Proof. Suppose that $a \sim _{l} b$ . Then $ab^{-1}b = ba^{-1}a$ by definition. But then $ab^{-1}= ba^{-1}ab^{-1}$ , which is an idempotent. Conversely, suppose that $ab^{-1}$ is an idempotent. Then $ab^{-1}b, ba^{-1}a \leq a,b$ . But $(ab^{-1}b)^{\ast } = (ba^{-1}a)^{\ast }$ . Thus, $ab^{-1}b = ba^{-1}a$ by Lemma 2.5.

Proof. (1) We are given that $a \sim _{l} b$ and $c \sim _{l} d$ . This means that $ab^{\ast } = ba^{\ast }$ and ${cd^{\ast } = dc^{\ast }}$ . Thus, $ab^{\ast }cd^{\ast } = ba^{\ast }dc^{\ast }$ . By Axiom (RR6), we have that $ac(bc)^{\ast }d^{\ast } = bd(ad)^{\ast }c^{\ast }$ . We focus on the left-hand side of the above equality. By Axiom (RR2), we have that $ac(bc)^{\ast }d^{\ast } = ac [(bc)^{\ast }d^{\ast }]^{\ast }$ . By Axiom (RR5), we have that $ac [(bc)^{\ast }d^{\ast }]^{\ast } = ac[bcd^{\ast }]^{\ast }$ . But $cd^{\ast } = dc^{\ast }$ . Thus, $ac[bcd^{\ast }]^{\ast } = ac[bdc^{\ast }]^{\ast }$ . Apply Axiom (RR5) to get $ac[bdc^{\ast }]^{\ast } = ac[(bd)^{\ast }c^{\ast }]^{\ast }$ . By Axiom (RR2), this is just $ac(bd)^{\ast }c^{\ast }$ . We now use Axiom (RR3) and Axiom (RR4) to get $ac(bd)^{\ast }c^{\ast } = ac (bd)^{\ast }$ . We have proved that $ac(bc)^{\ast }d^{\ast } = ac (bd)^{\ast }$ . We may similarly show that $bd(ad)^{\ast }c^{\ast } = bd(ac)^{\ast }$ . The result now follows.

(2) We have that $ab^{\ast } = ba^{\ast }$ and $x = ax^{\ast }$ and $y = by^{\ast }$ . We have that $xy^{\ast } = ax^{\ast }b^{\ast }y^{\ast } = ba^{\ast }x^{\ast }y^{\ast }$ and $yx^{\ast } = by^{\ast }a^{\ast }x^{\ast }$ . Since projections commute, we have shown that $x \sim _{l} y$ .

Proof. Suppose first that $a \sim _{l} b$ . Put $x = ab^{\ast } = ba^{\ast }$ . Clearly, $x \leq a,b$ and $x^{\ast } = a^{\ast }b^{\ast }$ . Suppose that $z \leq a,b$ . Then $z^{\ast } \leq a^{\ast }, b^{\ast }$ . By the definition of the natural partial order, we have that $z = az^{\ast } = bz^{\ast } \leq ab^{\ast } = x$ . We now prove the converse. We have that $a \wedge b \leq a$ and so $a \wedge b = a(a \wedge b)^{\ast } = ab^{\ast }$ . By symmetry, we have that $a \wedge b = ba^{\ast }$ . It follows that $ab^{\ast } = ba^{\ast }$ and so a and b are left compatible.

Proof. We have that $xa = ya$ . Thus, $xay = yay$ . It follows that $y = xy^{\ast }$ and so $y \leq x$ . By symmetry, $x \leq y$ and so $x = y$ .

Proof. (1) We have that $aa^{-1} = (a^{-1})^{\ast }$ and $a^{-1}a = a^{\ast }$ and $bb^{-1} = (b^{-1})^{\ast }$ and ${b^{-1}b = b^{\ast }}$ . Observe that $ab(b^{-1}a^{-1}) =(b^{-1}a^{-1})^{\ast }$ and $(b^{-1}a^{-1})ab = (ab)^{\ast }$ .

(2) Immediate from (1) above.

(3) Only one direction needs proving. Suppose that $a \leq b$ . Then $a = ba^{\ast }$ . Now, a has inverse $a^{-1}$ but $a^{\ast }b^{-1}$ is also an inverse. So, by the uniqueness of inverses guaranteed by Lemma 2.12, we have that $a^{-1} = a^{\ast }b^{-1}$ . Thus, $a^{-1} \leq b^{-1}$ and so $a^{-1} = b^{-1}(a^{-1})^{\ast }$ from the definition of the natural partial order. Taking inverses of both sides again, we get that $a = (a^{-1})^{\ast }b$ and so $a = aa^{-1}b$ , as required.

(4) By (1) above, it follows that $\mathsf {Inv}(S)$ is closed under products. Suppose now that $a \in \mathsf {Inv}(S)$ is an idempotent. Let $b \in S$ be such that $ab = b^{\ast }$ , $ba = a^{\ast }$ . Since a is an idempotent, we have that $a = (aba)(aba) = a(ba)(ab)a$ . But projections commute. Thus, $a = (a^{2}b)(ba^{2}) = (ab)(ba)$ . It follows that a is the product of two projections and so is itself a projection. Since the projections commute, we have proved that $\mathsf {Inv}(S)$ is an inverse semigroup.

(5) Suppose that $a \in \mathsf {Inv}(S)$ and that $b \leq a$ . We prove that $b \in \mathsf {Inv}(S)$ . By definition, $b = ab^{\ast }$ . Observe that $b^{\ast }a^{-1}b = a^{\ast }b^{\ast } = (ab^{\ast })^{\ast }$ and $bb^{\ast }a^{-1} = ab^{\ast }a^{-1} = (a^{-1})(b^{\ast }a^{-1})^{\ast } = (b^{\ast }a^{-1})^{\ast }$ . This proves that $b \in \mathsf {Inv}(S)$ .

Proof. It is enough to prove that if a and b are partial units that are compatible then $a \vee b$ is a partial unit. Since a and b are compatible, so too are $a^{-1}$ and $b^{-1}$ by Lemma 2.7. It follows that the element $a^{-1} \vee b^{-1}$ is defined. We calculate ${(a \vee b)(a^{-1} \vee b^{-1})}$ . This equals $aa^{-1} \vee ab^{-1} \vee ba^{-1} \vee bb^{-1}$ . By assumption, both $ab^{-1}$ and $ba^{-1}$ are idempotents. Observe that $ab^{-1} \leq aa^{-1}$ and $ba^{-1} \leq bb^{-1}$ Thus, ${(a \vee b)(a^{-1} \vee b^{-1}) = aa^{-1} \vee bb^{-1}}$ . This is a join of projections and so is a projection by Lemma 3.1. Dually, we have that $(a^{-1} \vee b^{-1})(a \vee b)$ is a projection itself.

Proof. It is immediate from the definition, that $\theta $ maps partial units to partial units. Since $\theta $ preserves left-compatible joins, it preserves compatible joins.

