$ as $j(e)\in E(S)$ and ${\varphi }$ is idempotent bijective. Since $te\leq t$ , the fact that j is order-preserving implies that $j(te)\leq j(t)$ and so

$$ \begin{align*} j(te)=j(t)j(te)^*j(te) = j(t)j(t)^*j(t)j(e)=j(t)j(e), \end{align*} $$

as required.

Proof. Suppose first that f exists. Then, since ${\varphi }$ is a local homeomorphism by Proposition 3.1, the mapping f is open. Let $U\in \Gamma _c({\mathscr H})$ and put $j(U)=f(U)$ . Then $j(U)$ is compact open. It is also a bisection since, if $f(h),f(h')\in f(U)=j(U)$ satisfy ${\mathop {\boldsymbol d}}(f(h))={\mathop {\boldsymbol d}}(f(h'))$ and $h,h'\in U$ , then ${\mathop {\boldsymbol d}}(h)={\mathop {\boldsymbol d}}(h')$ because ${\varphi }$ is a functor and ${\varphi }\circ f=1_{{\mathscr H}}$ . It follows that $h=h'$ because U is a bisection. Similarly, ${\mathop {\boldsymbol r}}|_{j(U)}$ is injective and so $j(U)\in \Gamma _c({\mathscr G})$ . Finally, j is order-preserving and preserves idempotents since $f({\mathscr H}^{(0)})\subseteq {\mathscr G}^{(0)}$ .

For the converse, it suffices by Theorem 3.4 to show that if ${\varphi }\colon S\to T$ is a surjective idempotent bijective homomorphism of Boolean inverse semigroups that admits an order-preserving section $j\colon T\to S$ with $j(E(T))\subseteq E(S)$ , then ${\mathcal G}({\varphi })\colon {\mathcal G}(S)\to {\mathcal G}(T)$ admits a continuous section f that preserves units. Put $f([t,\chi ]) =[j(t),\chi \circ {\varphi }]$ . This is well defined since, if $[t,\chi ]=[t',\chi ]$ , we can find $u\leq t,t'$ with $\chi (u^*u)=1$ and so $j(u)\leq j(t),j(t')$ as j is order-preserving and $\chi ({\varphi }({\kern2pt}j(u)^*j(u))) = \chi (u^*u)=1$ . Thus $[j(t),\chi \circ {\varphi }]=[j(t'),\chi \circ {\varphi }]$ . Trivially, ${\mathcal G}({\varphi })(f([t,\chi ]))= {\mathcal G}({\varphi })([j(t),\chi \circ {\varphi }])= [{\varphi }({\kern2pt}j(t)),\chi ]=[t,\chi ].$ Also, if $[e,\chi ]\in {\mathcal G}(T)^{(0)}$ with $e\in E(T)$ , then $f([e,\chi ]) = [j(e),\chi \circ {\varphi }]\in {\mathcal G}(S)^{(0)}$ since $j(e)\in E(S)$ .

For continuity, we show that $f^{-1}(D(s)) = \bigcup _{j(t)\leq s}D(t)$ . Indeed, if $j(t)\leq s$ , then $f([t,\chi ]) = [j(t),\chi \circ {\varphi }]=[s,\chi \circ {\varphi }]\in D(s)$ . Conversely, if $[j(t),\chi \circ {\varphi }]=f([t,\chi ])\in D(s)$ , then there exists $u\leq j(t),s$ with $\chi ({\varphi }(u^*u))=1$ . Put $e=u^*u$ and note that $j(t)e=u=se$ . Then ${\varphi }(u)\leq {\varphi }({\kern2pt}j(t))=t$ and $\chi ({\varphi }(u)^*{\varphi }(u)) =1$ , and so $[t,\chi ]=[{\varphi }(u),\chi ]\in D({\varphi }(u))$ . Since ${\varphi }$ is idempotent bijective and j is idempotent-preserving, $j({\varphi }(e))=e$ . Thus $t{\varphi }(e)={\varphi }({\kern2pt}j(t)u^*u) = {\varphi }(u)$ , and so $j({\varphi }(u))=j(t{\varphi }(e)) = j(t)j({\varphi }(e))=j(t)e=se\leq s$ by Lemma 3.7. This completes the proof.

Proof. Since ${\mathscr H}$ is Hausdorff, ${\mathscr H}^{(0)}$ is clopen in ${\mathscr H}$ . Therefore, ${\mathscr H}'={\mathscr H}\setminus {\mathscr H}^{(0)}$ is clopen. We show that every $\sigma $ -compact clopen subset X of ${\mathscr H}'$ admits a continuous section $s_X\colon X\to {\mathscr G}$ . Since X is a countable union of compact sets and each of these compact sets can be covered by finitely many compact open bisections of ${\mathscr H}$ contained in X, as ${\mathscr H}$ is ample and X is clopen, $X=\bigcup _{n=1}^\infty U_n$ , where the $U_n$ are compact open bisections of ${\mathscr H}$ . Put $V_1=U_1$ and $V_n = U_n\setminus \bigcup _{i=1}^{n-1}V_i$ . Then, since ${\mathscr H}$ is Hausdorff, each $V_n$ is a compact open bisection and, by construction, $\bigcup _{n=1}^\infty V_n=\bigcup _{n=1}^\infty U_n=X$ . Moreover, the $V_n$ are pairwise disjoint. By Remark 3.6, there is a continuous section $s_n\colon V_n\to {\mathscr G}$ of ${\varphi }$ and we can then define $s_X$ by $s_X|_{V_n}=s_n$ for $n\geq 1$ .

Next we use the well-known fact that a locally compact Hausdorff space is paracompact if and only if it can be partitioned into clopen $\sigma $ -compact subspaces to write ${\mathscr H}'=\coprod _{\alpha \in A}X_{\alpha }$ with each $X_{\alpha }$ clopen and $\sigma $ -compact. We now define $s\colon {\mathscr H}\to {\mathscr G}$ by $s|_{{\mathscr H}^{(0)}} = ({\varphi }|_{{\mathscr G}^{(0)}})^{-1}$ and $s|_{X_{\alpha }}=s_{X_{\alpha }}$ . This completes the proof.

Proof. Let $c\colon S\times S\to K$ be a $2$ -cocycle. Define $F\colon S\to K$ by $F(s) = c(ss^*,ss^*)^*$ and put $c'=c\delta F$ . Then, for $e\in E(S)$ , we have $c'(e,e) = c(e,e)F(e)(eF(e))F(e^2)^* = c(e,e)c(e,e)^*c(e,e)^*c(e,e)\in E(K)$ . Thus $c'$ is normalized.

