Does a map between topologies determine a map between sets?

Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be Hausdorff spaces. Consider a function \begin{equation*} \phi:\mathcal{B}\rightarrow \mathcal{A} \end{equation*} which preserves inclusion, arbitrary unions, finite intersections, and satifies $\phi(\emptyset)=\emptyset, \phi(Y)=\phi(X)$.

Does there exist $f: X\rightarrow Y$ such that $\phi= f^{-1}$ ?

I know that if such an $f$ exists it is uniquely determined by $\displaystyle f^{-1}(y)=\bigcap_{O\in \mathcal{B},y\in O} \phi(O)$. I also know this gives an effective definition for $f$ satisfying $f^{-1}=\phi$ if \begin{equation*} \bigcup_{y\in O}\left(\bigcap_{O'\in \mathcal{B},y\in O'}\phi(O')\right)=\phi(O) \end{equation*} for all open sets $O\subset Y$. But I don't know if this is necesarily the case.

I'm going to answer my own question, and I'm madly delighted to say the answer is yes, there always is such an $f$.

Note that for all $x\in X$ the set $\displaystyle N(x)=\bigcup_{O\in \mathcal{B}, x\notin \phi(O)}O$ has the form $Y\setminus\{y\}$. Indeed

• Suppose that $N(x)$ is all of $Y$. Then we would have $X=\phi(Y)=\phi(N(x))=\bigcup_{O\in \mathcal{B}, x\notin \phi(O)}\phi(O)$ and $x\notin X$ which is absurd.
• Suppose there were distinct $y_{1},y_{2}$ not in $N(x)$. Then there are two disjoint sets $O_{1},O_{2}$ containing $y_{1}$ and $y_{2}$ respectively. The sets $\phi(O_{1})$ and $\phi(O_{2})$ are disjoint so they cannot both contain $x$. Without loss of generality $x\notin \phi(O_{1})$ so that $O_{1}\subset N(x)$ and $y_{1}\in N(x)$ which is again absurd.

Define $f$ by letting $f(x)$ be the only element of $Y\setminus N(x)$. We have \begin{align*} &x\in f^{-1}(U) \\ \iff &U \not\subset N(x) \\ \iff &x\in\phi(U) \end{align*} hence $f$ has the desired property.

I don't know what interpretations there might be of this in terms or pointless topology, or topos theory (I suggest this because it seems to me the proof has a propositional logical flavour).