1. Suppose $E(\mathcal{S})$ is Hausdorff. Given $x_1\neq x_2$ in $\mathcal{S}^{(0)}$, choose $e_1,e_2\in E(\mathcal{S})$ with $\so(e_i)=x_i$, so $e_1\neq e_2$. Take any two open bisections $U_1,U_2$ with $e_i\in U_i$.

2. We have $\mathcal{S}^{(2)}=\left\{(a,b)\in\mathcal{S}\colon \so(a)=\ra(b)\right\}$, which is closed in $\mathcal{S}\times\mathcal{S}$ as $\so$ and $\ra$ are continuous and $\mathcal{S}^{(0)}$ is Hausdorff.

   If $\mu\colon\mathcal{S}^{(2)}\to\mathcal{S}$ denotes the (continuous) product, then for any two compact subsets $K,L\subseteq\mathcal{S}$, $(K\times L)\cap\mathcal{S}^{(2)}$ is closed in the compact $K\times L$, hence compact itself. Therefore $KL=\mu((K\times L)\cap\mathcal{S}^{(2)})$ is compact.

1. As $x\subseteq z$, we multiply both sides by $y^*y$ to obtain $x=xy^*y\subseteq zy^*y=y$, because $x\leq y\leq z$.

2. On one hand, $(a\cap b)c^*c\subseteq (a\cap b)b^*b=a\cap b\subseteq a$, and $(a\cap b)c^*c\subseteq bc^*c=c$, so $(a\cap b)c^*c$ is a $\subseteq$-lower bound of $\left\{a,c\right\}$. Simple computations prove that it is the greatest lower bound.

3. This was already proven in Lemma 4.11.

4. As $bx,by\subseteq ba$, item (b) yields $bx\cap by=bx(by)^*(by)$. As $x,y\leq a$, then $xy^*\in E(S)$, so we may commute $xy^*$ and $b^*b$ to obtain

   \begin{equation*} bx\cap by=b(xy^*)(b^*b)y=b(b^*b)(xy^*)y=b(x\cap y) \end{equation*}
   where we used item (b) in the last equality, as $x\cap y=xy^*y$.

5. By distributivity, Property ($\Sigma$-iii),

   \begin{equation*} (a\cup b)^*(a\cup b)=(a^*\cup b^*)(a\cup b)=(a^*a)\cup(a^*b)\cup(b^*a)\cup(b^*b). \end{equation*}
   But since $a^*b\subseteq a^*(a\cup b)=a^*a$, and similarly for $b^*a$, then
   \begin{equation*} (a\cup b)^*(a\cup b)=a^*a\cup b^*b. \end{equation*}

6. If $a\cap(b\cup c)$ exists, then item (e) implies that both $a\cap b$ and $a\cap c$ exist, Since both are $\subseteq a$, their $\subseteq$-join exists by Property ($\Sigma$-ii).

   Assume now that $(a\cap b)\cup(a\cap c)$ exists, and let us prove that it is the $\subseteq$-meet of $\left\{a,b\cup c\right\}$. It is a lower bound, since

   \begin{equation*} (a\cap b)\cup(a\cap c)\subseteq a\cup a=a\quad\text{and}\quad (a\cap b)\cup(a\cap c)\subseteq b\cup c. \end{equation*}
   To prove that it is the largest lower bound, suppose $p\subseteq a$ and $p\subseteq (b\cap c)$. In particular, $p\cap b=b$. As $a\cap b$ exists, item (e) implies $(a\cap b)p^*p=p\cap b=p$. Similarly, $(a\cap c)p^*p=p$. By distributivity, Property ($\Sigma$-iii), we have
   \begin{equation*} ((a\cap b)\cup(a\cap c))p^*p=((a\cap b)p^*p)\cap ((a\cap c)p^*p)=p\cap p=p, \end{equation*}
   that is, $p\leq(a\cap b)\cup(a\cap c)\leq b\cup c$. Since $p\subseteq b\cup c$, item (a) gives us $p\subseteq (a\cap b)\cup(a\cap c)$, as we wanted.