Proof. For $e\in E(S)$ , let $k_e$ denote the unique idempotent of K with $p(k_e)=e$ . By the $2$ -cocycle condition, the definition of a module and since c is normalized, $c(s,s^*s)= (sc(s^*s,s^*s))c(s,s^*s)=c(s,s^*s)c(s,s^*s)$ . Also, the $2$ -cocycle condition and the definition of a module yield

$$ \begin{align*} c(ss^*,s)^2=(ss^*c(ss^*,s))c(ss^*,s) =c(ss^*,ss^*)c(ss^*,s)=c(ss^*,s), \end{align*} $$

as c is normalized. Thus $c(s,s^*s),c(ss^*,s)$ are idempotents with image $ss^*$ under p and hence are equal as p is idempotent bijective. This yields (1).

For (2), we have $(sc(s^*,s))c(s,s^*s)=c(s,s^*)c(ss^*,s)$ . By the first item, we conclude that $c(s,s^*)=sc(s^*,s)$ .

For (3), we compute $c(e,ef)^2 = (ec(e,ef))c(e,ef) = c(e,e)c(e,ef)=c(e,ef)$ since c is normalized and

$$ \begin{align*} c(e,ef)=efc(e,ef)=ec(e,ef)=c(e,e)c(e,ef). \end{align*} $$

Similarly, we have

$$ \begin{align*} c(ef,e)=(efc(e,e))c(ef,e)=c(ef,e)c(ef,e) \end{align*} $$

as $efc(e,e)=k_{ef}$ because c is normalized. Since $p(c(e,ef))=ef=p(c(ef,e))$ and p is idempotent bijective, we deduce that $c(e,ef)=c(ef,e)$ .

To prove (4), observe that $(ec(f,ef))c(e,ef) =c(e,f)c(ef,ef) =c(e,f)$ since c is normalized, and so $c(e,f)\in E(K)$ by (3).

For (5), we have $(sc(s^*s,e))c(s,s^*se) = c(s,s^*s)c(s,e)$ . By (4), note that $sc(s^*s,e) = k_{ses^*}$ and, by (1), $c(s,s^*s) = k_{ss^*}\geq k_{ses^*}$ , and so the left-hand side is $c(s,s^*se)$ and the right-hand side is $c(s,e)$ .

For (6), we have $(ec(ss^*,s))c(e,s)=c(e,ss^*)c(ess^*,s)$ . As $c(ss^*,s)=k_{ss^*}$ , $c(e,ss^*)=k_{ess^*}$ by (1) and (4), and we deduce that $c(e,s)=c(ess^*,s)$ .

For (7), we have $c(e,es) = c(ess^*,es)\in E(K)$ and $c(se,e)=c(se,es^*s)\in E(K)$ by (6) and (1).

For (8), we note by the $2$ -cocycle condition that

$$ \begin{align*} (ec(s,s^*es))c(e,es)=c(e,s)c(es,s^*es). \end{align*} $$

But by (1), $c(es,s^*es)=k_{ess^*}$ and by (7), $c(e,es)=k_{ess^*}$ . Also, since $p(c(s,s^*es)) = ess^*\leq e$ , we have

$$ \begin{align*} ec(s,s^*es)=ess^*c(s,s^*es)=c(s,s^*es). \end{align*} $$

We conclude that $c(s,s^*es)=c(e,s)$ as required.

For (9), we note that $(ses^*c(s,e))c(ses^*,se) = c(ses^*,s)c(se,e)$ by the $2$ -cocycle condition. But $c(ses^*,se)=k_{ses^*}=c(se,e)$ by (1) and (7), and so we deduce, since $p(c(s,e)) = ses^*$ , that $c(s,e)=c(ses^*,s)$ .

To prove (10), we compute

$$ \begin{align*} (uc(t,s^*s))c(u,s) &= (uc(t,s^*s))c(u,ts^*s) = c(u,t)c(ut,s^*s)\\[-2pt] &= c(u,t)c(ut, t^*u^*uts^*s) \end{align*} $$

where the last equality uses (5). But $t^*u^*uts^*s=s^*st^*u^*uts^*s=s^*u^*us$ . Thus we have $(uc(t,s^*s))c(u,s)=c(u,t)c(ut,s^*u^*us)$ . Using $p(uc(t,s^*s)) = uss^*u^*=p(c(ut,s^*u^*us))$ , we deduce that

$$ \begin{align*} c(u,s)c(ut,s^*u^*us)^* = (uc(t,s^*s)^*)c(u,t), \end{align*} $$

as required.

We now turn to (11). From the $2$ -cocycle condition we obtain

$$ \begin{align*} (tc(u,u^*s^*su))c(t,s^*su) &= c(t,u)c(tu,u^*s^*su), \\[-2pt] (tc(s^*s,u))c(t,s^*su) &= c(t,s^*s)c(ts^*s,u) \\[-2pt] &= c(t,s^*s)c(s,u). \end{align*} $$

Note that $c(s^*s,u) = c(u,u^*s^*su)$ by (8). Therefore, $c(t,s^*s)c(s,u)= c(t,u)c(tu,u^*s^*su)$ . Then, using

$$ \begin{align*} p(c(t,s^*s))&=ss^*\geq suu^*s^*=p(c(tu,u^*s^*su))\\[-2pt] &=p(c(s,u)) = p(c(t,s^*s)^*c(t,u)), \end{align*} $$

we deduce that $c(s,u)c(tu,u^*s^*su)^* = c(t,s^*s)^*c(t,u)$ , as required.

Proof. By Theorem 3.4 and Proposition 3.5, all that remains to check is that $\Gamma _c(\iota )(\Gamma _c({\mathscr K}))=\Gamma _c({\varphi })^{-1}(E(\Gamma _c({\mathscr G})))$ and ${\mathcal G}(\iota )({\mathcal G}(K)) = {\mathcal G}({\varphi })^{-1} ({\mathcal G}(S)^{(0)})$ . For the first item, note that, since ${\mathscr G}^{(0)}$ is open, ${\varphi }^{-1}({\mathscr G}^{(0)})$ is an open subgroupoid of ${\mathscr H}$ and, hence, ample. Moreover, $\Gamma _c({\varphi })^{-1}(E(\Gamma _c({\mathscr G})))=\Gamma _c({\varphi ^{-1}}({\mathscr G}^{(0)}))$ . Since $\iota \colon {\mathscr K}\to {\varphi }^{-1}({\mathscr G}^{(0)})$ is surjective and open iso-unital, we deduce from Proposition 3.5 that

$$ \begin{align*} \Gamma(\iota)(\Gamma_c({\mathscr K})) = \Gamma_c({\varphi}^{-1}({\mathscr G}^{(0)}))=\Gamma_c({\varphi})^{-1}(E(\Gamma_c({\mathscr G}))), \end{align*} $$

as required.