Proof. We prove the case $n = 2$ directly. Suppose that $x \sim _{l} y$ . We prove first that ${x \vee y = x \vee y\overline {x^{\ast }}}$ . This follows from the fact that $x^{\ast } \vee \overline {x^\ast } = 1$ and that $yx^{\ast } = xy^{\ast }$ . It is clear that x and $y \overline {x^{\ast }}$ are left-orthogonal. We now prove the general case by induction. Let $a = \bigvee _{i=1}^{m+1} a_{i}$ be a left-compatible join. We can write this as

$$ \begin{align*} a = \bigg( \bigvee_{i=1}^{m} a_{i} \bigg) \vee a_{m+1}. \end{align*} $$

By induction, this is

$$ \begin{align*} a = \bigg( \bigvee_{i=1}^{m} b_{i} \bigg) \vee a_{m+1} \end{align*} $$

where $b_{i} \leq a_{i}$ and the join in the parentheses is a left-orthogonal one. Put $e = \bigvee _{i=1}^{m} b_{i}^{\ast }$ where we have used part (2) of Lemma 3.1. Then

$$ \begin{align*} a = \bigg( \bigvee_{i=1}^{m} b_{i} \bigg) \vee a_{m+1}\overline{e} \end{align*} $$

is a left-orthogonal join.

Proof. We first show that $\mathsf {R}(S)$ is a right restriction monoid. Using parts (1) and (3) of Lemma 4.1, the set $\mathsf {R}(S)$ is a semigroup under subset multiplication. The set of all order ideals of $\mathsf {Proj}(S)$ is a set of idempotents (later, projections) for $\mathsf {R}(S)$ . Observe that the set of all projections, $1^{\downarrow }$ , is an identity for $\mathsf {R}(S)$ , which is therefore a monoid. If we define a unary map on $\mathsf {R}(S)$ by $A \mapsto A^{\ast }$ then this is well defined by part (2) of Lemma 4.1. It remains to check that $\mathsf {R}(S)$ is a right restriction monoid with respect to these operations. The proofs of Axioms (RR1), (RR2) and (RR3) are immediate. To prove that Axiom (RR4) holds, it is immediate that $A \subseteq AA^{\ast }$ . The proof that the reverse inclusion holds follows from the fact that A is an order ideal. The proof that Axiom (RR5) holds is immediate. It remains to show that Axiom (RR6) holds. It is immediate that $A^{\ast }B \subseteq B(AB)^{\ast }$ by Axiom (RR6), which holds in S. We now prove the reverse inclusion. Let $b(ab_{1})^{\ast }$ be such that $b,b_{1} \in B$ and $a \in A$ . We prove that this is an element of $A^{\ast }B$ . Because $b,b_{1} \in B$ , an acceptable set, we have that $bb_{1}^{\ast } = b_{1}b^{\ast }$ . We have that

$$ \begin{align*}b(ab_{1})^{\ast} = b(ab_{1})^{\ast}b_{1}^{\ast} = bb_{1}^{\ast}(ab_{1})^{\ast} = b_{1}b^{\ast}(ab_{1})^{\ast} = b_{1}(ab_{1})^{\ast}b^{\ast}.\end{align*} $$

Thus,

$$ \begin{align*}b(ab_{1})^{\ast} = b_{1}(ab_{1})^{\ast}b^{\ast} = a^{\ast}b_{1}b^{\ast}.\end{align*} $$

But $b_{1}b^{\ast } \in B$ because B is an order ideal. We have therefore proved that $B(AB)^{\ast } \subseteq A^{\ast }B$ . This completes the proof that $\mathsf {R}(S)$ is a right restriction monoid.

• • Claim. If A and B are acceptable sets then $A \leq B$ in $\mathsf {R}(S)$ if and only if $A \subseteq B$ . To prove the claim, suppose first that $A \leq B$ . By definition $A = BA^{\ast }$ . Let $a \in A$ . Then $a = bc^{\ast }$ where $b \in B$ and $c \in A$ . But $a \leq b$ and $b \in B$ , an order ideal. It follows that $a \in B$ . We have proved that $A \subseteq B$ . We now prove the converse. Suppose that $A \subseteq B$ . We prove that $A = BA^{\ast }$ . Observe that $A \subseteq BA^{\ast }$ . Let $x \in BA^{\ast }$ . Then $x = bc^{\ast }$ where $c \in A$ . We have that $A \subseteq B$ and so $b \sim _{l} c$ . It follows that $bc^{\ast } = cb^{\ast }$ . Thus, $x = cb^{\ast }$ and so $x \leq c$ . But $c \in A$ and A is an order ideal and so $c \in A$ , as required.

• • Claim. If A and B are acceptable sets then $A \sim _{l} B$ in $\mathsf {R}(S)$ if and only if $A \cup B \in \mathsf {R}(S)$ . To prove the claim, suppose first that $A \cup B \in \mathsf {R}(S)$ . We prove that $A \sim _{l} B$ . In fact, we prove that $AB^{\ast } \subseteq BA^{\ast }$ and then appeal to symmetry. Let $ab^{\ast } \in AB^{\ast }$ . By assumption, $A \cup B$ is an acceptable set and so, in particular, $ab^{\ast } = ba^{\ast }$ . It follows that $ab^{\ast } \in BA^{\ast }$ . Suppose now that $A \sim _{l} B$ . We prove that $A \cup B \in \mathsf {R}(S)$ . It is enough to prove that if $a \in A$ and $b \in B$ then $a \sim _{l} b$ . We are given that $AB^{\ast } = BA^{\ast }$ . We have that $ab^{\ast } \in AB^{\ast }$ and so $ab^{\ast } = b_{1}a_{1}^{\ast }$ where $b_{1} \in B$ and $a_{1} \in A$ . By assumption, $a \sim _{l} a_{1}$ and $b \sim _{l} b_{1}$ . We claim that $ab^{\ast } \leq ba^{\ast }$ and symmetry delivers the result. To prove the claim, we use the fact that $ab^{\ast } = b_{1}a_{1}^{\ast }$ . Thus,

  $$ \begin{align*}ab^{\ast} = b_{1}a_{1}^{\ast}b^{\ast}a^{\ast} = b_{1}b^{\ast}a_{1}^{\ast}a^{\ast} = bb_{1}^{\ast}(a_{1}^{\ast}a^{\ast}).\end{align*} $$

With these two results, we can now prove that $\mathsf {R}(S)$ is a complete right restriction monoid. Let $\{A_{i} \colon i \in I\}$ be a left-compatible subset of $\mathsf {R}(S)$ . We claim that ${A = \bigcup _{i \in I} A_{i} \in \mathsf {R}(S)}$ . We therefore have to prove that A is acceptable. It is clearly an order ideal and so we have to show that any two elements of A are left-compatible. Without loss of generality, suppose that $a \in A_{i}$ and $b \in A_{j}$ . Then, by the above, $A_{i} \cup A_{j}$ is acceptable and so a and b are left-compatible. We have proved that A is an acceptable set. Also, by what we proved above, we have that $A_{i} \leq A$ for any $i \in I$ . Suppose that $A_{i} \leq B$ for any $i \in I$ where B is acceptable. Then $A_{i} \subseteq B$ for any $i \in I$ , by what we proved above. Thus, $A \subseteq B$ and so $A \leq B$ . We have therefore proved that $\mathsf {R}(S)$ has joins of left-compatible subsets. Now let $\{A_{i} \colon i \in I\}$ be a left-compatible subset of $\mathsf {R}(S)$ and let B be any acceptable set. We have to prove that $( \bigcup _{i \in I} A_{i} )B = \bigcup _{i \in I} A_{i}B$ . However, this is true on set-theoretic grounds alone. This completes the proof that $\mathsf {R}(S)$ is a complete right restriction monoid.

