Vietoris and open maps

The topological analogue of the powerset monad is the Vietoris hyperspace. Given a (compact Hausdorff) topological space XX, we define 𝒱(X)\mathcal{V}(X) to be the space of closed subsets of XX with the least topology containing U={CCU}andU={CCU}(open U). \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 LL, its Vietoris powerlocale 𝕍(L)\mathbb{V}(L) is defined as the free frame on the set {a,aaL}\{ \Box a, \Diamond a \mid a\in L\} subject to the following equations

(ab)=ab(ab)=ab0=01=1(iai)=i(ai)(iai)=i(ai)(for {ai}i directed)ab(ab)(ab)ab\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 XX is a compact Hausdorff space which is dual to the frame LL, then 𝒱(X)\mathcal{V}(X) is also dual to 𝕍(L)\mathbb{V}(L).

The points of the pointfree Vietoris

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

What Johnstone does not emphasise is that the mapping L𝕍(L)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:LMh\colon L \to M be a frame homomorphism between two compact regular frames and let f:XYf\colon X \to Y be the corresponding dual continuous map.

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

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

  1. f:XYf\colon X\to Y is an open map.
  2. f1:𝒱(Y)𝒱(X)f^{-1}\colon \mathcal{V}(Y)\to \mathcal{V}(X) is continuous.
  3. f1η:Y𝒱(X)f^{-1}\circ \eta\ \colon Y\to \mathcal{V}(X) is continuous (where η:Y𝒱(Y)\eta\colon Y\to \mathcal{V}(Y) is the map y{y}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:LMh\colon L \to M be a frame morphism. Then, the following are equivalent:

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

If, moreover, hh is the dual frame homomorphisms of a continuous map ff between two TDT_D spaces, then hh is open iff ff is open.

This means that if a frame homomorphism hh is open then it is both left and right adjoint: h!hh* h_! \dashv h \dashv h_* Recall also that, if hh is dual to a continuous map f:XYf\colon X\to Y between compact Hausdorff spaces, then we know that the map hh corresponds to taking the preimage f1[]f^{-1}[-] and h*h_* corresponds to taking forward image f[]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:LMh\colon L \to M be an open homomorphism between compact regular frames. Then, hh seen as a map 𝕍(L)𝕍(M)\mathbb{V}_\dagger(L) \to \mathbb{V}_\dagger(M) is continuous.

First we show that the preimage of x\Diamond_\dagger x, for xMx\in M, under the map hh is open: h1[x]={aLxh(a)}={aLh!(x)a}=h!(x) 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*a^* of aa can be equivalently expressed as {bab=1}\bigwedge \{ b \mid a \vee b = 1 \}. In other words, we have that ab=1a\vee b = 1 iff a*ba^* \leq b.

In the rest of the proof we show that h1[x]h^{-1}[\Box_\dagger x] is equal to h!(x*)*\Box_\dagger h_!(x^*)^*. Observe that ah1[x]iffh(a)x=1iffx*h(a)iffh!(x*)a. 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 ah1[x]a \in h^{-1}[\Diamond_\dagger x] then, because MM is compact and regular and because hh is a frame homomorphism, there exists an aMa'\in M rather bellow aa (i.e. a(a)*=1a \vee (a')^* = 1) such that h!(x*)ah_!(x^*) \leq a'. Therefore ah!(x*)*a(a)*=1. a \vee h_!(x^*)^* \geq a \vee (a')^* = 1. Conversely, if ah!(x*)*=1a \vee h_!(x^*)^* = 1, then ah!(x*)*=1a' \vee h_!(x^*)^* = 1 for some aa' rather below aa. Because hh is an open homomorphism, we see that ah!(x*)*=1iff(a)*h!(x*)*iffh!(h((a)*)x*)=(a)*h!(x*)0iffh((a)*)x*h(0)=0iffx*h((a)*)*iffxh((a)**)=xh((a)*)*=1\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 hh is a Heyting algebra homomorphism in the last “iff”).

Lastly, since aa' is rather bellow aa, it follows that (a)**a(a')^{* *} \leq a (and vice versa) and so xh(a)=1x \vee h(a) = 1. We have proved that h(a)x=1h(a)\vee x = 1 if and only if ah!(x*)*=1a \vee h_!(x^*)^* = 1, or in other words that h1[x]h^{-1}[\Box_\dagger x] is equal to h!(x*)*\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 (f1η)1(U)=f[U](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 𝒱(X)𝒱(Y)\mathcal{V}(X) \to \mathcal{V}(Y) (or equivalently 𝕍(L)𝕍(M)\mathbb{V}_\dagger(L) \to \mathbb{V}_\dagger(M)) that arise this way?

Question 2: If f:XYf\colon X\to Y is an open map, when is 𝒱(f):𝒱(X)𝒱(Y)\mathcal{V}(f)\colon\mathcal{V}(X) \to \mathcal{V}(Y) or f1:𝒱(Y)𝒱(X)f^{-1}\colon \mathcal{V}(Y) \to \mathcal{V}(X) also open?

Relationship to logic

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

hh has a left adjoint h!h_! iff ff 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!h_! exists then also the right adjoint h*h_* of hh exists. It can be expressed by h*(a)=¬h!(¬a)h_*(a) = \neg h_!(\neg a). The right adjoint represents universal quantification.

From what we’ve discussed above we see that if hh has a left adjoint then the extension of hh to the homomorphism between the corresponding frames 𝐼𝑑𝑙(A)𝐼𝑑𝑙(B)\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!f*f*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.

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