What is the Gelfand dual of an open surjection?

Question

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.)

Cf. Reference request for translating from Top to C*-alg

Answer 1

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 $\pi_!(f)(y)<q$ then $f(x)<q$ for all $x \in \pi^{-1}(y)$, hence the lemma applied to $U= \{x | f(x) <q\}$ 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.