Proof. (1) Suppose that $A \cdot B = A \cdot C$ , where $A,B,C$ are prime filters. We prove that $B = C$ . The dual result follows by symmetry. In fact, we prove $B \subseteq C$ . The reverse inclusion then follows by symmetry. Let $b \in B$ . Then, $b^{+} \in \mathbf {r}(B)$ . So, $b^{+} \in \mathbf {d}(A)$ , since $A \cdot B$ is defined in the category. There is therefore $a \in A$ such that $a^{\ast } \leq b^{+}$ . However, $ab \in A \cdot B$ and so $ab \in A \cdot C$ . Therefore, $a_{1}c \leq ab$ , where $a_{1} \in A$ and $c \in C$ . Since $a,a_{1} \in A$ , there exists $a_{2} \leq a,a_{1}$ , where $a_{2} \in A$ . Thus, $a_{2}c \leq ab$ . However, $a_{2} = aa_{2}^{\ast }$ . Thus, $aa_{2}^{\ast }c = ab(a_{2}c)^{\ast }$ . We are working in an ample semigroup, so $a^{\ast }a_{2}^{\ast }c = a^{\ast }b(a_{2}c)^{\ast }$ . Now, $a^{\ast }, a_{2}^{\ast } \in \mathbf {r}(C)$ . Thus, $a^{\ast }a_{2}^{\ast }c \in \mathbf {r}(C)\mathbf {r}(C)C \subseteq C$ . It follows that $a^{\ast }b(a_{2}c)^{\ast } \in C$ . Thus, $b \in C$ , as required.

(2) It is enough to check that Axiom (A7) holds. Suppose that $AB = AC$ , where $A,B,C$ are compact-open local bisections. We prove that $A^{\ast }B = A^{\ast }C$ . Let $\mathbf {d}(a)b \in A^{\ast }B$ , where $a \in A$ . Then, $ab \in AB$ . By assumption, $ab = a_{1}c_{1}$ , where $a_{1} \in A$ and $c_{1} \in C$ . Clearly, $\mathbf {r}(a) = \mathbf {r}(a_{1})$ . However, A is a local bisection. Thus, $a = a_{1}$ . Therefore, $ab = ac_{1}$ . The category C is cancellative; thus, $b = c_{1}$ . It follows that $\mathbf {d}(a)b \in A^{\ast }C$ , as required.

The proofs of parts (3) and (4) may be found in [Reference Lawson7, Ch. 9].

Proof. Let U be any nonempty open set in C and let $a \in U$ . Then, there is an open neighbourhood V of a on which $\mathbf {d}$ is a homeomorphism. Likewise, there is an open neighbourhood W of a on which $\mathbf {r}$ is a homeomorphism. The set $U \cap V \cap W$ is an open set containing a inside U, which is a local bisection. This proves that the open local bisections form a basis for C.

Proof. Suppose that A is an open local section and that $\mathbf {d}(A)$ is compact. We prove that A is compact. Suppose that $A \subseteq \bigcup _{i \in I}U_{i}$ , where $U_{i}$ are open sets of C. Then, $A = \bigcup _{i \in I}U_{i} \cap A$ . The sets $U_{i} \cap A$ are open and they are local bisections since they are contained in a local bisection. It follows that $\mathbf {d}(A) = \bigcup _{i \in I}\mathbf {d}(U_{i} \cap A)$ . By assumption, $\mathbf {d}(A)$ is compact. We may therefore write $\mathbf {d}(A) = \bigcup _{j = 1}^{m} \mathbf {d}(U_{j} \cap A)$ , relabelling if necessary. We claim that $A \subseteq \bigcup _{i = 1}^{m}U_{i}$ . Let $x \in A$ . Then, $\mathbf {d}(x) \in \mathbf {d}(A)$ and so $\mathbf {d}(x) \in \mathbf {d}(U_{j} \cap A)$ for some j. However, $U_{j} \cap A$ is a local section. Thus, there is a unique element $y \in U_{j} \cap A$ such that $\mathbf {d}(y) = \mathbf {d}(x)$ . However, $x,y \in A$ , which is a local bisection. It follows that $x = y$ and so $x \in U_{j} \cap A \subseteq U_{j}$ . The proof of the converse follows from the fact that C is étale