7. Suppose $a\subseteq b$. Let $e=b^*b\setminus a^*a$, i.e., $a^*a\cup e=b^*b$ and $a^*a\cap e=0$. Multiplying both equalities by $b$ on the left on both sides and using distributivity (item (c) and Property ($\Sigma$-iii)) yields $a\cup(be)=b$ and $a\cap(be)=0$, so $be$ is the complement of $a$ relative to $b$.

8. Since both $a$ and $b\cap c$ are $\subseteq$-bounded above by $a\cup b$, then $a\cup(b\cap c)$ is defined by Property ($\Sigma$-ii), and is a $\subseteq$-lower bound of $a\cup b$. Symmetrically, changing the roles of $b$ and $c$, we see that it is a $\subseteq$-lower bound of $\left\{a\cup b,a\cup c\right\}$. Again, let us prove that it is the largest one. Suppose $p\subseteq a\cup b,a\cup c$. Let us verify some cases:

   1. If $p\subseteq a$, then of course $p\subseteq a\cup(b\cap c)$.

   2. Suppose $p\cap a=0$. Since both $p$ and $a$ are $\subseteq a\cup b$, then this means that $0=ap^*p$, so $p=(a\cup b)p^*p=ap^*p\cup bp^*p=bp^*p$, and thus $p\leq bp^*p$. Both $p$ and $b$ are $\subseteq a\cup b$, so $p\subseteq b$. Similarly, $p\subseteq c$, so $p\subseteq b\cap c\subseteq a\cup(b\cap c)$.

   3. For a general lower bound $p$ of $\left\{a\cup b,a\cup c\right\}$, we apply the first case to $p\cap a$ (which exists by item (c), since $a,p\subseteq a\cup b$) and the second one to $p\setminus(p\cap a)$, to conclude that

      \begin{equation*} p=(p\cap a)\cup(p\setminus(p\cap a))\subseteq a\cup(b\cap c). \end{equation*}

   The last case proves that $a\cup(b\cap c)$ is the meet of $(a\cup b)\cap(a\cup c)$, as we wanted.

9. If $a\subseteq b$ and $c\in S$, then it is easy enough, with the previous items, to verify that $c(b\setminus a)$ satisfies the properties of the complement of $ca$ in $cb$, so $c(b\setminus a)=(cb)\setminus(ca)$.

10. Similarly to the previous item, it is only a matter of verifying that $(b\setminus a_1)\cap(b\setminus a_2)$ satisfies the defining properties of the complement $b\setminus (a_1\cup a_2)$, so these elements coincide, and similarly $b\setminus (a_1\cap a_2)=(b\setminus a_1)\cup(b\setminus a_2)$.

We have $t\leq a$ and $t^*\leq a^*$. By ($\Sigma$-vi), consider $z,w\in S$ such that $t\leq z,w\subseteq a$ and

As for the uniqueness of $p$, suppose that $q$ is another element with the same property. Then $p^*p,q^*q\subseteq a^*a$, and we have $0=p(a^*a\setminus p^*p)=0$, hence $x(a^*a\setminus p^*p)=0$ and $q(a^*a\setminus p^*p)=0$ by the given properties of $p$ and $q$. Therefore

If $0\not\in e\mathfrak{F}$, then the set $(e\mathfrak{F})^{\uparrow}=\left\{u\in E:ef\leq u\text{ for some }f\in\mathfrak{F}\right\}$ is a proper filter containing $\mathfrak{F}\cup\left\{e\right\}$. If $\mathfrak{F}$ is maximal, then $\mathfrak{F}=(e\mathfrak{F})^{\uparrow}$, which contains $e$. This proves (1)$\Rightarrow$(2).

Conversely, if (2) is valid, consider a proper filter $\mathcal{G}$ containing $\mathfrak{F}$. As $0\not\in\mathfrak{G}$, then every $e\in G$ satisfies the condition of (2), so $\mathfrak{G}\subseteq\mathfrak{F}$. Hence $\mathfrak{F}$ is valid, i.e., (1).