It remains to prove that the function $\iota $ is a homomorphism of right restriction monoids. It is immediate that it is an embedding. Observe that $a^{\downarrow }b^{\downarrow } = (ab)^{\downarrow }$ ; this is true since if $a' \leq a$ and $b' \leq b$ then $a'b' \leq ab$ and if $x \leq ab$ then $x = a(bx^{\ast }) = a'b'$ where $a' = a$ and $b'= bx^{\ast } \leq b$ . We therefore have a homomorphism of semigroups that is also a monoid homomorphism since the identity of $\mathsf {R}(S)$ is $1^{\downarrow }$ . Thus, we finish if we show that $\iota $ is a homomorphism of right restriction monoids. This requires us to show that $\iota (a^{\ast }) = \iota (a)^{\ast }$ . Let $x \in \iota (a^{\ast })$ . Then $x \leq a^{\ast }$ and so is a projection. Consider the element $ax$ . Because x is a projection, we have that $ax \leq a$ and so $ax \in \iota (a)$ . But $(ax)^{\ast } = a^{\ast }x = x$ . It follows that $x \in \iota (a)^{\ast }$ . On the other hand, let $b^{\ast } \in \iota (a)^{\ast }$ where $b \leq a$ . It follows that $b^{\ast } \leq a^{\ast }$ and so $b^{\ast } \in \iota (a^{\ast })$ .

Proof. Let T be a complete right restriction monoid and let $\alpha \colon S \rightarrow T$ be a monoid homomorphism of right restriction monoids. Define $\beta \colon \mathsf {R}(S) \rightarrow T$ by $\beta (A) = \bigvee _{a \in A} \alpha (a)$ . This makes sense since the elements of A are pairwise left-compatible and left-compatibility is preserved by homomorphisms of right restriction semigroups. We now calculate $\beta (\iota (a))$ . By definition, this is $\beta (a^{\downarrow })$ , which is $\bigvee _{x \leq a} \alpha (x)$ . Observe that $x \leq a$ implies that $\alpha (x) \leq \alpha (a)$ . It follows that $\beta (\iota (a)) = \alpha (a)$ . We show that $\beta $ is a right restriction monoid homomorphism. It is immediate from the definitions that this is a monoid homomorphism. We need to prove that it is a homomorphism of right restriction semigroups. Let A be an acceptable set. Then, by definition,

$$ \begin{align*}\beta (A^{\ast}) = \bigvee_{a^{\ast} \in A^{\ast}} \alpha (a^{\ast}).\end{align*} $$

But $\alpha $ is a homomorphism of right restriction semigroups. Thus, $\alpha (a^{\ast }) = \alpha (a)^{\ast }$ . Now apply part (2) of Lemma 3.1 to get

$$ \begin{align*} \beta (A^{\ast}) = \bigg(\bigvee_{a^{\ast} \in A^{\ast}} \alpha (a) \bigg)^{\ast}. \end{align*} $$

We now use the fact that $a \in A$ if and only if $a^{\ast } \in A^{\ast }$ . This gives us

$$ \begin{align*} \beta (A^{\ast}) = \bigg( \bigvee_{a \in A} \alpha (a) \bigg)^{\ast} = \beta (A)^{\ast}. \end{align*} $$

We have therefore shown that $\beta $ is a homomorphism of right restriction semigroups. We show that $\beta $ preserves arbitrary left-compatible joins. Let $\{A_{i} \colon i \in I \}$ be a left-compatible set in $\mathsf {R}(S)$ . Put $A = \bigcup _{i \in I} A_{i}$ , the join of the $A_{i}$ in $\mathsf {R}(S)$ . By definition,

$$ \begin{align*} \beta (A) = \bigvee_{a \in A} \alpha (a) = \bigvee_{a \in A_{i}, i \in I} \alpha (a) = \bigvee_{i \in I}\bigg( \bigvee_{a \in A_{i}} \alpha (a)\bigg). \end{align*} $$

But this is equal to

$$ \begin{align*} \bigvee_{i \in I} \beta (A_{i}). \end{align*} $$

We finish off by proving the categorical property we need. Observe that for any acceptable set A we have that $A = \bigcup _{a \in A} a^{\downarrow }$ , because A is an order ideal. It now follows that if $\beta '$ is a homomorphism of complete right restriction monoids such that $\alpha = \beta ' \iota $ then $\beta ' = \beta $ .

Proof. Suppose that $\theta (a)$ is a projection. Thus, $\theta (a)^{\ast } = \theta (a)$ . Observe that $\theta (a) \sim _{l} \theta (a^{\ast })$ . Thus, $a \sim _{l} a^{\ast }$ , since $\theta $ is projection pure. It follows from this that $a = a^{\ast }$ .

Proof. By (N1), we have that $a \leq \nu (a)$ and $b \leq \nu (b)$ . In particular, $ab \leq \nu (a)\nu (b)$ . Thus, by (N2), we have that $\nu (ab) \leq \nu (\nu (a)\nu (b))$ . But by (N4), we have that $\nu (a)\nu (b) \leq \nu (ab)$ . Thus, by (N2), we have that $\nu (\nu (a)\nu (b)) \leq \nu ^{2}(ab)$ . But by (N3), we have that $\nu (\nu (a)\nu (b)) \leq \nu (ab)$ . We have therefore proved that $\nu (ab) = \nu (\nu (a) \nu (b))$ . The other cases are proved similarly.

Proof. By Lemma 4.5, $(S_{\nu },\cdot )$ is a semigroup. It is, in fact, a monoid with identity $\nu (1)$ since $a \cdot \nu (1) = \nu (a \nu (1))$ where $a \in S_{\nu }$ . By Lemma 4.5, we have that $\nu (a \nu (1)) = \nu (a1) = \nu (a) = a$ . We have proved that $\nu (1)$ is a right identity. It is a left identity by symmetry.

We now prove that $S_{\nu }$ is a right restriction monoid. Put

$$ \begin{align*}\mathsf {Proj} (S_{\nu }) = \{\nu (e) \colon e \in \mathsf {Proj}(S)\}.\end{align*} $$

This is a set of projections of S by Axiom (N5). Thus, $\nu (1)$ is a projection. If $a \in S_{\nu }$ , define

$$ \begin{align*}a^{\circ} = \nu (a^{\ast}).\end{align*} $$

This will be our ‘star’ operation. It is a projection by (N5). We now show that the axioms for a right restriction semigroup hold. Let $a,b \in S_{\nu }$ .

