Open maps of spaces vs open maps of locales

Question

A continuous function between topological spaces $f:X\to Y$ is called open, if $f[U]\in\mathcal{O}(Y)$ for all $U\in\mathcal{O}(X)$.

A morphism of locales $f:X\to Y$ is called open, if the associated morphism of frames $f^*:\mathcal{O}(Y)\to\mathcal{O}(X)$ has a monotone left adjoint $f_!$ which satisfies the Frobenius condition $f_!(U)\wedge V\leq f_!(U\wedge f^*(V))$ for all $U\in\mathcal{O}(X)$ and $V\in\mathcal{O}(Y)$.

It is easy to see that every continuous $f$ that is open in the topological sense is also open in the localic sense, but the converse is false in general. For example, the maps $1\to\nabla(2)$ and $\mathsf{cofin}(\mathbb{N})\to\mathsf{sobrify}(\mathsf{cofin}(\mathbb{N}))$ are open in the localic, but not in the topological sense, since they're not surjective, but the induced locale-morphisms are iso.

These two examples involve non-sober spaces, so the natural thing to ask is: Do the two notions coincide for sober spaces? Or more generally, does the canonical functor from locales to spaces preserve open maps?

Or are there other criteria for when the two notions coincide? For example, I think I have convinced myself that continuous maps that are open in the localic sense are open in the topological sense whenever the codomain is $T_1$.

Answer 1

Here is a counterexample with sober spaces. Let $X=\mathbb{N}\times\mathbb{N}$, equipped with a topology defined as follows. Fix an injection $i:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$. Say that a set $U\subseteq X$ is open if for each $(a,b)\in U$, $U$ contains $(i(a,b),c)$ for all but finitely many $c\in\mathbb{N}$.

I claim that $X$ is Hausdorff, and thus sober. Let $p,q\in X$ be distinct points. Start with $U_0=\{p\}$ and $V_0=\{q\}$. Recursively define $U_{n+1}=U_n\cup ((i[U_n]\times\mathbb{N})\setminus\{q\})$ and $V_{n+1}=V_n\cup((i[V_n]\times\mathbb{N})\setminus\{p\})$. It is clear that $U=\bigcup U_n$ and $V=\bigcup V_n$ are open, since if $(a,b)\in U_n$ then $U_{n+1}$ contains $(i(a,b),c)$ for all but finitely many $c$, and similarly for $V$. I claim that $U_n$ and $V_n$ are disjoint for all $n$, and thus $U$ and $V$ are disjoint. Indeed, note that $U_n$ has the property that if $(a,b)\in U_n$ then either $(a,b)=p$ or $a=i(x)$ for some $x\in U_{n-1}$, and similarly for $V_n$. Thus $U_n$ and $V_n$ are disjoint by induction on $n$: if we already know $U_{n-1}$ and $V_{n-1}$ are disjoint, then all the points of $U_n$ (except $p$) have distinct first coordinate from all the points of $V_n$ (except $q$).

Now let $Y=\mathbb{N}\cup\{g\}$ where $\mathbb{N}$ has the cofinite topology and $g$ is the generic point. Define $f:X\to Y$ by $f(a,b)=b$. For any nonempty open $U\subseteq X$, $f[U]$ is cofinite in $\mathbb{N}$, and so there is a smallest open set of $Y$ that contains it, namely $f_!(U)=f[U]\cup\{g\}$. To verify the Frobenius condition, suppose $x\in f_!(U)\cap V$ for $U$ open in $X$ and $V$ open in $Y$. If $x\in\mathbb{N}$, then $(a,x)\in U$ for some $a\in U$ and then this $(a,x)$ is in $U\cap f^{-1}(V)$, witnessing that $x\in f_!(U\cap f^{-1}(V))$. If $x=g$, then note that $U$ and $V$ are both nonempty; let $(a,b)\in U$. Then $(i(a,b),c)\in U$ for all but finitely many $c$, and also $(i(a,b),c)\in f^{-1}(V)$ for all but finitely many $c$ since $V$ is cofinite. Thus $U\cap f^{-1}(V)$ is nonempty, and thus $g\in f_!(U\cap f^{-1}(V))$.

So, the map $f:X\to Y$ is open in the localic sense. However, it is not open in the topological sense, because $f[X]=\mathbb{N}$ is not open in $Y$.

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On the other hand, you are correct that the two notions coincide if $Y$ is $T_1$. First, the existence of the left adjoint $f_!$ just means that for each $U\in\mathcal{O}(X)$, there is a smallest open set $f_!(U)\in\mathcal{O}(Y)$ that contains $f[U]$. If $Y$ is $T_1$, this immediately implies $f_!(U)=f[U]$ so $f$ is open in the topological sense: if $x\in f_!(U)\setminus f[U]$ then $f_!(U)\setminus\{x\}$ still contains $f[U]$ and is open, contradicting the definition of $f_!(U)$.

A bit more generally, they coincide if $Y$ is $T_D$, meaning that for each $y\in Y$, $\{y\}$ is open as a subset of $\overline{\{y\}}$. To prove this, note that a similar argument as in the paragraph above shows that every point of $f_!(U)$ must be a generization of a point of $f[U]$. So, to show that $f$ is open in the topological sense, it suffices to show that $f[U]$ is specialization-open for any $U\in\mathcal{O}(X)$. Suppose $y\in Y$ is a generization of a point of $f[U]$. Since $Y$ is $T_D$, there exists $V\in \mathcal{O}(Y)$ that contains $y$ but does not contain any other specialization of $y$. Now the Frobenius condition says $y\in f_!(U)\cap V\subseteq f_!(U\cap f^{-1}(V))$, so there is a point $x\in U\cap f^{-1}(V)$ such that $y$ is a generization of $f(x)$. Since $f(x)\in V$, this means $f(x)=y$, so $y\in f[U]$, as desired.

Answer 2

Sketches of an Elephant notes (before Lemma C1.5.3)

Of course, we say a [locale morphism] $f: X \to Y$ is open if, for every open sublocale $U$ of $X$, its image $f_!(U)$ in $\mathrm{Sub}(Y)$ is an open sublocale of $Y$....this is not quite equivalent to the usual notion of openness for continuous maps of spaces; it is implied by the latter, but in the opposite direction it only implies that the set-theoretic image of each open subset of $X$ is 'almost open' in $Y$, in the sense that its subclosure is open. Even when X and Y are both sober, this does not suffice to make all such images open..., but it obviously does so if $Y$ is a $T_D$-space.

The distinction here seems to correlate with the distinction between subspaces and sublocales. The latter extends to a distinction between image of a subspace and image of a sublocale, which are invoked in the definition of open map of spaces and locales, respectively.