If $e_1\in\mathfrak{F}$ we are done. If $e_1\not\in\mathfrak{F}$, then by Lemma 5.10 there exists $f\in\mathfrak{F}$ such that $e_1f=0$. For any other $f'\in\mathfrak{F}$, as $e_1\cup e_2\in\mathfrak{F}$, we have

First we prove that $\Omega(E(S))$ is Hausdorff. Suppose that $F\neq G$ in $\Omega(E(S))$. As $F$ and $G$ are ultrafilters, Lemma 5.10 allows us to take $f\in F$ and $g\in G$ such that $fg=0$, so $X[f]$ and $X[g]$ are disjoint neighbourhoods of $\mathfrak{F}$ and $\mathfrak{G}$, respectively.

To prove that $\Omega(E(S))$ is locally compact and zero-dimensional, it suffices to prove that each basic open set $X[e]$ is compact. Suppose that $X[e]=\bigcup_{i\in I}X[r_i]$. Since $X[er_i]=X[e]\cap X[r_i]$, we may assume that $r_i\leq e$. Using interpolation, let $s_i=e|r_i$, i.e.,

Suppose otherwise, and let us arrive at a contradiction. For every finite subset $F\subseteq I$, let $s(F)=\bigcup_{i\in F}s_i$. Then $e\setminus s(F)\neq 0$ for all $F$. Using Lemma 5.5,

First note that, as $\mathcal{S}^{(0)}$ is Hausdorff, then $\psi(x)\subseteq\psi(y)$ if and only if $x=y$.

We thus only need to prove that every proper filter $\mathfrak{F}$ of $E(\mathbf{KB}(\mathcal{S}))$ is contained in $\psi(x)$ for some $x$. As $\mathfrak{F}$ is downwards directed, then the family $\left\{\so(U):U\in\mathfrak{F}\right\}$ of compact subsets of the Hausdorff space $\mathcal{S}^{(0)}$ has the finite intersection property, so Cantor's Intersection Theorem implies that its intersection is nonempty. Any element $x\in\bigcap_{U\in\mathfrak{F}}\so(U)$ satisfies $\mathfrak{F}\subseteq\psi(x)$.

This in fact implies that $\psi(x)$ is an ultrafilter for any $x\in\mathcal{S}^{(0)}$: By Zorn's lemma and the previous paragraph, there are an ultrafilter $\mathfrak{F}$ and $y\in\mathcal{S}^{(0)}$ such that $\psi(x)\subseteq\mathfrak{F}\subseteq\psi(y)$. By the first paragraph of the proof and maximality of $\mathcal{F}$, $\psi(x)=\psi(y)=\mathfrak{F}$ is an ultrafilter. The argument in the previous paragraph implies the statement about $\psi^{-1}$.

Suppose $A\in\mathbf{KB}(\mathcal{T})$. As $\phi$ is continuous and proper, then $\phi^{-1}(A)$ is open and compact. Let us prove that it is a bisection of $\mathcal{S}$.

Suppose $a,b\in\phi^{-1}(A)$ and $\so(a)=\so(b)$. Then $\so(\phi(a))=\so(\phi(b))$, andas $\phi(a)$ and $\phi(b)$ belong to the bisection $A$ then $\phi(a)=\phi(b)$. As $\phi$ is star-injective then $a=b$. Thus the source map is injective on $\phi^{-1}(A)$, and similarly the range map is injective on $\phi^{-1}(A)$.