• • (RR1) holds: $(a^{\circ })^{\circ } = \nu (\nu (a^{\ast })^{\ast }) = \nu (\nu (a^{\ast })) = \nu (a^{\ast }) = a^{\circ }$ by (N5) and (N3).

• • (RR2) holds: $(a^{\circ } \cdot b^{\circ })^{\circ } = (\nu (a^{\circ }b^{\circ }))^{\circ }$ . This is equal to $\nu (\nu (a^{\ast })\nu (b^{\ast }))^{\circ } = \nu (a^{\ast }b^{\ast })^{\circ }$ using Lemma 4.5. This is equal to $\nu (\nu (a^{\ast }b^{\ast })^{\ast }) = \nu (a^{\ast }b^{\ast })$ using (N6). Whereas $a^{\circ } \cdot b^{\circ } = \nu (a^{\circ }b^{\circ }) = \nu (\nu (a^{\ast }) \nu (b^{\ast })) = \nu (a^{\ast }b^{\ast })$ using Lemma 4.5.

• • (RR3) holds: $a^{\circ } \cdot b^{\circ } = \nu (a^{\circ }b^{\circ }) = \nu (b^{\circ }a^{\circ }) = b^{\circ } \cdot a^{\circ }$ where we have used (N5).

• • (RR4) holds: $a \cdot a^{\circ } = \nu (aa^{\circ }) = \nu (a\nu (a^{\ast })) = \nu (aa^{\ast }) = \nu (a) = a$ by Lemma 4.5.

• • (RR5) holds: $(a \cdot b)^{\circ } = \nu (\nu (ab)^{\ast }) = \nu ((ab)^{\ast })$ by (N6). On the other hand, $(a^{\circ } \cdot b)^{\circ } = \nu ( \nu (\nu (a^{\ast })b)^{\ast }) = \nu (\nu (a^{\ast }b)^{\ast }) = \nu ((a^{\ast }b)^{\ast }) = \nu ((ab)^{\ast })$ by (N3) and (N6).

• • (RR6) holds: $b^{\circ } \cdot a = \nu (b^{\circ }a) = \nu (\nu (b^{\ast })a) = \nu (b^{\ast }a)$ by Lemma 4.5. Whereas $a \cdot (b \cdot a)^{\circ } = \nu (a(b \cdot a)^{\circ }) = \nu (a \nu (ba)^{\circ }) = \nu (a \nu (\nu (ba)^{\ast })) = \nu (a(ba)^{\ast })$ by (N6) and Lemma 4.5.

Thus, we have proved that $(S_{\nu },\cdot )$ is a right restriction monoid. We now establish a number of claims.

• • Claim. $a \preceq b$ if and only if $a \leq b$ . Let $a,b \in S_{\nu }$ . Denote the natural partial order on $S_{\nu }$ by $\preceq $ . To prove the claim, suppose first that $a \preceq b$ . Then $a = b \cdot a^{\circ }$ . Thus, $a = \nu (ba^{\circ }) =\, \nu (b \nu (a^{\ast })) = \nu (ba^{\ast })$ using Lemma 4.5. But $ba^{\ast } \leq b$ and so $\nu (ba^{\ast }) \leq \nu (b) = b$ by (N2). Thus, $a \leq b$ . Suppose now that $a \leq b$ . This means that $a = ba^{\ast }$ . Thus, $a = \nu (ba^{\ast }) = \nu (b \nu (a^{\ast })) = b \cdot a^{\circ }$ by Lemma 4.5. Whence we have proved that $a \preceq b$ .

• • Claim. $a \sim _{l} b$ in $S_{\nu }$ if and only if $a \sim _{l} b$ in S. To prove the claim, suppose, first, that $a \sim _{l} b$ in $S_{\nu }$ . Then $a \cdot b^{\circ } = b \cdot a^{\circ }$ . This means that $\nu (a \nu (b^\ast )) = \nu (b \nu (a^{\ast }))$ . Consequently, we have that $\nu (ab^{\ast }) = \nu (ba^{\ast })$ by Lemma 4.5. But $ab^{\ast } \leq \nu (ab^{\ast })$ by (N1). Similarly, $ba^{\ast } \leq \nu (ba^{\ast })$ . It follows that $ab^{\ast } \sim _{l} ba^{\ast }$ by Lemma 2.5. Thus, $ab^{\ast }(ba^{\ast })^{\ast } = ba^{\ast }(ab^{\ast })^{\ast }$ . Whence $ab^{\ast } = ba^{\ast }$ and so $a \sim _{l} b$ in S. Now suppose that $a \sim _{l} b$ in S. This means that $ab^{\ast } = ba^{\ast }$ . But $a \cdot \nu (b^{\ast }) = a \cdot b^{\circ }$ and $a \cdot \nu (b^{\ast }) = \nu (a \nu (b^{\ast })) = \nu (ab^{\ast })$ by Lemma 4.5. It follows that $a \sim _{l} b$ in S.

• • Claim. If $X = \{a_{i} \colon i \in I\}$ is a left-compatible set in $S_{\nu }$ then the join in $S_{\nu }$ of X exists, it is denoted by $\bigsqcup _{i \in I} a_{i}$ , and is equal to $\nu ( \bigvee _{i \in I} a_{i} )$ . To prove the claim we observe that, by the above, this is a left-compatible set in S. It therefore has a join a in S. We claim that $\nu (a)$ is the join of X in $S_{\nu }$ . It is an element of $S_{\nu }$ by (N2). We have that $a_{i} \leq a$ for all i. Thus, $a_{i} \leq \nu (a)$ for all i by (N2). Let $a_{i} \leq b$ for all i where $b \in S_{\nu }$ . This means that $a_{i} \leq b$ in S. Thus, $a \leq b$ . It follows that $\nu (a) \leq b$ in $S_{\nu }$ . Thus, the join of X exists in $S_{\nu }$ . It follows that all joins of left-compatible subsets of $S_{\nu }$ exist.

• • Claim. $\nu ( \bigvee _{i \in I} \nu (a_{i}) ) = \nu (\bigvee _{i \in I} a_{i})$ , where the $a_{i}$ are arbitrary elements of S, which form a left-compatible set. To prove the claim, we observe that we have that ${a_{i} \leq \bigvee _{i \in I} a_{i}}$ . Thus, $\nu (a_{i}) \leq \nu ( \bigvee _{i \in I} a_{i} )$ by (N2). Thus, $\bigvee _{i \in I} \nu (a_{i}) \leq \nu ( \bigvee _{i \in I} a_{i} )$ . Whence $\nu ( \bigvee _{i \in I} \nu (a_{i}) ) \leq \nu ( \bigvee _{i \in I} a_{i} )$ using (N2). To prove the reverse inequality, we start with $a_{i} \leq \nu (a_{i})$ by (N1). It follows that $\bigvee _{i \in I} a_{i} \leq \bigvee _{i \in I} \nu (a_{i})$ and so $\nu ( \bigvee _{i \in I} a_{i} ) \leq \nu ( \bigvee _{i \in I} \nu (a_{i}) )$ .