For the second item, let $[k,\chi ]\in {\mathcal G}(K)$ . Then

$$ \begin{align*} {\mathcal G}({\varphi}){\mathcal G}(\iota)([k,\chi]) = [{\varphi}\iota(k),\chi\circ ({\varphi}\iota|_{E(K)})^{-1}]\in {\mathcal G}(S)^{(0)} \end{align*} $$

since ${\varphi }\iota (k)\in E(S)$ . Conversely, if $[{\varphi }(t),\chi \circ ({\varphi }|_{E(T)})^{-1}]={\mathcal G}({\varphi })([t,\chi ])\in {\mathcal G}(S)^{(0)}$ , then there is an idempotent $f\in E(S)$ with $f\leq {\varphi }(t)$ and

$$ \begin{align*} \chi\circ ({\varphi}|_{E(S)})^{-1}(f)=1. \end{align*} $$

Putting $e=({\varphi }|_{E(S)})^{-1} (f)$ , we have that $\chi (e)=1$ and ${\varphi }(te)={\varphi }(t)f=f$ . Since ${\chi (et^*te)=1}$ , we deduce that $[t,\chi ]=[te,\chi ]$ . As $te\in {\varphi ^{-1}}(E(S))$ , we have that $te=\iota (k)$ for some $k\in K$ . Then $[k,\chi \circ \iota |_{E(K)}]\in {\mathcal G}(K)$ (as $\chi \iota (k^*k) = \chi (et^*te)=1$ ) and ${\mathcal G}(K)(\iota )([k,\chi \circ \iota |_{E(K)}]) = [te,\chi ]=[t,\chi ]$ . Thus ${\mathcal G}(\iota )({\mathcal G}(K))= {\mathcal G}({\varphi })^{-1}({\mathcal G}(S)^{(0)})$ , as required. This completes the proof.

Proof. Without loss of generality, we may assume that $\iota $ is an inclusion. Suppose that ${\mathscr G}$ is Hausdorff and $U\in \Gamma _c({\mathscr H})$ . Then ${\varphi }(U)\in \Gamma _c({\mathscr G})$ has a unique maximum idempotent $V\subseteq {\varphi }(U)$ . Since $V\subseteq {\mathscr G}^{(0)}$ , we have ${\varphi ^{-1}}(V)\subseteq {\mathscr K}$ . Also, since V is clopen (as ${\mathscr G}$ is Hausdorff), ${\varphi ^{-1}}(V)\cap U\subseteq {\mathscr K}$ is clopen and, hence, compact (since U is compact). Thus $W={\varphi ^{-1}}(V)\cap U\in \Gamma _c({\mathscr K})$ and $W\subseteq U$ . Moreover, if $W'\in \Gamma _c({\mathscr K})$ with $W'\subseteq U$ , then ${\varphi }(W')\subseteq {\varphi }(U)\cap {\mathscr G}^{(0)}=V$ and so $W'\subseteq W$ . Thus W is the maximum element of $\Gamma _c({\mathscr K})$ below U.

For the converse, by Proposition 3.5, every element of $\Gamma _c({\mathscr G})$ is of the form ${\varphi }(U)$ for some $U\in \Gamma _c({\mathscr H})$ . Let W be the maximum element of $\Gamma _c({\mathscr K})$ below U. Then ${\varphi }(W)\subseteq {\mathscr G}^{(0)}$ is an idempotent and ${\varphi }(W)\subseteq {\varphi }(U)$ . Suppose that $V\subseteq {\mathscr G}^{(0)}$ is compact open with $V\subseteq {\varphi }(U)$ . Let $x\in V$ and let $h\in U$ satisfy ${\varphi }(h)=x$ . Since $U\cap {\varphi ^{-1}}(V)$ is an open subset of ${\mathscr K}$ containing h, we can find a compact open bisection $W'\in \Gamma _c({\mathscr K})$ with $h\in W'\subseteq U\cap {\varphi ^{-1}}(V)\subseteq {\mathscr K}$ . Then $W'\subseteq W$ by choice of W and so ${\varphi }(W')\subseteq {\varphi }(W)$ . As $x={\varphi }(h)\in {\varphi }(W')$ , we deduce that $x\in {\varphi }(W)$ and, hence, $V\subseteq {\varphi }(W)$ . This shows that ${\varphi }(W)$ is the maximum idempotent below ${\varphi }(U)$ and, hence, ${\mathscr G}$ is Hausdorff.

Suppose now that ${\mathscr G}$ and ${\mathscr K}$ are Hausdorff. We continue to assume that $\iota $ is an inclusion. If $U\in \Gamma _c({\mathscr H})$ , then there is a unique maximum element V of $\Gamma _c({\mathscr K})$ with $V\subseteq U$ by the above paragraph. Since ${\mathscr K}$ is Hausdorff, there is a unique maximum idempotent $W\subseteq V$ in $\Gamma _c({\mathscr K})$ . Note that $E(\Gamma _c({\mathscr K})) = E(\Gamma _c({\mathscr H}))$ as ${\mathscr K}^{(0)}={\mathscr H}^{(0)}$ . Thus any idempotent $U'\subseteq U$ belongs to $\Gamma _c({\mathscr K})$ , and, hence, is below V, and therefore below W. Thus W is the maximum idempotent below U. It follows that ${\mathscr H}$ is Hausdorff.

