# 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.

• If there is such an $f$, then for any $y$, the set of $X_y=X\setminus\phi(Y\setminus\{y\})$ is the set of values $x\in X$ such that $f(x)=y$. The the question is, is it possible for two $y_1\neq y_2$ to have $X_{y_1}\cap X_{y_2}\neq \emptyset$, or, alternatively, is it possible that sucn an $f$ is not continuous? Commented Feb 9, 2016 at 21:42
• +1, nice question. Where did you get this? :) Commented Feb 9, 2016 at 22:00
• @IvoTerek I made it up ! My motivations come from category theory which I'm studying at the moment. Commented Feb 10, 2016 at 0:13
• @ThomasAndrews Because $Y\setminus \{y_{1}\} \cup Y\setminus \{y_{2}\}=Y$ you have $X_{y_{1}}\cap X_{y_{2}}=\emptyset$. What seems more difficult to me is that $\bigcup X_{y}=X$ since I do not assume $\phi$ preserves arbitrary intersections. Commented Feb 10, 2016 at 0:16
• This might be related to the fact that $T_2$ spaces are sober, and the category of sober spaces is equivalent to that of spatial locales, that is, (the opposite of) lattices of open sets with maps as you defined. Check this Wikipedia article. Commented Feb 10, 2016 at 3:19

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.