• • Claim. $( \bigsqcup _{i \in I} a_{i} ) \cdot b = \bigsqcup _{i \in I} a_{i} \cdot b$ in $S_{\nu }$ . This is straightforward to prove: we have that $( \bigsqcup _{i \in I} a_{i} ) \cdot b = \nu ( \bigvee _{i \in I} a_{i}b )$ and $\bigsqcup _{i \in I} a_{i} \cdot b = \nu ( \bigvee _{i \in I} \nu (a_{i}b))$ .

The result now follows by what we proved above.

Proof. Define $\alpha \colon S_{\nu } \rightarrow T$ by $\alpha (s) = \theta (s)$ . Then $\alpha (s_{1} \cdot s_{2}) = \theta (\nu (s_{1}s_{2})) = \theta (s_{1}s_{2}) = \alpha (s_{1}) \alpha (s_{2})$ . Thus, $\alpha $ is a semigroup homomorphism. Suppose that $\alpha (s_{1}) = \alpha (s_{2})$ . Then $\theta (s_{1}) = \theta (s_{2})$ . It follows that $\nu (s_{1}) = \nu (s_{2})$ . Thus, $s_{1} = s_{2}$ , since both elements are assumed closed. We have shown that $\alpha $ is an injective homomorphism. Let $t \in T$ . Then since $\theta $ is assumed surjective, there is $s \in S$ such that $\theta (s) = t$ . But $\theta = \theta \nu $ . Thus, $\theta (\nu (s)) = t$ . But $\nu (s)$ is closed. We have shown that $\alpha $ is an isomorphism of semigroups. Let s be closed. Then

$$ \begin{align*}\alpha (s^{\circ}) = \alpha (\nu (s^{\ast})) = \theta (\nu (s^{\ast})) = \theta (s^{\ast}) = \theta (s)^{\ast} = \alpha (s)^{\ast}.\end{align*} $$

Thus, $\alpha $ is an isomorphism of right restriction monoids. We now refer the reader to Proposition 4.6. Finally, suppose that $a_{i}$ , $i \in I$ , is a set of left-compatible elements in $S_{\nu }$ . We calculate $\alpha (\bigsqcup _{i \in I} a_{i})$ . But $\bigsqcup _{i \in I} a_{i} = \nu (\bigvee _{i \in I} a_{i})$ . It now follows that $\alpha $ preserves any left-compatible joins. Thus, $\alpha $ is an isomorphism of complete right restriction monoids.

Proof. (1) Suppose that a is finite. We prove that $a^{\ast }$ is finite. Suppose that $a^{\ast } \leq \bigvee _{i \in I} b_{i}$ . Then from the definition of the natural partial order, we have that $a^{\ast } = ( \bigvee _{i \in I} b_{i} )a^{\ast }$ . Thus, $a^{\ast } = \bigvee _{i \in I} b_{i}a^{\ast }$ . By part (6) of Lemma 2.4, each $b_{i}a^{\ast }$ is a projection. From Axiom (RR4), we have that $a = \bigvee _{i \in I} ab_{i}a^{\ast }$ . But a is finite. Thus, relabelling if necessary, $a = \bigvee _{i = 1}^{n} ab_{i}a^{\ast }$ . Now apply part (2) of Lemma 3.1 to deduce that $a^{\ast } = \bigvee _{i = 1}^{n} (ab_{i}a^{\ast })^{\ast }$ . But by Axiom (RR5), we have that $(ab_{i}a^{\ast })^{\ast } = (a^{\ast }b_{i}a^{\ast })^{\ast }$ . But this is just $a^{\ast }b_{i}a^{\ast }$ since $b_{i}a^{\ast }$ is a projection. Thus, $a^{\ast } = \bigvee _{i = 1}^{n} a^{\ast }b_{i}a^{\ast }$ . But $a^{\ast }b_{i}a^{\ast } \leq b_{i}$ . We have therefore proved that $a^{\ast }$ is finite. We now prove the converse. Suppose that $a^{\ast }$ is finite. We prove that a is finite. Suppose that $a \leq \bigvee _{i \in I} b_{i}$ . From the definition of the natural partial order, we have that $a = \bigvee _{i \in I} b_{i}a_{i}^{\ast }$ . Thus, $b_{i}a^{\ast } \leq a$ . This means that $b_{i}a^{\ast } = ab_{i}^{\ast }a^{\ast }$ . From $a = \bigvee _{i \in I} b_{i}a_{i}^{\ast }$ we deduce that $a^{\ast } = \bigvee _{i \in I} b_{i}^{\ast }a_{i}^{\ast }$ by part (2) of Lemma 3.1. But $a^{\ast }$ is finite. Thus, relabelling if necessary, we have that $a^{\ast } = \bigvee _{i = 1}^{n} b_{i}^{\ast }a_{i}^{\ast }$ . Multiply on both sides on the left by a to obtain $a = \bigvee _{i = 1}^{n} ab_{i}^{\ast }a_{i}^{\ast }$ . With what we said above, this proves that a is finite.

(2) Suppose that a and b are both finite and $a \sim _{l} b$ . We prove that $a \vee b$ is finite. Suppose that $a \vee b \leq \bigvee _{i \in I} a_{i}$ . Since $a \leq a \vee b$ there is a finite subset $I_{1}$ of I such that $a \leq \bigvee _{i \in I_{1}} a_{i}$ . Likewise, there is a finite subset $I_{2}$ of I such that $b \leq \bigvee _{i \in I_{2}} a_{i}$ . It follows that $a \vee b \leq \bigvee _{i \in I_{1} \cup I_{2}} a_{i}$ , and so $a \vee b$ is finite.

(3) This now follows from (1) and (2) above.

Proof. (1) Let $a \in S$ . Then, under the assumption that S is étale, we may write

$$ \begin{align*}a = \bigvee_{i=1}^{m} a_{i},\end{align*} $$

where the $a_{i}$ are partial units such that $a_{i} \sim _{l} a_{j}$ . From $a_{i} \sim _{l} a_{j}$ , it follows that $\theta (a_{i}) \sim _{l} \theta (a_{j})$ and so $\bigvee _{i=1}^{m} \theta (a_{i})$ is defined in T. This means that we must define

$$ \begin{align*}\phi (a) = \bigvee_{i=1}^{m} \theta (a_{i}).\end{align*} $$

Observe that this establishes uniqueness. Of course, this appears to depend on our choice of partial units $a_{i}$ . We prove that this is not the case. Accordingly, suppose that $a = \bigvee _{j=1}^{n} b_{j}$ , where the $b_{j}$ are partial units. Then

$$ \begin{align*}\bigvee_{i=1}^{m} a_{i} = \bigvee_{j=1}^{n} b_{j}.\end{align*} $$