We now need to prove that $\mathbb{K}(\phi)\colon A\mapsto\phi^{-1}(A)$ is a semigroup homomorphism, i.e., that $\phi^{-1}(AB)=\phi^{-1}(A)\phi^{-1}(B)$ for all $A,B\in\mathbf{KB}(\mathcal{T})$. The inclusion $\phi^{-1}(A)\phi^{-1}(B)\subseteq\phi^{-1}(AB)$ is immediate as $\phi$ is a semigroupoid homomorphism. Conversely, suppose that $z\in\phi^{-1}(AB)$. Then there exist $(a,b)\in (A\times B)\cap\mathcal{T}^{(2)}$ such that $\phi(z)=ab$, and in particular $\so(\phi(z))=\so(b)$. As $\phi$ is star-surjective, there exists $p_b\in\mathcal{S}$ with $\so(p_b)=\so(z)$ and $\phi(p_b)=b$. Then $\so(\phi(pp_b^*))=\so(b^*)=\so(a)$, so again since $\phi$ is star-surjective then there exists $p_a\in\mathcal{S}$ with $\so(p_a)=\so(p_b^*)$ and $\phi(p_a)=a$.

Then $\so(p_ap_b)=\so(p_b)=\so(z)$, and $\phi(p_ap_b)=ab=\phi(x)$. As $\phi$ is star-injective then $x=p_ap_b\in\phi^{-1}(A)\cap\phi^{-1}(B)$.

The first non-trivial property that we need to prove for $\mathbb{K}(\phi)$, to conclude that it is a morphism of Σ-ordered inverse semigroups, is properness. Suppose $K\in\mathbf{KB}(\mathcal{S})$. Then $\phi(K)$ is compact in $\mathcal{T}$, so we may cover it by finitely many bisections $A_1,\ldots,A_n\in\mathbf{KB}(\mathcal{T})$. Setting $K_i=\phi^{-1}(A_i)\cap K$, we have $K=\bigcup_{i=1}^n K_i$ and $K_i\subseteq\mathbb{K}(\phi)(A_i)$. This proves that $\mathbb{K}(\phi)$ is proper.

The fact that $\mathbb{K}(\phi)$ preserves interpolators it equivalent to the equality

Choose any $t\in\mathfrak{F}$, so as $\theta$ is proper, there exist $t_1,\ldots,t_n\in E(T)$ and $s_1,\ldots,s_n\in E(S)$ such that $t=\bigcup_{i=1}^n t_i$ and $t_i\subseteq\phi(s_i)$ for each $i$. As $\mathfrak{F}$ is $\subseteq$-prime (Lemma 5.12), for some $i$ we have $t_i\in\mathfrak{F}$, so $s_i\in\theta|_{E(S)}^{-1}(\mathfrak{F})$.

Of course, $\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$ is upwards closed, does not contain $0$, and contains $\theta^{-1}(\mathfrak{F})$, so to prove that it is a filter we need to prove that it is closed under products. For this, fix an arbitrary $p\in\theta|_{E(S)}^{-1}(\mathfrak{F})$. Suppose that $f,g\in\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$. Then $fp\leq p$, so we may consider the interpolator $q_f\defeq p|(fp)$.

Since $\theta(p)=\theta(q_f)\cup\theta(p\setminus q_f)$, one of $q_f$ or $(p\setminus q_f)$ belongs to $\theta^{-1}(\mathfrak{F})$ (by Lemma 5.12). We have $fp(p\setminus q_f)\leq q_f(p\setminus q_f)=q_f\cap(p\setminus q_f)=0$, and because $f\in\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$ then $p(p\setminus q_f)\not\in\theta^{-1}(\mathfrak{F})$, so $(p\setminus q_f)\not\in\theta^{-1}(\mathfrak{F})$ as well.

We therefore have $q_f\in\theta^{-1}(\mathfrak{F})$. Suppose now, in order to obtain a contradiction, that $fg\not\in\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$, so that we may find $e\in\theta^{-1}(\mathfrak{F})$ such that $fge=0$. Then $fp(eg)=0$, implying $q_f(eg)=0$, contradicting the facts that $e,q_f\in\theta^{-1}(\mathfrak{F})$ and $g\in\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$.

From all of this, we conclude that $\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$ is a proper filter of $E(S)$. In fact, its definition makes it clear, by Lemma 5.10, that it is an ultrafilter in $E(S)$, i.e., an element of $\Omega(E(S))$.