Proof. Clearly $\widetilde {A}$ is a commutative inverse semigroup with $f^*(x) = f(x)^\ast $ giving the inversion. Note that

$$ \begin{align*} {\mathop{\mathrm{supp}}}(fg) = {\mathop{\mathrm{supp}}}(f)\cap {\mathop{\mathrm{supp}}}(g)={\mathop{\mathrm{supp}}}(f){\mathop{\mathrm{supp}}}(g) \end{align*} $$

since A is a group. The map $\gamma $ is a homomorphism because

$$ \begin{align*} U_fU_g &= \{(f(x),x)(g(x),x)\mid x\in {\mathop{\mathrm{supp}}}(f)\cap {\mathop{\mathrm{supp}}}(g)\} \\ &= \{(f(x)g(x),x)\mid x\in {\mathop{\mathrm{supp}}}(fg)\}=U_{fg}. \end{align*} $$

In addition, $\gamma $ is injective since ${\mathop {\mathrm {supp}}}(f)={\mathop {\boldsymbol d}}(U_f)$ and $f = \pi \circ ({\mathop {\boldsymbol d}}|_{U_f})^{-1}$ on its support, where $\pi $ is the projection to A. Conversely, given $U\in \Gamma _c(A\times {\mathscr G}^{(0)})$ , we can define f by

$$ \begin{align*} f(x) = \begin{cases}\pi\circ ({\mathop{\boldsymbol d}}|_U)^{-1} (x) & \text{if } x\in {\mathop{\boldsymbol d}}(U),\\ 0 & \text{otherwise},\end{cases} \end{align*} $$

and $U=U_f$ . Thus $\gamma $ is an isomorphism.

Proof. The idempotents of $\widetilde {A}$ are the characteristic functions $1_U$ with $U\subseteq {\mathscr G}^{(0)}$ compact open and $p(1_U)=U$ . Since ${\mathop {\mathrm {supp}}}(fg)={\mathop {\mathrm {supp}}}(f)\cap {\mathop {\mathrm {supp}}}(g)={\mathop {\mathrm {supp}}}(f){\mathop {\mathrm {supp}}}(g)$ , we have that p is an idempotent bijective semigroup homomorphism. If $W\subseteq {\mathscr G}^{(0)}$ is compact open and $U\in \Gamma _c({\mathscr G})$ , then $UWU^*$ consists of those $x\in {\mathscr G}^{(0)}$ such that there is an arrow $g\colon y\to x$ in U with $y\in W$ . The support of $Uf$ consists of those $x\in {\mathscr G}^{(0)}$ such that there is an arrow $g\colon y\to x$ with $f(y)\neq 0$ . Hence, ${\mathop {\mathrm {supp}}}(Uf) \!= U{\mathop {\mathrm {supp}}}(f)U^*$ . Also notice that if $U\subseteq {\mathscr G}^{(0)}$ is compact open, then $Uf \!=\! 1_Uf$ . It follows that $p(f)f= 1_{{\mathop {\mathrm {supp}}}(f)}f=f$ . It remains to check that $(U,f)\mapsto Uf$ is a semigroup action by endomorphisms.

Clearly, $U(fg)=(Uf)(Ug)$ from the definition. If $U,V\in \Gamma _c({\mathscr G})$ , then $x\in {\mathop {\mathrm {supp}}}(U(Vf))$ if and only if there is $g\colon y\to x$ with $g\in U$ and $h\in V$ with $h\colon z\to y$ for some z such that $f(z) \neq 0$ , in which case $(U(Vf))(x)\!=\!f(z)$ . But $(UV)f(x)$ is nonzero if and only if there is $k\colon z\to x$ in $UV$ for some z such that $f(z)\neq 0$ , in which case $(UV)f(x) \!=\! f(z)$ . But then $k=gh$ for unique $g\in U$ and $h\in V$ with, say, $g\colon y\to x$ and $h\colon z\to y$ . We deduce that $(UV)f(x) = U(Vf(x))$ . This concludes the proof that $\widetilde {A}$ is a $\Gamma _c({\mathscr G})$ -module.

Proof. Note that ${\varphi }\iota (\gamma (f))={\varphi }(\iota (U_f)) = {\mathop {\mathrm {supp}}}(f)$ by construction. So we just need to check that $\Gamma _c({\mathscr G})$ -equivariance is equivalent to centrality. Fix a section $j\colon \Gamma _c({\mathscr G})\to \Gamma _c(\Sigma )$ of $\Gamma _c({\varphi })$ with $j|_{E(\Gamma _c({\mathscr G}))}= (\Gamma _c({\varphi })|_{E(\Gamma _c(\Sigma ))})^{-1}$ . Then the action of $V\in \Gamma _c({\mathscr G})$ on $\gamma (f)=U_f$ is given by

$$ \begin{align*} VU_f = \iota^{-1}({\kern2pt}j(V)\iota(U_f)j(V)^*). \end{align*} $$

This set consists of all $(a,x)$ such that there exist $y\in {\mathop {\mathrm {supp}}}(f)$ and $s\in j(V)$ with ${\varphi }(s)\colon y\to x$ and $(a,x)=\iota ^{-1}(s\iota (f(y),y)s^{-1})$ . On the other hand, $(Vf)(x)\neq 0$ if and only if there exist $g\in V$ and $y\in {\mathop {\mathrm {supp}}}(f)$ , with $g\colon y\to x$ ; and then $(Vf)(x)=f(y)$ . Since ${\varphi }|_{j(V)}\colon j(V)\to V$ is a homeomorphism by Proposition 3.1, this is equivalent to there being $s\in j(V)$ and $y\in {\mathop {\mathrm {supp}}}(f)$ with ${\varphi }(s)\colon y\to x$ and then $(Vf)(x)=f(y)$ . Thus $VU_f=U_{Vf}=\gamma (Vf)$ if and only if $\iota (f({\mathop {\boldsymbol d}}({\varphi }(s))),{\mathop {\boldsymbol r}}({\varphi }(s)))=s\iota (f({\mathop {\boldsymbol d}}({\varphi }(s))),{\mathop {\boldsymbol d}}({\varphi }(s)))s^{-1}$ for all $s\in j(V)$ with ${\mathop {\boldsymbol d}}({\varphi }(s))\in {\mathop {\mathrm {supp}}}(f)$ . This equivalent to

(3-6) $$ \begin{align} \iota(f({\mathop{\boldsymbol d}}({\varphi}(s))),{\mathop{\boldsymbol r}}({\varphi}(s)))s= s\iota(f({\mathop{\boldsymbol d}}({\varphi}(s))),{\mathop{\boldsymbol d}}({\varphi}(s))) \end{align} $$

for all $s\in j(V)$ with ${\mathop {\boldsymbol d}}({\varphi }(s))\in {\mathop {\mathrm {supp}}}(f)$ , which is satisfied if the extension is central. Hence, if the extension is central, $\gamma $ is an isomorphism of modules.

