# Vietoris and open maps

The topological analogue of the powerset monad is the Vietoris hyperspace. Given a (compact Hausdorff) topological space $X$, we define $\mathcal{V}(X)$ to be the space of closed subsets of $X$ with the least topology containing $\Box U = \{ C \mid C \subseteq U \} \quad\text{and}\quad \Diamond U = \{ C \mid C \cap U \neq \emptyset \}\quad (\forall \text{open }U).$ The pointfree analogue was introduced by Johnstone. Given a frame $L$, its Vietoris powerlocale $\mathbb{V}(L)$ is defined as the free frame on the set $\{ \Box a, \Diamond a \mid a\in L\}$ subject to the following equations

$\begin{align*} \Diamond (a \vee b) = \Diamond a \vee \Diamond b & \qquad \Box (a \wedge b) = \Box a \wedge \Box b\\ \Diamond 0 = 0 & \qquad \Box 1 = 1\\ \Diamond (\bigvee_i a_i) = \bigvee_i (\Diamond a_i) & \qquad \Box (\bigvee_i a_i) = \bigvee_i (\Box a_i) \quad (\text{for } \{a_i\}_i \text{ directed}) \\ \Box a \wedge \Diamond b \leq \Diamond (a \wedge b) & \qquad \Box (a \vee b) \leq \Box a\vee \Diamond b \end{align*}$

Johnstone then proved that, if $X$ is a compact Hausdorff space which is dual to the frame $L$, then $\mathcal{V}(X)$ is also dual to $\mathbb{V}(L)$.

## The points of the pointfree Vietoris

In fact, Johnstone showed that the dual space of the Vietoris frame $\mathbb{V}(L)$ can be computed directly (and constructively) without talking about the points of $L$. Assuming that $L$ is compact and regular and that $X$ is its topological dual, the Vietoris space $\mathcal{V}(X)$ is homeomorphic to the space $\mathbb{V}_\dagger(L) = (L, \tau_{\mathbb{V}L})$ where the topology $\tau_{\mathbb{V}L}$ is generated by the elements of the form $\Box_\dagger x = \{ a\in L \mid x \vee a = 1 \} \quad\text{and}\quad \Diamond_\dagger x = \{ a\in L \mid x \nleq a \},$ for every $x\in L$. The intuition behind this is that instead of the space of closed set $\mathcal{V}(X)$ we have the space of opens $\mathbb{V}_\dagger(L)$ and then the expressions $K \subseteq U$ and $K \cap U \neq \emptyset$ are just rewritten accordingly in the language of open sets.

What Johnstone does not emphasise is that the mapping $L \mapsto \mathbb{V}_\dagger(L)$ extends to a functor (although, as I found much later, his Lemma 4.7 (ii) in *Stone spaces* gives a clue). Let $h\colon L \to M$ be a frame homomorphism between two compact regular frames and let $f\colon X \to Y$ be the corresponding dual continuous map.

Then, as I showed in Section 4.4 of my thesis, the map $\mathcal{V}(f)\colon \mathcal{V}(X) \to \mathcal{V}(Y)$ sending $K$ to $f[X]$ can be equivalently expressed as the map $h_*\colon \mathbb{V}_\dagger(M) \to \mathbb{V}_\dagger(L)$ where $h_*$ is the right adjoint to $h$, i.e. the localic map corresponding to $h$.

*Remark:* A similar formula can be also used for stably compact frames. The only difference is that we need to take the dual topology into account. For details, see Remark 4.4.1 and Lemma 4.4.6 in my thesis.

## Open maps

It was proved by Celia Borlido and Mai Gehrke that the preimage mapping is continuous w.r.t. the Vietoris topology precisely when the underlying map is open:

Let $f\colon X\to Y$ be a continuous map between compact Hausdorff spaces. The following statements are equivalent:

- $f\colon X\to Y$ is an open map.
- $f^{-1}\colon \mathcal{V}(Y)\to \mathcal{V}(X)$ is continuous.
- $f^{-1}\circ \eta\ \colon Y\to \mathcal{V}(X)$ is continuous (where $\eta\colon Y\to \mathcal{V}(Y)$ is the map $y \mapsto \{y\}$).

In the following I will give a pointfree proof of “(1) $\Rightarrow$ (2)”. Let us first recall recall some facts about open frame homomorphisms (note that the third characterisation has deep connections with logic):

Let $h\colon L \to M$ be a frame morphism. Then, the following are equivalent:

- $h_*$ maps open sublocales to open sublocales
- $h$ is a Heyting algebra homomorphism
- $h$ has a left adjoint $h_!$ satisfying the Frobenius identity: $h_!(a \wedge h(b)) = h_!(a) \wedge b$.
- $h$ has a left adjoint $h_!$ satisfying the identity: $h_*(a \to h(b)) = h_!(a) \to b$.

If, moreover, $h$ is the dual frame homomorphisms of a continuous map $f$ between two $T_D$ spaces, then $h$ is open iff $f$ is open.

This means that if a frame homomorphism $h$ is open then it is both left and right adjoint: $h_! \dashv h \dashv h_*$ Recall also that, if $h$ is dual to a continuous map $f\colon X\to Y$ between compact Hausdorff spaces, then we know that the map $h$ corresponds to taking the preimage $f^{-1}[-]$ and $h_*$ corresponds to taking forward image $f[-]$. With this in mind, let us now state and prove the pointfree version of the implication “(1) $\Rightarrow$ (2)” in Theorem 1:

Let $h\colon L \to M$ be an open homomorphism between compact regular frames. Then, $h$ seen as a map $\mathbb{V}_\dagger(L) \to \mathbb{V}_\dagger(M)$ is continuous.