Proof. By Lemma 3.3, it is enough to consider the case where U is an open local bisection in C. Let $x \in U$ . There is an open set V containing x such that V and $\mathbf {d}(V)$ are homeomorphic by $\mathbf {d}$ . Thus, $U \cap V$ contains a and is homeomorphic to $\mathbf {d}(U \cap V)$ under $\mathbf {d}$ . However, the clopen subsets of the space of identities form a basis for the topology on the space of identities. Thus, there is an open set $Y \subseteq U \cap V$ containing x such that $\mathbf {d}(Y)$ is clopen. However, the closed subsets of Hausdorff spaces are compact and so Y is compact by Lemma 3.4.

Proof. We regard C as a subcategory of G in what follows to ease notation. We use the fact that a basis $\beta $ for the topology in C consists of open local bisections.

An element g of G can be written $a^{-1}b$ , where $a,b \in C$ . By assumption, $a \in U \in \beta $ and $b \in V \in \beta $ for some U and V. It follows that $g \in U^{-1}V$ . Thus, every element of G is in some element of $\beta ^{-1} \beta $ .

Suppose that $g \in U_{1}^{-1}V_{1} \cap U_{2}^{-1}V_{2}$ , where $U_{1}, U_{2}, V_{1}, V_{2} \in \beta $ . Then, $g = a_{1}^{-1}b_{1} = a_{2}^{-1}b_{2}$ , where $a_{1} \in U_{1}$ , $a_{2} \in U_{2}$ , $b_{1} \in V_{1}$ and $b_{2} \in V_{2}$ . Observe that $\mathbf {d}(b_{1}) = \mathbf {d}(b_{2})$ and $\mathbf {d}(a_{1}) = \mathbf {d}(a_{2})$ . Since C is right reversible, we may find elements $a,a' \in C$ such that $ab_{1} = a'b_{2} = r$ (say) and $aa_{1} = a'a_{2} = s$ (say). (That the elements $a,a'$ are the same in each case follows from the two different ways that g can be written). Choose $A,B \in \beta $ such that A contains a and B contains $a'$ . Using the fact that the product of open local bisections is an open local bisection, $AV_{1} \cap BV_{2} = X$ is an open local bisection containing r and $AU_{1} \cap BU_{2} = Y$ is an open local bisection containing s. Thus, $X, Y \in \beta $ . Observe that $s^{-1}r \in Y^{-1}X$ , which simplifies to g. We have shown that $g \in Y^{-1}X$ . Now, let $h \in Y^{-1}X$ be any element. Then, we can write $h = (au)^{-1}(a_{1}v)$ , where $a \in A$ , $u \in U_{1}$ , $a_{1} \in A$ and $v \in V_{1}$ . Thus, $h = u^{-1}a^{-1}a_{1}v$ . However, $a,a_{1} \in A$ , a local bisection, and $\mathbf {r}(a) = \mathbf {r}(a_{1})$ . It follows that $a = a_{1}$ . We deduce that $h \in U_{1}^{-1}V_{1}$ . We may similarly show that $h \in U_{2}^{-1}V_{2}$ . We have therefore proved that $g \in Y^{-1}X \subseteq U_{1}^{-1}V_{1} \cap U_{2}^{-1}V_{2}$ .

We have therefore proved that $\beta ^{-1}\beta $ really is a basis.

We now prove that each element of $\beta ^{-1}\beta $ is a local bisection. Let $g,h \in U^{-1}V$ such that $\mathbf {d}(g) = \mathbf {d}(h)$ . We may write $g = u_{1}^{-1}v_{1}$ and $h = u_{2}^{-1}v_{2}$ . Thus, $\mathbf {d}(v_{1}) = \mathbf {d}(v_{2})$ . It follows that $v_{1} = v_{2}$ . From this, we deduce that $\mathbf {r}(u_{1}) = \mathbf {r}(u_{2})$ and so $u_{1} = u_{2}$ . It follows that $g = h$ . Symmetry delivers the result.