Conversely, if $\gamma $ is a module isomorphism and $s\in \Sigma $ , $a\in A$ , then we can choose a bisection V containing ${\varphi }(s)$ and we may assume that $j(V)$ was chosen to contain s. We can define $f = a1_{{\mathop {\boldsymbol d}}(V)}$ . Then, by (3-6), the equivariance of $\gamma $ implies that $\iota (a,{\mathop {\boldsymbol r}}({\varphi }(s)))s= \iota (f({\mathop {\boldsymbol d}}({\varphi }(s))),{\mathop {\boldsymbol r}}({\varphi }(s)))s =s\iota (f({\mathop {\boldsymbol d}} ({\varphi }(s))),{\mathop {\boldsymbol d}}({\varphi }(s))) = s\iota (a,{\mathop {\boldsymbol d}}({\varphi }(s)))$ . We conclude that the extension is central.

Proof. Choose a section $j\colon S\to T$ with $j|_{E(S)} = ({\varphi }|_{E(T)})^{-1}$ . We can then turn K into an S-module (with $p={\varphi }\iota \colon K\to E(S)$ ); denote by $c\colon S\times S\to K$ the corresponding normalized $2$ -cocycle. Since $E(T)\cong E(S)$ is a Boolean algebra, we just need to show that if $t,u\in T$ are orthogonal, then they have a join. In fact, it is enough to show they have a common upper bound y for then, by [Reference Donsig, Fuller and Pitts21, Proposition 1.4.18], one has $y(t^*t\vee u^*u) = yt^*t\vee yu^*u = t\vee u$ . Put $\widetilde {t}={\varphi }(t)$ and $\widetilde {u}={\varphi }(u)$ . Then the fact that $t,u$ are orthogonal is equivalent to $t^*t,u^*u$ and $tt^*,uu^*$ being orthogonal. This in turn is equivalent to $\widetilde {t},\widetilde {u}$ being orthogonal since ${\varphi }$ is idempotent bijective. Thus $x=\widetilde {t}\vee \widetilde {u}$ exists. We can write $t=\iota (k_t)j(\widetilde {t})$ and $u=\iota (k_u)j({\kern1pt}\widetilde {u}{\kern1pt})$ for unique $k_t,k_u\in K$ with $p(k_t) = \widetilde {t}{\kern4pt}\widetilde {t}^*$ and $p(k_u)=\widetilde {u}{\kern2pt}\widetilde {u}^*$ . Then $p(k_tc(x,\widetilde {t}^*{\kern2pt}\widetilde {t}{\kern1pt})^*) = \widetilde {t}{\kern4pt}\widetilde {t} ^*$ and $p(k_uc(x,\widetilde {u}^*{\kern2pt}\widetilde {u})^*)=\widetilde {u}{\kern2pt}\widetilde {u}^*$ and, hence, since p is idempotent bijective, $k=k_tc(x,\widetilde {t}^*{\kern2pt}\widetilde {t}{\kern1pt})^* \vee k_uc(x,\widetilde {u}^*{\kern2pt}\widetilde {u})^*$ is defined and $p(k)=\widetilde {t}{\kern4pt}\widetilde {t}^*\vee \widetilde {u}{\kern2pt}\widetilde {u}^*= xx^*$ (since p is additive, being idempotent bijective). Let $y=\iota (k)j(x)$ . We claim that $t,u\leq y$ . The argument is symmetric, so we just give it for t.

Note that $\iota (K)$ centralizes $E(T)$ as $\iota (k)e=\iota (kf)=\iota (fk)=e\iota (k)$ , where $f\in E(K)$ is unique with $\iota (f)=e$ (using that $\iota $ is idempotent bijective). We then have

$$ \begin{align*} yt^*t = \iota(k)j(x)j({\kern1pt}\widetilde{t}^*{\kern2pt}\widetilde{t}{\kern1pt})=\iota(k)\iota(c(x,\widetilde{t}^*{\kern2pt}\widetilde{t}{\kern1pt}))j(x\widetilde{t}^*{\kern2pt}\widetilde{t}{\kern1pt}) = \iota(kc(x,\widetilde{t}^*{\kern2pt}\widetilde{t}{\kern1pt}))j({\kern1pt}\widetilde{t}{\kern1pt}). \end{align*} $$

Therefore, it suffices to show that $kc(x,\widetilde {t}^*{\kern2pt}\widetilde {t}{\kern1pt}) = k_t$ . Note that $p(c(x,\widetilde {t}^*{\kern2pt}\widetilde {t}{\kern1pt}))= \widetilde {t}^*{\kern2pt}\widetilde {t}$ and so $k_uc(x,\widetilde {u}^*{\kern2pt}\widetilde {u}{\kern1pt})^*c(x,\widetilde {t}^*{\kern2pt}\widetilde {t}{\kern1pt}) =0$ . Therefore,

$$ \begin{align*} kc(x,\widetilde{t}^*{\kern2pt}\widetilde{t}{\kern1pt}) = k_tc(x, \widetilde{t}^*{\kern2pt}\widetilde{t}{\kern1pt})^*c(x, \widetilde{t}^*{\kern2pt}\widetilde{t}{\kern1pt}) =k_t, \end{align*} $$

as required, since multiplication distributes over joins in a Boolean inverse semigroup. This completes the proof.

Proof. By Theorem 3.19, we just need to show that the cohomology class corresponding to the equivalence class of $\Sigma \oplus \Sigma '$ is the product of the cohomology classes corresponding to $\Sigma $ and $\Sigma '$ . Let $j\colon \Gamma _c({\mathscr G})\to \Gamma _c(\Sigma )$ and $j'\colon \Gamma _c({\mathscr G})\to \Gamma _c(\Sigma ')$ be set-theoretic sections with j and $j'$ preserving idempotents. Then the cohomology classes corresponding to $\Sigma $ and $\Sigma '$ are given by normalized $2$ -cycles $c,c'$ , respectively, where $j(U)j(V) = \iota (\gamma (c(U,V)))j(UV)$ and $j'(U)j'(V)=\iota '(\gamma (c'(U,V)))j'(UV)$ where $\gamma \colon \widetilde {A}\to \Gamma _c(A\times {\mathscr G}^{(0)})$ is the isomorphism of Proposition 3.16. Define $\widetilde {j}\colon \Gamma _c({\mathscr G})\to \Gamma _c(\Sigma \oplus \Sigma ')$ by

$$ \begin{align*} \widetilde{j}(U) = \{[g,g']\mid g\in j(U), g'\in j'(U)\}. \end{align*} $$