To prove that $\mathbb{P}(\theta)^{(0)}$ is continuous and proper, it is sufficient to prove that $\left(\mathbb{P}(\theta)^{(0)}\right)^{-1}(X[e])=X[\theta(e)]$, as it shows that preimages of basic compact open subsets of $\Omega(E(\mathbf{KB}(\mathcal{S})))$ are compact and open in $\Omega(E(\mathbf{KB}(\mathcal{T})))$. The inclusion $X[\theta(e)]\subseteq\left(\mathbb{P}(\theta)^{(0)}\right)^{-1}(X[e])$ is simple enough, however the converse is harder. Namely, we need to prove that if $e\in\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$, then $\theta(e)\in\mathfrak{F}$.

Suppose that this was not the case, so there exists $a\in\mathfrak{F}$ with $\theta(e)a=0$. By Property 5.4($\Sigma$-v), the join $\theta(e)\cup a$ exists. As $\theta$ is proper, the same considerations as in the proof that $\theta^{-1}(\mathfrak{F})$ is nonempty allow us to assume that $\theta(e)\cup a\subseteq\theta(g)$, and in particular $\theta(e)=\theta(eg)$. Letting $p=g|(eg)$, we have $\theta(p)=\theta(g)|\theta(e)$. In particular, $\theta(e)a=0$ implies $\theta(p)a=0$, so $\theta(p)\not\in\mathfrak{F}$. On the other hand, $e(g\setminus p)=ge(g\setminus p)\leq p(g\setminus p)=0$, and since $e\in\mathbb{P}(\theta)^{(0)}(\mathfrak{F})$ then $g\setminus p\not\in\theta^{-1}(\mathfrak{F})$, i.e., $\theta(g\setminus p)\not\in\mathfrak{F}$.

This is a contradiction, as neither $\theta(p)$ nor $\theta(g\setminus p)$ belong to the ultrafilter $\mathfrak{A}$ but $\theta(g)=\theta(p)\cup\theta(g\setminus p)$ does.

We therefore conclude that $\left(\mathbb{P}(\theta)^{(0)}\right)^{-1}(X[e])=X[\theta(e)]$, as we wanted.

Since $\theta$ is proper, choose $t_1,\ldots,t_n\in T$ and $s_1,\ldots,s_n\in S$ such that $t=\bigcup_{i=1}^n t_i$ and $t_i\subseteq \theta(s_i)$ for each $i$. As $t^*t=\bigcup_{i=1}^n t_i^*t_i$ belongs to $\mathfrak{F}$, we have $t_i^*t_i\in\mathfrak{F}$ for some $i$, and of course

For the second part, first note that if $[\theta(s),\mathfrak{F}]$ is a well-defined element of $\mathbb{P}(T,\subseteq)$, i.e., if $\mathfrak{F}\in X[\theta(s)^*\theta(s)]$ then the definition of $\mathbb{P}(\theta)^{(0)}$ readily yields $\mathbb{P}(\theta)^{(0)}(\mathfrak{F})\in[s^*s]$, so $[s,\mathbb{P}(\theta)^{(0)}(\mathfrak{F})]$ is a well-defined element of $\mathbb{P}(S,\subseteq)$.

Now assume that $[\theta(s_1),\mathfrak{F}]=[\theta(s_2),\mathfrak{F}]$, i.e., there exists $t\subseteq\theta(s_1),\theta(s_2)$ such that $t^*t\in\mathfrak{F}$. Property (iv) of morphisms of Σ-ordered inverse semigroups yields $u\subseteq s_1,s_2$ such that $t\subseteq\theta(u)$. In particular $\mathfrak{F}\in X[\theta(u)^*\theta(u)]$, so the previous paragraph gives us $\mathbb{P}(\theta)^{(0)}(\mathfrak{F})\in X[u^*u]$. As $u\subseteq s_1,s_2$, we conclude that $[s_1,\mathbb{P}(\theta)^{(0)}(\mathfrak{F})]=[s_2,\mathbb{P}(\theta)^{(0)}(\mathfrak{F})]$.