Proof. We proved in Lemma 3.7 that $\beta ^{-1} \beta $ is a basis for a topology on G. However, the topology on G has, as a basis, local bisections. From this, it follows that G will be étale. It is clear that inversion is a homeomorphism. It remains to show that the partial multiplication in G is continuous. To that end, let $g_{1}h_{1} \in U^{-1}V$ be any element where $U,V \in \beta $ . Let $g_{1} = a_{1}^{-1}b_{1}$ and $h_{1} = a_{2}^{-1}b_{2}$ , where $a_{1},b_{1},a_{2},b_{2} \in C$ . By right reversibilty, let p and q be such that $pb_{1} = qa_{2} = r$ . Then, $g_{1}h_{1} = (pa_{1})^{-1}(qb_{2})$ . By assumption, $g_{1}h_{1} = u^{-1}v$ , where $u \in U$ and $v \in V$ . From the fact that we are dealing with a groupoid of fractions, it follows that there are elements $a,b \in C$ such that $apa_{1} = bu$ and $aqb_{2} =~bv$ . Let Z and B be open local bisections of C such that $apb_{1} \in Z$ and $b \in B$ . Consider, first, the set $(BU)^{-1}Z$ . We have $apa_{1} \in BU$ . Thus, $a_{1}^{-1}p^{-1}a^{-1} \in (BU)^{-1}$ . By definition, $apb_{1} \in Z$ . It follows that $g_{1} \in (BU)^{-1}Z$ . Now, we consider $Z^{-1}BV$ . By definition, $apb_{1} \in Z$ . However, $pb_{1} = qa_{2}$ . Thus, $aqa_{2} \in Z$ . Whence, $a_{2}^{-1}q^{-1}a^{-1} \in Z^{-1}$ . However, $aqb_{2} \in BV$ . It follows that $h_{1} \in Z^{-1}BV$ . We now calculate the product $[(BU)^{-1}Z][Z^{-1}BV]$ . This is contained in $U^{-1}V$ because if X is a local bisection, both $XX^{-1}$ and $X^{-1}X$ consist of identities. This is enough to show that the partial multiplication in the groupoid G is continuous.

It only remains to show that C is an open subset of G. The set of identities $C_{o}$ is an open set in C. It is therefore an open local bisection. Thus, $\beta \subseteq \beta \beta ^{-1}$ . This means that C is an open subset of its groupoid of fractions.

Proof of Theorem 3.6.

Let C be a Boolean cancellative right reversible category whose groupoid of fractions is G. We have seen that G is an étale category. It is Boolean because C is a wide subcategory of G. Because C is open in G, it follows that a compact subset of C is also compact in G. Let $\beta $ be a basis for the topology on C consisting of all compact-open local bisections of C. The elements of $\beta ^{-1}\beta $ are compact-open local bisections of G. It only remains to show that $\beta ^{-1}\beta $ is a basis for the topology on G. We have proved that a basis for the topology on G consists of the subsets of the form $U^{-1}V$ , where U and V are open local bisections of C. However, $\beta $ is a basis for the topology on C. Thus, U is a union of elements of $\beta $ . Likewise, V is a union of elements of $\beta $ . It follows that $U^{-1}V$ is a union of elements of $\beta ^{-1}\beta $ . This proves that $\beta ^{-1}\beta $ is a basis for the topology on G.

Proof. It is immediate that C is cancellative and, because C is open in G, it inherits the main features of the topology of G. Thus, C is an étale topological category. Let X be an open local bisection of G containing the element g. Let $g = a^{-1}b$ , where $a,b \in C$ . Let A be any open local bisection of C that contains a. Then, $b \in AX$ . Thus, $Y = AX \cap C$ is nonempty, since it contains b, and is an open local bisection of C. Thus, $g \in A^{-1}Y$ . We prove that $A^{-1}Y \subseteq X$ . Let $h \in A^{-1}Y$ . Then, $h = a_{1}^{-1}a_{2}x$ , where $a_{1},a_{2} \in A$ and $x \in X$ . However, A is a local bisection and so $a_{1} = a_{2}$ . It follows that $h \in X$ . We have proved that each open local bisection of G is a union of open local bisections of the form $U^{-1}V$ , where U and V are open local bisections of C.