This is a compact open set since it is the image of $j(U)\times _{{\varphi },{\varphi }'} j'(U)$ , which is compact open in $\Sigma \times _{{\varphi },{\varphi }'}\Sigma '$ . It is a bisection because if ${\mathop {\boldsymbol d}}([g,g']) = {\mathop {\boldsymbol d}}([h,h'])$ for some $g,h\in j(U)$ and $g',h'\in j'(U)$ , then ${\mathop {\boldsymbol d}} (g)={\mathop {\boldsymbol d}}(h)$ and ${\mathop {\boldsymbol d}}(g')={\mathop {\boldsymbol d}}(h')$ , and so $g=g'$ and $h=h'$ . Similarly, ${\mathop {\boldsymbol r}}|_{\widetilde {j}(U)}$ is injective. By construction, $\rho ({\kern2pt}\widetilde {j}(U))=U$ . Also, if $U\subseteq {\mathscr G}^{(0)}$ , then $\widetilde {j}(U)\subseteq (\Sigma \oplus \Sigma ')^{(0)}$ . Let us compute the normalized $2$ -cocycle $\widetilde {c}$ corresponding to $\widetilde {j}$ .

Let $U,V\in \Gamma _c({\mathscr G})$ . Then $\widetilde {j}(U)\widetilde {j}(V)({\kern2pt}\widetilde {j}(UV))^*$ consists of all composable products $[g,g'][h,h'][k^{-1}, (k')^{-1}]$ , for $g\in j(U)$ , $h\in j(V)$ , $k\in j(UV)$ , $g'\in j'(U)$ , $h'\in j'(V)$ and $k'\in j'(UV)$ . Then since $U,V$ are bisections, it follows from composability that ${\varphi }(g){\varphi }(h)={\varphi }(k)$ and ${\varphi }'(g'){\varphi }'(h')={\varphi }'(k')$ . Therefore, $gh=ak$ and $g'h'=a'k'$ for some $a,a'\in A$ . But, using $j(U)j(V) = \iota (\gamma (c(U,V)))j(UV)$ and $j'(U)j'(V)=\iota '(\gamma (c'(U,V)))j'(UV)$ , we deduce that $a=c(U,V)({\mathop {\boldsymbol r}}(gh))$ and $a'=c'(U,V)({\mathop {\boldsymbol r}}(gh))$ . Thus

$$ \begin{align*} [g,g'][h,h'][k^{-1}, (k')^{-1}] &= [c(U,V)({\mathop{\boldsymbol r}}(gh)){\mathop{\boldsymbol r}}(gh),c'(U,V)({\mathop{\boldsymbol r}}(gh)){\mathop{\boldsymbol r}}(gh)]\\ &= [{\mathop{\boldsymbol r}}(gh),(c(U,V)c'(U,V))({\mathop{\boldsymbol r}}(gh)){\mathop{\boldsymbol r}}(gh)]\\ &= \kappa((c(U,V)c'(U,V))({\mathop{\boldsymbol r}}(gh)),{\mathop{\boldsymbol r}}(gh)). \end{align*} $$

It follows that $\widetilde {c}(U,V) = c(U,V)c'(U,V)$ , as required.

Proof. For ${\varphi }\in {\mathop {\mathrm {End}}}_c(R)$ , we put ${\varphi }(R)=Re_{{\varphi }}$ , where $e_{{\varphi }}\in E(Z(R))$ . First we claim that if ${\varphi }\in {\mathop {\mathrm {End}}}_c(R)$ and $z\in Z(R)$ , then ${\varphi }(z)\in Z(R)$ . Indeed, if $r\in R$ , then $re_{{\varphi }}\in Re_{{\varphi }}={\varphi }(R)$ and so $re_{{\varphi }} = {\varphi }(r')$ for some $r'\in R$ . Thus

$$ \begin{align*} {\varphi }(z)r &= ({\varphi }(z)e_{{\varphi }})r = {\varphi }(z)(re_{{\varphi }})={\varphi }(z){\varphi }(r') = {\varphi }(zr')={\varphi }(r'z) = {\varphi }(r'){\varphi }(z) \\ &= re_{{\varphi }}{\varphi }(z)=r{\varphi }(z). \end{align*} $$

So if ${\varphi },\psi \in {\mathop {\mathrm {End}}}_c(R)$ , then ${\varphi }(\psi (R)) = {\varphi }(Re_{\psi }) = {\varphi }(R){\varphi }(e_{\psi }) = Re_{{\varphi }}{\varphi }(e_{\psi })$ and $e_{{\varphi }}{\varphi }(e_{\psi })$ is a central idempotent as ${\varphi }(Z(R))\subseteq Z(R)$ .

Clearly, ${\varphi }_e(r)=re$ , where e is a central idempotent is a proper endomorphism with image $Re$ . Conversely, if ${\varphi }\in {\mathop {\mathrm {End}}}_c(R)$ is an idempotent, then ${\varphi }$ fixes ${\varphi }(R) = Re_{{\varphi }}$ . So if $r\in R$ , then ${\varphi }(r)={\varphi }(r)e_{{\varphi }}={\varphi }(r){\varphi }(e_{{\varphi }}) = {\varphi }(re_{{\varphi }}) =re_{{\varphi }}$ , as required. This completes the proof.

Proof. Now, $\alpha _s(R) = \alpha _{ss^*}(\alpha _s(R))\subseteq R1_{ss*}$ . On the other hand, since $\alpha _{ss^*}(r) = r1_{ss^*}$ by Proposition 4.1,

$$ \begin{align*}r1_{ss^*} = \alpha_{ss^*}(r) = \alpha_s(\alpha_{s^*}(r))\in \alpha_s(R).\end{align*} $$

This establishes the first item. The second follows from the first because $\alpha _s(R1_e) = \alpha _s\alpha _e(R)=\alpha _{se}(R)=R1_{ses^*}$ .

Proof. Given $r,r'\in (Z(R)1_e)^\times $ and $u,u'\in (Z(R)1_f)^\times $ satisfying $rr'=1_e$ and $uu'=1_f$ , we have $ru,u'r'\in Z(R)1_{ef}$ and $ruu'r' = r1_fr'=rr'1_f=1_e1_f=1_{ef}$ . Thus $ru,u'r'\in \widetilde {R}$ and $(ru)(u'r')(ru)=ru$ . Therefore, $\widetilde {R}\subseteq Z(R)$ is a commutative von Neumann regular semigroup and, hence, inverse. Clearly, $E(\widetilde {R}) =\{1_e\mid e\in E(S)\}$ .