First we prove that $U=X[e]$ for some $e\in E(S)$. As $U$ is compact-open, we may write it as a finite union of basic open subsets of $\Omega(E(S))$, $U=\bigcup_{i=1}^n X[e_i]$, where $e_i\in E(S)$. Proceeding inductively, it is enough to assume $n=2$.

For $i=1,2$, let $f_i=e_i|(e_1e_2)$, so $X[f_i]=X[e_1e_2]=X[e_1]\cap X[e_2]$. We have $X[e_i]=X[f_i]\cup X[e_i\setminus f_i]$, and the latter sets are disjoint, so $X[e_i\setminus f_i]=X[e_i]\setminus X[f_i]$. We thus have a partition

Therefore, as we can assume $U=X[e]$. Since $U\subseteq X[s^*s]$ then $U=X[e]\cap X[s^*s]=X[es^*s]$, i.e., we may assume $U=X[e]$ with $e\leq s^*s$. Let $z=s|(se)$, so

First we write $U$ in terms of the basic open sets: $U=\bigcup_{i=1}^n[s_i,U_i']$, each $U_i'$ being compact-open and hence clopen in $\Omega(E(S))$.

Let $U_1=U_1'$ and for $i\geq 2$, $U_i=U_i'\setminus\bigcup_{j=1}^{i-1}U_j$. Then we have disjoint compact-open bisection $[s_i,U_i]$ (as their sources are disjoint), and also $\bigcup_{i=1}^n[s_i,U_i]\subseteq U$. However, the source map is injective on $U$, and $\so(\bigcup_{i=1}^n[s_i,U_i])=\bigcup_{i=1}^n U_i=\bigcup_{i=1}^n U_i'=\so(U)$, and thus we must have $U=\bigcup_{i=1}^n[s_i,U_i]$. By Lemma 5.19, we may assume $U_i=\dom(\kappa_{s_i})$.

As the sets $[a_i,U_i]$ are disjoint, and contained in a bisection, then their sources and ranges are also disjoint. It follows that $a_i^*a_j=a_ia_j^*=0$ whenever $i\neq j$, and thus the supremum $a=\bigcup_{i=1}^n a_i$ exists. It readily follows that $X[a^*a]=\bigcup_{i=1}^n U_i$, and thus that $[a,X[a^*a]]=\bigcup_{i=1}^n[a_i,U_i]=U$.

Of course $s\subseteq t$ implies $\kappa_{(S,\subseteq)}(s)\subseteq\kappa_{(S,\subseteq)}(t)$, so we deal with the converse implication.

Assume that $\kappa_{(S,\subseteq)}(s)\subseteq\kappa_{(S,\subseteq)}(t)$, i.e., $[s,X[s^*s]]\subseteq[t,X[t^*t]]$. For every $\mathfrak{F}\in X[s^*s]$, we have $[s,\mathfrak{F}]=[t,\mathfrak{F}]$, so we may find $a\subseteq s,t$ with $\mathfrak{F}\in X[a^*a]$. This means that $\left\{X[a^*a]:a\subseteq s,t\right\}$ is an open cover of $X[s^*s]$, so we may find a finite subcover, i.e., $a_1,\ldots,a_n\subseteq s,t$ such that $X[s^*s]=\bigcup_{i=1}^nX[a_i^*a_i]$. Letting $a=\bigcup_{i=1}^n a_i$, we have $a\subseteq s,t$, and $X[a^*a]=X[s^*s]$. Thus we just need to prove that $a=s$.

Indeed, if $a\neq s$ then $s\setminus a\neq 0$, so $X[(s\setminus a)^*(s\setminus a)]$ is a nonempty subset of $X[s^*s]$ which does not intersect $X[a^*a]=X[s^*s]$, a contradiction.