First we show that the preimage of $\Diamond_\dagger x$, for $x\in M$, under the map $h$ is open: $h^{-1}[\Diamond_\dagger x] = \{ a\in L \mid x \nleq h(a) \} = \{ a\in L \mid h_!(x) \nleq a \} = \Diamond_\dagger h_!(x)$ Next, recall that in a regular frame the pseudocomplement $a^*$ of $a$ can be equivalently expressed as $\bigwedge \{ b \mid a \vee b = 1 \}$. In other words, we have that $a\vee b = 1$ iff $a^* \leq b$.

In the rest of the proof we show that $h^{-1}[\Box_\dagger x]$ is equal to $\Box_\dagger h_!(x^*)^*$. Observe that $a \in h^{-1}[\Box_\dagger x] \quad\text{iff}\quad h(a) \vee x = 1 \quad\text{iff}\quad x^* \leq h(a) \quad\text{iff}\quad h_!(x^*) \leq a.$ Therefore, if $a \in h^{-1}[\Diamond_\dagger x]$ then, because $M$ is compact and regular and because $h$ is a frame homomorphism, there exists an $a'\in M$ *rather bellow* $a$ (i.e. $a \vee (a')^* = 1$) such that $h_!(x^*) \leq a'$. Therefore $a \vee h_!(x^*)^* \geq a \vee (a')^* = 1.$ Conversely, if $a \vee h_!(x^*)^* = 1$, then $a' \vee h_!(x^*)^* = 1$ for some $a'$ rather below $a$. Because $h$ is an open homomorphism, we see that $\begin{align}
a' \vee h_!(x^*)^* = 1
&\quad\text{iff}\quad (a')^* \leq h_!(x^*)^* \\
&\quad\text{iff}\quad h_!(h((a')^*)\wedge x^*) = (a')^* \wedge h_!(x^*) \leq 0 \\
&\quad\text{iff}\quad h((a')^*)\wedge x^* \leq h(0) = 0 \\
&\quad\text{iff}\quad x^* \leq h((a')^*)^* \\
&\quad\text{iff}\quad x \vee h((a')^{**}) = x \vee h((a')^*)^* = 1
\end{align}$ (We used the Frobenius identity in the second “iff” and the fact that $h$ is a Heyting algebra homomorphism in the last “iff”).

Lastly, since $a'$ is rather bellow $a$, it follows that $(a')^{* *} \leq a$ (and vice versa) and so $x \vee h(a) = 1$. We have proved that $h(a)\vee x = 1$ if and only if $a \vee h_!(x^*)^* = 1$, or in other words that $h^{-1}[\Box_\dagger x]$ is equal to $\Box_\dagger h_!(x^*)^*$.

Showing the opposite direction seems a bit more elaborate but I believe that it should be also doable. Assuming the Axiom of Choice we know that it is true as it follows from Theorem 1. Furthermore, the proof of the opposite direction is classically completely straightforward. Indeed, in Theorem 1, (1) => (2) follows automatically and to show (3) => (1) it is enough to observe that $(f^{-1} \circ \eta)^{-1}(\Diamond U) = f[U]$ which is open by (2). Anyway, I’ll leave it for the next time :-).

**Question 1:** Can we characterise the continuous maps $\mathcal{V}(X) \to \mathcal{V}(Y)$ (or equivalently $\mathbb{V}_\dagger(L) \to \mathbb{V}_\dagger(M)$) that arise this way?

**Question 2:** If $f\colon X\to Y$ is an open map, when is $\mathcal{V}(f)\colon\mathcal{V}(X) \to \mathcal{V}(Y)$ or $f^{-1}\colon \mathcal{V}(Y) \to \mathcal{V}(X)$ also open?

## Relationship to logic

Let $h\colon A \to B$ be a homomorphism of Boolean algebras and $f\colon X \to Y$ its continuous dual. Then, as Celia and Mai also proved,

$h$ has a left adjoint $h_!$ iff $f$ is open.

The left adjoint can be thought to express *existential quantification* (cf. hyperdoctrines for details). Note that it is not difficult to show that if the left adjoint $h_!$ exists then also the right adjoint $h_*$ of $h$ exists. It can be expressed by $h_*(a) = \neg h_!(\neg a)$. The right adjoint represents *universal quantification*.

From what we’ve discussed above we see that if $h$ has a left adjoint then the extension of $h$ to the homomorphism between the corresponding frames $\mathit{Idl}(A) \to \mathit{Idl}(B)$ automatically has a left adjoint which satisfies the Frobenius identity.

*Remark 1:* Geometric morphisms between toposes also introduce a double adjunction $f_! \dashv f^* \dashv f_*$, which then takes care of dependent quantifiers in topos type theory. See ncatlab for references.

*Remark 2:* Another property required by hyperdoctrines is the Beck-Chevalley condition. This is a global property of hyperdoctrines, as it requires commutativity of certain squares. The Beck-Chevalley condition together with the Frobenius identity make sure that subsitution is well-behaved with respect to existential quantification, see e.g. here.