Thus, $a_{i} \leq \bigvee _{j=1}^{n} b_{j}$ . It follows by Lemma 3.1 that $a_{i} = \bigvee _{j=1}^{n} a_{i} \wedge b_{j}$ . Now $a_{i} \wedge b_{j}$ is a partial unit by Lemma 3.3. In addition, $a_{i} \wedge b_{j} \leq a_{i}$ . It follows that these elements are pairwise compatible (not merely left-compatible). Thus, the join is an honest-to-goodness join in $\mathsf {Inv}(S)$ . It follows that $\theta (a_{i}) = \bigvee _{j=1}^{n} \theta (a_{i} \wedge b_{j})$ . But the elements $a_{i}$ and $b_{j}$ are partial units and we know that they are $\sim _{l}$ -related (because they are all below a). It follows that the meet $a_{i} \wedge b_{j}$ is algebraic by Lemma 2.10 and the remark that follows. Thus, $\theta (a_{i} \wedge b_{j}) = \theta (a_{i}) \wedge \theta (b_{j})$ and so $\theta (a_{i}) = \bigvee _{j=1}^{n} \theta (a_{i}) \wedge \theta (b_{j})$ . By Lemma 3.1, we deduce that $\theta (a_{i}) \leq \bigvee _{j=1}^{n} \theta (b_{j})$ . Thus, $\bigvee _{i=1}^{m} \theta (a_{i}) \leq \bigvee _{j=1}^{n} \theta (b_{j})$ . By symmetry we get equality. This proves that $\phi $ is a well-defined function extending $\theta $ . The fact that $\phi $ is a semigroup homomorphism follows from the observation that if $a = \bigvee _{i=1}^{m} a_{i}$ and $b = \bigvee _{j=1}^{n}b_{j}$ , where the $a_{i}$ and $b_{j}$ are partial units, then $ab = \bigvee _{i,j} a_{i}b_{j}$ , where each product $a_{i}b_{j}$ is a partial unit by Lemma 3.3. Suppose that $a = \bigvee _{i=1}^{m}a_{i}$ , where the $a_{i}$ are partial units. Then we claim that $a^{\ast } = \bigvee _{i=1}^{m}a_{i}^{-1}a_{i}$ . From $a = \bigvee _{i=1}^{m}a_{i}$ we get that $a^{\ast } = \bigvee _{i=1}^{m}a_{i}^{\ast }$ by part (2) of Lemma 3.1. But $a_{i}^{\ast } = a_{i}^{-1}a_{i}$ by Lemma 2.12 and the sentences that follow. We have proved that $\phi $ is a homomorphism of right restriction monoids.

(2) Suppose now that $\theta $ is a bijection. We prove that $\phi $ is a bijection. Let $a,b \in S$ such that $\phi (a) = \phi (b)$ . We can write

$$ \begin{align*}a = \bigvee_{i=1}^{m} a_{i}\quad \text{and} \quad b = \bigvee_{j=1}^{n} b_{j},\end{align*} $$

where the $a_{i}$ and $b_{j}$ are partial units. By assumption,

$$ \begin{align*}\bigvee_{i=1}^{m} \theta (a_{i}) = \bigvee_{j=1}^{n} \theta (b_{j}).\end{align*} $$

Thus, $\theta (a_{i}) \leq \bigvee _{j=1}^{n} \theta (b_{j})$ . It follows by Lemma 3.1 that $\theta (a_{i}) = \bigvee _{j=1}^{n} \theta (a_{i}) \wedge \theta (b_{j})$ . The element $\theta (a_{i}) \wedge \theta (b_{j})$ is a partial unit by Lemma 2.13. Since $\theta $ is an isomorphism (and so an order isomorphism) it follows that $a_{i} \wedge b_{j}$ exists and ${\theta (a_{i}) \wedge \theta (b_{j}) = \theta (a_{i} \wedge b_{j})}$ . Again, since $\theta $ is an isomorphism, $\bigvee _{j=1}^{n} \theta (a_{i} \wedge b_{j}) = \theta (\bigvee _{j=1}^{n} a_{i} \wedge b_{j} )$ . It follows that ${a_{i} = \bigvee _{j=1}^{n} a_{i} \wedge b_{j}}$ . Whence $a_{i} \leq b$ . This means that $a \leq b$ . By symmetry we have that $a = b$ and we have proved that $\phi $ is injective. We now prove that $\phi $ is surjective. Let ${a' \in T}$ . Then $a' = \bigvee _{i=1}^{m} a_{i}'$ where the $a_{i}'$ are partial units that are pairwise left-compatible. Let $a_{i}$ be the partial unit of S such that $\theta (a_{i}) = a_{i}'$ . The elements $a_{i}$ are also left-compatible. Put $a = \bigvee _{i=1}^{m} a_{i}$ . Then, by construction, $\phi (a) = a'$ . We have therefore proved that $\phi $ is an isomorphism of right restriction monoids.

Proof. (1) $a^{\downarrow }$ is an order ideal by construction, and $\mathsf {Inv}(S)$ is an order ideal by part (2) of Lemma 3.3. Thus, $[a]$ is an order ideal. The fact that this set is left-compatible follows by Lemma 2.5.

(2) Suppose that $\{a_{1},\ldots , a_{n}\}$ is a compatible set of partial units each less than or equal to a. Then their join is less than or equal to a. This join is a partial unit by Lemma 3.3.

(3) By the definition of étale, we may write $a = \bigvee _{i=1}^{n} b_{i}$ , where the $b_{i}$ are a finite left-compatible set of partial units. Let c be any partial unit $c \leq a$ . Then $c = \bigvee _{i=1}^{n} c \wedge b_{i}$ by Lemma 3.1. By part (2) of Lemma 3.3, each $c \wedge b_{i}$ is a partial unit and $c \wedge b_{i} \leq b_{i}$ .

(4) We prove first that $a \leq b$ ; the result then follows by symmetry. By definition, $a = \bigvee _{i=1}^{m} a_{i}$ where the $a_{i} \in a^{\downarrow } \cap \mathsf {Inv}(S)$ and so are elements of $b^{\downarrow } \cap \mathsf {Inv}(S)$ . It follows that $a_{i} \leq b$ from which it follows that $a \leq b$ .

Proof. (1) We prove that $A^{\vee }$ is acceptable. We prove first that $A^{\vee }$ is an order ideal. A typical element of $A^{\vee }$ has the form $\bigvee _{i = 1}^{m} a_{i}$ where $\{a_{1}, \ldots , a_{m}\}$ is a compatible subset of A. Suppose that $x \leq \bigvee _{i = 1}^{m} a_{i}$ . So, by Lemma 3.1, we have that $x = \bigvee _{i = 1}^{m} a_{i} \wedge x$ . Because A is an order ideal, we have that $a_{i} \wedge x \in A$ . Thus, $x \in A^{\vee }$ .