Proof. Suppose first that $\mathsf {C}(S)$ is right reversible. We show that condition (C) holds. Let $F \subseteq ~\mathsf {E}(S)$ be a prime filter and let $a, b \in S$ such that $a^{\ast }b^{\ast } \in F$ . Put $A = (aF)^{\uparrow }$ and $B = (bF)^{\uparrow }$ . Both A and B are prime filters with $\mathbf {d}(A) = F^{\uparrow } = \mathbf {d}(B)$ . Thus, by right reversibility, there are prime filters C and D such that $\exists C \cdot A$ and $\exists B \cdot D$ and $C \cdot A = D \cdot B$ . We may write $C = (c \mathbf {r}(A))^{\uparrow }$ , where $c^{\ast } \in \mathbf {r}(A)$ , and $D = (d \mathbf {r}(B))^{\uparrow }$ , where $d^{\ast } \in \mathbf {r}(B)$ . Then, $\mathbf {r}(A) = ((aF)^{+})^{\uparrow }$ and $\mathbf {r}(B) = ((bF)^{+})^{\uparrow }$ . It follows that $(ae)^{+} \leq c^{\ast }$ for some $e \in F$ and $(bf)^{+} \leq d^{\ast }$ for some $f \in F$ . We may therefore put $a_{1} = ae$ and $b_{1} = bf$ . We can write $C \cdot A = (caF)^{\uparrow }$ and $D \cdot B = (dbF)^{\uparrow }$ . Since $\mathbf {d}(C \cdot A) = \mathbf {d}(D \cdot B)$ , the equality $C \cdot A = D \cdot B$ follows from the fact that $C \cdot A \cap D \cdot B \neq \varnothing $ . This means that $cae = dbe$ for some $e \in F$ .

We now prove the converse. Suppose that condition (C) holds. We prove that $\mathsf {C}(S)$ is right reversible. Let A and B be prime filters such that $\mathbf {d}(A) = \mathbf {d}(B)$ . Then, $A = (aF)^{\uparrow }$ , where $a^{\ast } \in F$ , and $B = (bF)^{\uparrow }$ , where $b^{\ast } \in F$ . We now invoke condition (C). There are elements c and d having various properties. Put $C = (c\mathbf {r}(A))^{\uparrow }$ and $D = (d\mathbf {r}(B))^{\uparrow }$ . Now, $\mathbf {r}(A) = ((aF)^{+})^{\uparrow }$ . However, $a_{1} = ae$ , where $e \in F$ . Thus, $a_{1}^{+} \in \mathbf {r}(A)$ . It follows that $c^{\ast } \in \mathbf {r}(A)$ . Thus, $c \in C$ . Similarly, $d \in D$ . Thus, $C \cdot A = (ca \mathbf {d}(A))^{\uparrow }$ and $D \cdot B = (db\mathbf {d}(A))$ . There is some $e \in F$ such that $cae = dbe$ . It follows that $C \cdot A \cap D \cdot B \neq \varnothing $ . Thus, in fact, $C \cdot A = D \cdot B$ and so $\mathsf {C}(S)$ is right reversible.

Proof. Because $a^{\ast }b^{\ast } \neq 0$ , there is a prime filter F in $\mathsf {E}(S)$ containing $a^{\ast }b^{\ast }$ . Invoking condition (C), there are elements c and d, and an idempotent $e \in F$ such that $cae = dbe$ . We now apply Axiom (A6) to get $(c(ae)^{+})a = (d(be)^{+})b$ , which is nonzero since it lives in a prime filter.

