# What is the Gelfand dual of an open surjection?

Does anyone have a handy characterisation of open continuous surjections $$X \to Y$$ in terms of the corresponding injective $$*$$-homomorphism $$C(Y) \to C(X)$$? (I'm only interested in the case where $$X$$ and $$Y$$ are compact Hausdorff spaces, but the question could be suitably modified for locally compact Hausdorff spaces.)

• A guess: this is equivalent to the fact that the inclusion of positive elements $C(Y)^+ \to C(X)^+$ admits a left adjoint (in the sens of the order relation). To give some context: I know from my work on the constructive Gelfand duality that (constructively) $X \to *$ is an open map if and only if the norm of each $f \in C(X)$ is continuous (instead of semi-continuous). I think, that should translate to the claim I made above. But I would need to think more about it, and probably there might be a direct proof not using all these ideas. Jun 7, 2021 at 17:12

After more thought, I think the correct statement is the following:

Theorem: Let $$\pi : X \to Y$$ be a continuous map between compact Hausdorff spaces. Then the following condition are equivalents:

(a) $$\pi$$ is an open map.

(b) The map $$\pi^* : C(Y)^+ \to C(X)^+$$ has a left adjoint which is $$C(Y)^+$$-linear, i.e. there is a (automatically unique) map $$\pi_! : C(X)^+ \to C(Y)^+$$ such that for all $$f \in C(X)^+$$ and $$h \in C(Y)^+$$ one has $$\pi_!(f) \leqslant h \Leftrightarrow f \leqslant \pi^*(h)$$ and $$\pi_!(f \pi^*(h)) = \pi_!(f) h$$.

Note: By $$C(X)^+$$ I mean the subset of selfadjoint positive elements, i.e. continuous function with values in non-negative real numbers.

Before proving the theorem, let me a recall:

Lemma : Given a continuous map $$\pi:X \to Y$$ between comapct Hausdorff space, if $$y \in Y$$ and $$U$$ is an open containing the fiber $$\pi^{-1}(y)$$, then there exists a neighborhood $$V$$ of $$y$$ such that $$\pi^{-1}(V) \subset U$$.

Proof: It feels like it is a form of openness, but it is actually related to properness. Indeed, let $$K$$ be the closed complement of $$U$$, $$K$$ is compact, so $$f(K)$$ is a compact of $$Y$$ not containing $$y$$, hence the complement $$V$$ of $$f(K)$$ is an open neighborhood of $$y$$, and by construction, its preimage is included in $$U$$.

Proof: (a) $$\Rightarrow$$ (b)

For $$f$$ a positive functions on $$X$$, We define $$\pi_!(f)$$ as a function $$Y \to \mathbb{R}$$ by:

$$\pi_!(f)(y) = \sup_{x \in \pi^{-1}(y)} f(x)$$

The sup is taken in the poset of positive real numbers, so it is $$0$$ if the fiber is empty, it is always finite by compactness of the fiber, so that $$\pi_!(f)(y)$$ is indeed a positive real number for all $$y$$.

It is immediate that $$\pi_!(f \pi^*(h))= \pi_!(f) h$$ and $$\pi_!(f) \leqslant h$$ if and only if $$f \leqslant \pi^*(h)$$, what is not so clear is whether $$\pi_!(f)$$ is indeed an element of $$C(X)^+$$, i.e. is continuous.

The map $$\pi_!(f)$$ is always an upper semi-continuous functions. It only rely on the lemma above: If $$\pi_!(f)(y) then $$f(x) for all $$x \in \pi^{-1}(y)$$, hence the lemma applied to $$U= \{x | f(x) immediately implies that there is a neighborhood $$V$$ of $$y$$ such that $$\pi^{-1}(V)$$ is included in $$U$$, and hence $$\pi_!(f)(t) \leqslant q$$ for all $$t \in V$$.

The fact that $$\pi_!(f)$$ is lower semi-continuous is an immediate consequence of the openness of $$\pi$$: if $$\pi_!(f)(y)>q$$, it means there is an element $$x_0 \in \pi^{-1}(y)$$ such that $$f(x_0)>q$$, hence there is a neighborhood of $$x_0$$ such that $$f(x)>q$$, the image of this neighborhood is a neighborhood of $$y$$ on which $$\pi_!(f)>q$$.

Together they implies that $$\pi_!(f)$$ is continuous which concludes the proof.

(b) $$\Rightarrow$$ (a)

Assume we have such a map $$\pi_!$$. For $$f$$ a continuous positive function, I write $$\{f>0\}$$ for the open subset $$\{ x | f(x) >0 \}$$, these forms a basis of the topology, so it is enough to check that the direct image of these by $$\pi$$ are open subsets. We will show that $$\pi_!$$ has to be of the form defined in the previous part of the proof. It then imediately follows that $$\pi\{f>0\} = \{ \pi_! f >0 \}$$ and this concludes the proof.

First one show that $$\pi_!(f)(y) \geqslant \sup_{x \in \pi^{-1} y} f(x)$$.

Indeed, assume that $$\pi_!(f)(y) \leqslant a$$. Then $$\forall \epsilon>0$$ there exists a positive function $$\chi$$ such that $$\chi(y)=1$$ and $$\chi \pi_!(f) \leqslant (a+\epsilon)\chi$$ (take $$\chi$$ to be $$1$$ in a small neighborhood of $$x$$, and $$0$$ away from $$x$$).

The adjunction formula give you that $$\pi^*(\chi) f \leqslant (a+ \epsilon)\pi^*(\chi)$$.

Now for any $$x \in \pi^{-1}(y)$$, evaluating the previous inequality at $$x$$ gives $$f(x) \leqslant a+ \epsilon$$, hence $$\sup_{x \in \pi^{-1}(y)} f(x) \leqslant a$$.

Conversely, assume that $$\sup_{x \in \pi^{-1} y} f(x) \leqslant a$$, i.e. $$f(x) \leqslant a$$ everywhere on the fiber of $$a$$, for all $$\epsilon>0$$ there is an open neighborhood $$U$$ of $$\pi^{-1}(y)$$ on which $$f(x) < a + \epsilon$$.

Applying the lemma again, we get an open neighborhood $$V$$ of $$y$$ such that $$\pi^{-1}(V)$$ is included in $$U$$. Take $$\chi$$ such that $$\chi(y)=1$$ and $$\chi$$ is $$0$$ outside of $$V$$ and run the same argument as above, you have that $$\pi^*(\chi) f \leqslant a+\epsilon$$, hence $$\pi^*(\chi) f \leqslant \pi^*(a+ \epsilon)$$, hence $$\pi_!(\pi^*(\chi) f) \leqslant a+ \epsilon$$, and finally $$\chi \pi_!(f) \leqslant a + \epsilon$$

evaluating at $$y$$ gives $$\pi_!(f)(y) \leqslant a+ \epsilon$$, which conclude the proof.

• Great, thank you! Jun 8, 2021 at 20:37