We prove next that $A^{\vee }$ is left-compatible. Suppose that $x,y \in A^{\vee }$ . Then $x = \bigvee _{i=1}^{m}x_{i}$ and $y = \bigvee _{j=1}^{n} y_{j}$ where $x_{i}, y_{j} \in A$ . We are working inside a Boolean inverse monoid and so $y^{-1} = \bigvee _{j=1}^{n} y_{j}^{-1}$ . By assumption, $x_{i} \sim _{l} y_{j}$ for all i and j. Thus, by Lemma 2.7, we have that $x_{i}y_{j}^{-1}$ is an idempotent. But $xy^{-1} = \bigvee _{i,j}x_{i}y_{j}^{-1}$ . In a Boolean inverse monoid the finite join of idempotents is an idempotent and so $xy^{-1}$ is an idempotent. By Lemma 2.7, we have proved that x and y are left-compatible. Thus, $A^{\vee }$ is acceptable. It is immediate that $A \subseteq A^{\vee }$ .

(2) The function is well defined by part (1) above. The proof that the properties (N1), (N2) and (N3) of a nucleus hold are immediate. It is easy to verify that (N4) and (N5) hold. We prove that (N6) holds. Only one direction needs proving. Let ${x \in ((A^{\vee })^{\ast })^{\vee }}$ . Then $x = \bigvee _{i=1}^{m} x_{i}$ where $x_{i} \in (A^{\vee })^{\ast }$ . Now each $x_{i} = (\bigvee _{j_{i}} y_{j_{i}})^{\ast }$ , where $y_{j_{i}} \in A$ . By part (2) of Lemma 3.1, we have that $x_{i} = \bigvee _{j_{i}} y_{j_{i}}^{\ast }$ . Thus, $x = \bigvee _{i}\bigvee _{j_{i}} y_{j_{i}}^{\ast }$ where $y_{j_{i}} \in A$ . Whence $x \in (A^{\ast })^{\vee }$ .

Proof. The proof of Proposition 4.6 will be used below; particularly the form taken by the natural partial order and the join.

We work in the complete right restriction monoid $\mathsf {R}(S)_{\nu }$ defined above. Observe that for each $a \in S$ , the set $a^{\downarrow }$ belongs to $\mathsf {R}(S)_{\nu }$ . We show that, in fact, each element of the form $a^{\downarrow }$ , where $a \in S$ , is finite (in the technical sense). Suppose that $a^{\downarrow } \leq \bigsqcup _{i \in I} B_{i}$ where $B_{i} \in \mathsf {R}(S)_{\nu }$ . Then $a^{\downarrow } \subseteq \nu ( \bigvee _{i \in I} B_{i})$ where $B_{i}$ are $\nu $ -closed elements of $\mathsf {R}(S)$ . Thus, $a = \bigvee _{j=1}^{n} b_{j}$ where $b_{j} \in B_{j}$ , relabelling if necessary. This implies that $a^{\downarrow } \leq \bigsqcup _{j = 1}^{n} B_{i}$ . This proves that $a^{\downarrow }$ is finite.

Put $A = \{a_{1}, \ldots , a_{m}\}$ . Then $A^{\downarrow }$ is an acceptable set, using Lemma 2.9, and so a well-defined element of $\mathsf {R}(S)$ . Observe that if $X = (\{a_{1}, \ldots , a_{m}\}^{\downarrow })^{\vee }$ , then $X = a_{1}^{\downarrow } \sqcup \cdots \sqcup a_{1}^{\downarrow }$ and so is finite, since any finite join of such elements is again finite by part (2) of Lemma 4.8.

We now show that all finite elements have this form. Suppose that X is a finite element of $\mathsf {R}(S)_{\nu }$ . Thus, X is a closed acceptable subset. We may write $X = \bigsqcup _{a \in A} a^{\downarrow }$ . But we have assumed that X is finite. Thus, $X = \bigsqcup _{i=1}^{m} a_{i}^{\downarrow }$ for some m.

Proof. According to Lemma 5.6, we have a description of the finite elements. Let $A = (\{a_{1}, \ldots , a_{m}\}^{\downarrow })^{\vee }$ , where $\{a_{1}, \ldots , a_{m}\}$ is a left-compatible subset, and let ${B = (\{b_{1}, \ldots , b_{m}\}^{\downarrow })^{\vee }}$ , where $\{b_{1}, \ldots , b_{m}\}$ is a left-compatible subset, be finite elements of $\mathsf {R}(S)_{\nu }$ . Their product, $A \cdot B$ is just $(\{a_{1}, \ldots , a_{m}\}^{\downarrow }\{b_{1}, \ldots , b_{m}\}^{\downarrow })^{\vee }$ , where we have used the definition of the product in $\mathsf {R}(S)_{\nu }$ and Lemma 4.5. Now we apply Lemma 2.9 to deduce that $a_{i}b_{j} \sim _{l} a_{s}b_{t}$ . It follows that $A \cdot B$ is also finite (in our technical sense).

Proof. The product of two finite elements is finite by Lemma 5.7. It follows by part (3) of Lemma 4.8 that $\mathsf {Etale}(S)$ is a distributive right restriction monoid. If A is finite (in our technical sense) then $A^{\circ }$ is finite by Lemma 4.8. However, $A^{\circ }$ is of the form $(\{a_{1}^{\ast },\ldots , a_{m}^{\ast }\}^{\downarrow })^{\vee }$ , which is equal to $(a_{1}^{\ast } \vee \cdots \vee a_{m}^{\ast })^{\downarrow }$ . We deduce that the projections of $\mathsf {Etale}(S)$ are the principal order ideals of the idempotents of S and so form a meet-semilattice isomorphic with the set of idempotents of S, which is a Boolean algebra. It follows that $\mathsf {Etale}(S)$ is, in fact, a Boolean right restriction monoid.