If $s\in S$ and $r,r'\in (Z(R)1_e)^\times $ satisfy $rr'=1_e$ , then $\alpha _s(r)\alpha _s(r') = \alpha _s(1_e) = 1_{ses^*}$ , and so $\alpha _s(r)\in (Z(R)1_{ses^*}{\kern-1pt})^\times $ (using Propositions 4.1 and 4.3).

Proof. For the first item, it suffices to check associativity. We calculate

$$ \begin{align*} ((r_1\delta_s)(r_2\delta_t))r_3\delta_u &= r_1(sr_2)c(s,t)(str_3)c(st,u) \delta_{stu}, \\ r_1\delta_s(r_2\delta_t r_3\delta_u) &= r_1s(r_2(tr_3)c(t,u))c(s,tu)\delta_{stu}\\ &= r_1(sr_2)(str_3)(sc(t,u))c(s,tu)\delta_{stu}. \end{align*} $$

Since c takes values in the center of R, associativity follows from c being a $2$ -cocycle.

The second item is more technical. Consider $s\leq t\in S$ and $r\in R$ . If $u\in S$ and $a\in R$ , then

$$ \begin{align*}a\delta_u(r\delta_s-rc(t,s^*s)^*\delta_t)= a(ur)c(u,s)\delta_{su} - a(ur)(uc(t,s^*s)^*)c(u,t)\delta_{ut}.\end{align*} $$

Since $su\leq ut$ , it suffices to show that $c(u,s)c(ut, s^*u^*us)^* = (uc(t,s^*s)^*)c(u,t)$ , which is the content of Proposition 3.11(10). On the other hand,

$$ \begin{align*}(r\delta_s-rc(t,s^*s)^*\delta_t)a\delta_u = r(sa)c(s,u)\delta_{su}-rc(t,s^*s)^*(ta)c(t,u)\delta_{tu}.\end{align*} $$

Note that $c(t,s^*s)\in R1_{ss^*}$ and so

$$ \begin{align*}c(t,s^*s)^*(ta) = c(t,s^*s)^*1_{ss^*}(ta) = c(t,s^*s)^*(ta)1_{ss^*} = c(t,s^*s)^*(ss^*(ta)) = c(t,s^*s)^*(sa).\end{align*} $$

Thus it suffices to show that $r(sa)c(s,u)\delta _{su}-r(sa)c(t,s^*s)^*c(t,u)\delta _{tu}\in I$ . Since $su\leq tu$ , it suffices to show that $c(s,u)c(tu,u^*s^*su) = c(t,s^*s)^*c(t,u)$ , which follows from Proposition 3.11(11). This completes the proof that I is an ideal.

If $\alpha $ is zero-preserving, then, since $0\leq 0$ , we have $r\delta _0-rc(0,0)^*\delta _0\in I$ . But $c(0,0)=0$ as $1_0=0$ and so $(Z(R)1_0)^\times =0$ . Thus $r\delta _0\in I$ .

Proof. Let $F\colon S\to \widetilde {R}$ be a mapping with $F(s)\in (Z(R)1_{ss^*}{\kern-1pt})^\times $ , for all $s\in S$ , and with $c=c'\cdot \delta F$ . Then $c(s,t) = c'(s,t)F(s)(sF(t))F(st)^*$ . Since $c,c'$ are normalized, $c(e,e)=1_e=c'(e,e)$ for $e\in E(S)$ . Thus

$$ \begin{align*} F(e)= c'(e,e)F(e)(eF(e))F(ee)^* = c(e,e)=1_e \end{align*} $$

and F is idempotent-preserving.

We put $A=\bigoplus _{s\in S}R\delta _s$ with product $r\delta _s\cdot r'\delta _t = r(sr')c(s,t)\delta _{st}$ and let I be the additive subgroup generated by all $r\delta _s-rc(t,s^*s)^*\delta _t$ with $s\leq t$ . Similarly, we let $A'=\bigoplus _{s\in S}R\delta _s$ with multiplication $r\delta _s\cdot r'\delta _t = r(sr')c'(s,t)\delta _{st}$ and $I'$ be the additive subgroup generated by all $r\delta _s-rc'(t,s^*s)^*\delta _t$ with $s\leq t$ . We first construct a homomorphism $\Phi \colon A\to A'$ with $\Phi (I)\subseteq I'$ . Define $\Phi (r\delta _s) = rF(s)\delta _s$ . Note that $\Phi (r\delta _s\cdot r'\delta _t) = r(sr')c(s,t)F(st)\delta _{st}$ , whereas $\Phi (r\delta _s)\Phi (r'\delta _t) = rF(s)\delta _s\cdot r'F(t)\delta _t = r(sr')F(s)(sF(t))c'(s,t)\delta _{st}$ . But $F(st)^*F(st) = 1_{stt^*s^*}$ and $c'(s,t)\in R1_{stt^*s^*}$ , whence

$$ \begin{align*} c'(s,t) = F(st)^*F(st)c'(s,t). \end{align*} $$

Therefore,

$$ \begin{align*} \Phi(r\delta_s)\Phi(r'\delta_s) &= r(sr')F(s)(sF(t))F(st)^*c'(s,t)F(st)\delta_{st} \\ &= r(sr')c(s,t)F(st)\delta_{st}=\Phi(r\delta_s\cdot r'\delta_t). \end{align*} $$

We conclude that $\Phi $ is a homomorphism.

Suppose now that $s\leq t$ and note that

$$ \begin{align*} \Phi (r\delta _s-rc(t,s^*s)^*\delta _t)\! = rF(s)\delta _s\! - rc(t,s^*s)^* F(t)\delta _t. \end{align*} $$

But

$$ \begin{align*} c(t,s^*s)^*=c'(t,s^*s)^*F(t)^*(tF(s^*s))^*F(ts^*s) = c'(t,s^*s)^*F(t)^*1_{ss^*}F(s) \end{align*} $$

as $F(s^*s) =1_{s^*s}$ and $ts^*s=s$ . Thus, using $1_{tt^*}\geq 1_{ss^*}$ , we have

$$ \begin{align*} rc(t,s^*s)^*F(t)=rF(s)c'(t,s^*s)^*F(t)^*F(t)1_{ss^*} = rF(s)c'(t,s^*s)^*. \end{align*} $$