Proof. We define first a functor $\iota \colon \mathsf {C}(S) \rightarrow \mathsf {C}(T)$ , which is an embedding where $\iota (\mathsf {C}(S))$ is a wide subcategory of $\mathsf {C}(T)$ . This will depend only on the fact that S is a full submonoid of T. This will involve comparing prime filters in S with prime filters in T. Let A be a prime filter of S. Then,

$$ \begin{align*}A^{u} = \{t \in T \colon a \leq t \text{ for some } a \in A\}\end{align*} $$

is a prime filter in T that contains elements of S. Conversely, let B be a prime filter in T that contains elements of S. Then,

$$ \begin{align*}B^{d} = B \cap S\end{align*} $$

is a prime filter in S. Observe that $(A^{u})^{d} = A$ and $(B^{d})^{u} = B$ . We have therefore defined a bijection between the prime filters in S and the prime filters in T that contain elements of S. We therefore define the functor $\iota $ by $A \mapsto A^{u}$ . We may now prove the proposition.

Suppose that each nonzero element of T can be written as a finite join of elements of the form $a^{-1}b$ , where $a,b \in S$ . Let X be any prime filter of T. Let $t \in X$ be any element, necessarily nonzero. Then, from the fact that we are dealing with a prime filter and given how t can be written, there exists $a^{-1}b \in X$ for some $a,b \in S$ . Then, $X = A \cdot B$ , where $a^{-1} \in A$ and $b \in B$ . We may therefore write $X = ((A)^{-1})^{-1} \cdot B$ , where $A = (a^{-1}\mathbf {r}(B))^{\uparrow }$ and $B =(b \mathbf {d}(X))^{\uparrow }$ . Observe that $a \in A^{-1}$ and so both $A^{-1}$ and B contain elements of S. It follows that both $A^{-1}$ and B are in the image of $\iota $ .

To prove the converse, let t be any nonzero element of T. By assumption, any prime filter X of T containing t must have the form $A^{-1} \cdot B$ , where A and B are prime filters of T that have nonempty intersections with S. Let $a \in A \cap S$ and let $b \in B \cap S$ . Thus, $a^{-1}b \in A^{-1} \cdot B = X$ . Now, X is a prime filter that contains both t and $a^{-1}b$ . However, any element less than $a^{-1}b$ has the form $a_{1}^{-1}b_{1}$ , where $a_{1}, b_{1} \in S$ ; this follows from the fact that S is an order ideal of T. Whence, $X_{t} = \bigcup _{a^{-1}b \leq t, a,b \in S} X_{a^{-1}b}$ . By compactness, $t = \bigvee _{i=1}^{n}a_{i}^{-1}b_{i}$ , where $a_{i},b_{i} \in S$ .

Proof. Let S be a Boolean ample monoid satisfying condition (C). By Theorem 2.1 and part (1) of Proposition 2.4, the category $\mathsf {C}(S)$ is cancellative. By Proposition 4.1, the cancellative category $\mathsf {C}(S)$ is right reversible because condition (C) holds. Thus, by Theorem 3.1, there is a groupoid G which contains $\mathsf {C}(S)$ as a wide subcategory. However, by Theorem 3.6, G is, in fact, a Boolean groupoid. Thus, $\mathsf {C}(S) \subseteq G$ as an open and wide subcategory; we proved openness in Proposition 3.8. It follows that every compact-open local bisection of $\mathsf {C}(S)$ is a compact-open local bisection of G. Thus, $\mathsf {KB}(\mathsf {C}(S))$ is a full submonoid of $\mathsf {KB}(G)$ ; the fact that we have a full submonoid is a result of the fact that the category $\mathsf {C}(S)$ is wide in G. By part (1) of Theorem 2.2, S is isomorphic to $\mathsf {KB}(\mathsf {C}(S))$ and we identify these two monoids. By part (4) of Proposition 2.4, $\mathsf {KB}(G)$ is a Boolean inverse monoid T, say. We have therefore shown that S is a full submonoid of the Boolean inverse monoid T. Finally, by Proposition 4.3, every nonzero element of T is a finite join of elements of the form $a^{-1}b$ , where $a,b \in S$ .