Elements of the form $a^{\downarrow }$ , for some $a \in S$ , are partial units because $a^{\downarrow }(a^{-1})^{\downarrow } = (aa^{-1})^{\downarrow }$ and $(a^{-1})^{\downarrow }a^{\downarrow } = (a^{-1}a)^{\downarrow }$ . However, every element of $\mathsf {Etale}(S)$ can be written $\bigsqcup _{i=1}^{m} a_{i}^{\downarrow }$ , for some left-compatible subset $\{a_{1},\ldots , a_{m}\}$ . This immediately implies that $\mathsf {Etale}(S)$ is étale. We now prove that every partial unit of $\mathsf {Etale}(S)$ is of the form $a^{\downarrow }$ , from which it follows that the partial units of $\mathsf {Etale}(S)$ form a semigroup isomorphic to S. Let $A = (\{a_{1},\ldots , a_{m}\}^{\downarrow })^{\vee }$ be a partial unit of $\mathsf {Etale}(S)$ . Then there is an element $X = (\{x_{1},\ldots , x_{n}\}^{\downarrow })^{\vee }$ of $\mathsf {Etale}(S)$ such that we have $A \cdot X = (x_{1}^{\ast } \vee \cdots \vee x_{n}^{\ast })^{\downarrow }$ and ${X \cdot A = (a_{1}^{\ast } \vee \cdots \vee a_{m}^{\ast })^{\downarrow }}$ . Choose any $a_{i}$ . Then $a_{i}^{\ast } \in X \cdot A$ . It follows that $a_{i}^{\ast } = \bigvee _{j=1}^{s}y_{j}b_{j}$ where $y_{j} \in \{x_{1},\ldots , x_{n}\}^{\downarrow }$ and $b_{j} \in \{a_{1},\ldots , a_{m}\}^{\downarrow }$ . Since each $b_{j} \in \{a_{1},\ldots , a_{m}\}^{\downarrow }$ , we may write $b_{j} = b_{j}b_{j}^{-1}a_{j}'$ where $a_{j}'$ is one of the elements $a_{1},\ldots , a_{m}$ . It is therefore immediate that $y_{j}b_{j} = (y_{j}b_{j}b_{j}^{-1})a_{j}'$ . But $y_{j}b_{j}b_{j}^{-1} \in \{x_{1},\ldots , x_{n}\}^{\downarrow }$ since this is an order ideal. Thus, we can write (relabelling if necessary) $a_{i}^{-1}a_{i} = \bigvee _{j=1}^{s} y_{j}a_{j}'$ where $y_{j} \in \{x_{1},\ldots , x_{n}\}^{\downarrow }$ and the $a_{j}'$ is one of the elements $a_{1},\ldots , a_{m}$ . It follows that $a_{i}^{-1} = \bigvee _{j=1}^{s} y_{j}(a_{j}'a_{i}^{-1})$ since $a_{i}^{-1} = a_{i}^{-1}a_{i}a_{i}^{-1}$ . Because $\{a_{1},\ldots , a_{m}\}$ is a left-compatible set the element $a_{j}'a_{i}^{-1}$ is always a projection/idempotent by Lemma 2.7. Thus, $y_{j}(a_{j}'a_{i}^{-1}) \in \{x_{1},\ldots , x_{n}\}^{\downarrow }$ . We have therefore shown that $a_{i}^{-1} \in X$ . The elements of X are left-compatible. It follows that $a_{i}^{-1}$ and $a_{j}^{-1}$ are left-compatible, and so $a_{i}$ and $a_{j}$ are right-compatible by Lemma 2.8. Thus, $a_{i}$ and $a_{j}$ are compatible. It follows that $A = (\{a_{1},\ldots , a_{m}\}^{\downarrow })^{\vee } = (a_{1} \vee \cdots \vee a_{m})^{\downarrow }$ .

Proof. Suppose first that $I = \mathsf {Inv}(S)$ is $0$ -simplifying. Let e be any nonzero projection. Then $IeI$ is an ideal of I and $(IeI)^{\vee }$ is an additive ideal of I. By assumption, $(IeI)^{\vee } = I$ . Since $1 \in I$ , it follows that $1 = \bigvee _{i=1}^{n} e_{i}$ for some idempotents/projections $e_{i} \in IeI$ . Observe that the join $e_{1} \vee e_{2} \overline {e_{1}} \vee \cdots \vee e_{n} \overline {e_{1}} \, \overline {e_{2}} \cdots \, \overline {e_{n-1}}$ is also $1$ and that each term belongs to $IeI$ since this is an order ideal. It follows that we lose no generality in assuming that $1 = \bigvee _{i=1}^{n} e_{i}$ is an orthogonal join. By assumption, $e_{i} = a_{i}eb_{i}$ where $a_{i}, b_{i} \in \mathsf {Inv}(S)$ . Define $x_{i} = e\mathbf {d}(a_{i})b_{i}$ . Observe that $x_{i} = ex_{i}$ . A simple calculation shows that $x_{i}^{-1}x_{i} = e_{i}$ . Thus, $1 = \bigvee _{i=1}^{n} x_{i}^{-1}ex_{i}$ . Observe that the elements $x_{i}$ are pairwise left-orthogonal. Put $a = \bigvee _{i=1}^{n} x_{i} \in S$ . But $a^{\ast } = 1$ . We now calculate $(ea)^{\ast }$ . But this is just $1$ . We now prove the converse. Let J be any nontrivial additive ideal of $\mathsf {Inv}(S)$ and let e be a nonzero projection in J. Then, by assumption, there is an element $a \in S$ such that $(ea)^{\ast } = 1$ . By assumption, S is étale. Thus, there are partial units $a_{i}$ such that $a = \bigvee _{i=1}^{n} a_{i}$ . Thus, $1 = (ea)^{\ast } = \bigvee _{i=1}^{n} (ea_{i})^{\ast }$ . It follows that $1 \in J$ and so $J = \mathsf {Inv}(S)$ . We have proved that $\mathsf {Inv}(S)$ is $0$ -simplifying.

Proof. Suppose first that left-compatible elements are compatible in S. Then it is immediate from the way $\mathsf {Etale}(S)$ is constructed that $\mathsf {Etale}(S) \cong S$ . We now prove the converse. Suppose that $\mathsf {Etale}(T) \cong S$ . Thus, we are given an étale Boolean right restriction monoid T such that $T \cong \mathsf {Inv}(S)$ . Thus, T is an étale Boolean right restriction monoid that is also inverse. We prove first that $T = \mathsf {Inv}(T)$ . We therefore need to prove that every element of T is a partial unit. Let $a \in T$ . Then there is a unique element $b \in T$ such that $a = aba$ and $b = bab$ . By assumption,

$$ \begin{align*}a = \bigvee_{i=1}^{m} a_{i}\quad \text{and} \quad b = \bigvee_{j=1}^{m} b_{j},\end{align*} $$

where the $a_{i}$ and $b_{j}$ are partial units. It follows that

$$ \begin{align*}ab = \bigvee_{1 \leq i \leq m, 1 \leq j \leq n} a_{i}b_{j},\end{align*} $$

where $a_{i}b_{j}$ is a partial unit by part (1) of Lemma 2.13. But $ab$ is an idempotent and so $a_{i}b_{j}$ is an idempotent. By part (3) of Lemma 2.13, it follows that $a_{i}b_{j}$ is a projection. By part (2) of Lemma 3.1, it follows that $ab$ is a projection. By symmetry, $ba$ is a projection. Thus, a is a partial unit. We have therefore proved that $T = \mathsf {Inv}(T)$ . We can now finish off the proof. Suppose that $a,b \in T$ are such that $a \sim _{l} b$ . Then $a \vee b$ exists and, by assumption, is a partial unit. We have that $a,b \leq a \vee b$ . It follows by part (2) of Lemma 2.13 that $a = aa^{-1}(a \vee b)$ . By Lemma 3.1, we have that $a = a \vee aa^{-1}b$ . Thus, in particular, $aa^{-1}b \leq a$ . But $bb^{-1}a \leq a$ . Now check that $(aa^{-1}b)^{\ast } = (bb^{-1}a)^{\ast }$ , using the fact that projections commute. By part (1) of Lemma 2.5, we have that $aa^{-1}b = bb^{-1}a$ . Thus, $a \sim _{r} b$ . We have therefore proved that $a \sim b$ .