Therefore, $ rF(s)\delta _s - rc(t,s^*s)^*F(t)\delta _t=rF(s)\delta _s-rF(s)c'(t,s^*s)^*\delta _t\in I'$ , and so we have a well-defined homomorphism ${\varphi }\colon R\rtimes _{\alpha ,c}S\to R\rtimes _{\alpha ,c'} S$ given by

$$ \begin{align*} {\varphi }(r\delta _s+I) = \Phi (r\delta _s)+I' = rF(s)\delta _s+I'. \end{align*} $$

Similarly, we have a well-defined homomorphism $\psi \colon R\rtimes _{\alpha ,c'}S\to R\rtimes _{\alpha ,c} S$ given by $\psi (r\delta _s+I') = rF(s)^*\delta _s+I$ . We show that these two homomorphisms are inverse to each other. By symmetry, it suffices to consider $\psi \circ {\varphi }$ . Now, $\psi {\varphi }(r\delta _s+I) = \psi (rF(s)\delta _s+I') = rF(s)F(s)^*\delta _s+I = r1_{ss^*}\delta _s+I$ . But since $s\leq s$ and $c(s,s^*s)= 1_{ss^*}$ by Proposition 3.11(1), it follows that $r\delta _s-r1_{ss^*}\delta _s = r\delta _s-rc(s,s^*s)^*\delta _s\in I$ . Thus $\psi {\varphi }(r\delta _s+I) = r\delta _s+I$ . This completes the proof.

Proof. Let A be the additive subgroup generated by the $r\delta _e+I$ with $r\in R$ and $e\in E(S)$ . Since $r\delta _e\cdot r'\delta _f+I = r(er')c(e,f)\delta _{ef}+I$ , it is clear that R is a subring. Suppose that $r\in R$ . Since the action is nondegenerate, we may write $r=\sum _{i=1}^n r_i$ with $r_i\in R1_{e_i}$ and $e_i\in E(S)$ for $i=1,\ldots , n$ . Define $\rho (r) = \sum _{i=1}^n r_i\delta _{e_i}+I$ . We claim that $\rho $ is a well-defined homomorphism from R to A. To show that $\rho $ is well defined, it suffices to show that if $0=\sum _{i=1}^n r_i$ with $r_i\in R1_{e_i}$ , then $\sum _{i=1}^nr_i\delta _{e_i}\in I$ . We proceed by induction on n, with the case $n=1$ being trivial, as then $r_1=0$ and so $r_1\delta _{e_1}=0\in I$ . Suppose now that the result is true for $n-1$ and $0=\sum _{i=1}^n r_i$ with $r_i\in R1_{e_i}$ and $n\geq 2$ . Then $r_n = -\sum _{i=1}^{n-1}r_i$ and so $r_n=r_n1_{e_n} = -\sum _{i=1}^{n-1}r_i1_{e_n}$ . Note that $r_i-r_i1_{e_n}\in R1_{e_i}$ and $0=\sum _{i=1}^{n-1} (r_i-r_i1_{e_n})$ . By induction, we deduce that $\sum _{i=1}^{n-1} (r_i-r_i1_{e_n})\delta _{e_i}\in I$ , that is,

(4-1) $$ \begin{align} \sum_{i=1}^{n-1}r_i\delta_{e_i}+I = \sum_{i=1}^{n-1}r_i1_{e_n}\delta_{e_i}+I. \end{align} $$

Since $e_ie_n\leq e_i$ , we have $r_i\delta _{e_ie_n}-r_i1_{e_n}\delta _{e_i}\in I$ , as $r_ic(e_i, e_ie_n)^* = r_i1_{e_i}1_{e_n}=r_i1_{e_n}$ by Proposition 3.11(4). On the other hand, since $e_ie_n\leq e_n$ , we have $r_i\delta _{e_ie_n} - r_i1_{e_n}\delta _{e_n}\in I$ as $r_ic(e_n,e_ie_n)^* = r_i1_{e_i}1_{e_n} = r_i1_{e_n}$ . Therefore, $r_i1_{e_n}\delta _{e_i}+I=r_i1_{e_n}\delta _{e_n}+I$ . Thus, we deduce from (4-1) that

$$ \begin{align*} \sum_{i=1}^{n-1}r_i\delta_{e_i}+I = \sum_{i=1}^{n-1}r_i1_{e_n}\delta_{e_n}+I = -r_n\delta_{e_n}+I. \end{align*} $$

It follows that $\sum _{i=1}^nr_i\delta _{e_i}\in I$ , as required.

To verify that $\rho $ is a homomorphism, let $r=\sum _{i=1}^n r_i$ and $r'=\sum _{j=1}^m r^{\prime }_j$ , where $r_i\in R1_{e_i}$ and $r_j\in R1_{f_j}$ with $e_i, f_j\in E(S)$ for $1\leq i\leq n$ , $1\leq j\leq m$ . Then $rr' = \sum _{i=1}^n\sum _{j=1}^m r_ir^{\prime }_j$ and $r_ir^{\prime }_j\in R1_{e_if_j}$ . Therefore, $\rho (rr') = \sum _{i=1}^n\sum _{j=1}^m r_ir^{\prime }_j\delta _{e_if_j}+I$ . On the other hand,

$$ \begin{align*} \rho(r)\rho(r') =\sum_{i=1}^nr_i\delta_{e_i}\sum_{j=1}^mr_j\delta_{f_j}+I=\sum_{i=1}^n\sum_{j=1}^m r_i(e_ir_j)c(e_i,f_j)\delta_{e_if_j}+I. \end{align*} $$

But $e_ir_j = r_j1_{e_i}$ and $c(e_i,f_j) =1_{e_if_j}$ by Proposition 3.11(4) and so

$$ \begin{align*} r_i(e_ir_j)c(e_i,f_j)=r_ir_j1_{e_i}1_{e_if_j} = r_ir_j. \end{align*} $$

Therefore, $\rho $ is a homomorphism.

To check $\rho $ is surjective, notice that, since $e\leq e$ , we have $r\delta _e+I = rc(e,e)^*\delta _e+I = r1_e\delta _e+I = \rho (r1_e)$ . Suppose that $\tau $ , as in the final statement, exists. If $r\in R$ with $r=\sum _{i=1}^n r_i$ and $r_i\in R1_{e_i}$ , for $i=1,\ldots , n$ , then $\tau ({\kern1pt}\rho (r)) = \tau (\sum _{i=1}^n r_i\delta _{e_i}+I)=\sum _{i=1}^n r_i = r$ and so $\rho $ is injective.