Can we build a continuous function from "fibers"/preimages defined over a topological base?

Question

I am looking for some proof insight or literature references for a statement which, if it's actually true, is probably a pretty trivial thing. I hope the question has not been asked here before, and that my notation is not too strange. Any help is appreciated!

Let $X, Y$ be topological spaces. Let $A$ be the collection of open sets of $X$, let $B$ be the collection of open sets of $Y$, and let $B^*$ be a base for $Y$. Let $F^*: B^* \to A$ be a function such that

1. $F^*(Y) = X$.
2. $F^*(\emptyset) = \emptyset$.
3. $F^*(U\cap V) = F^*(U) \cap F^*(V)$ for $U, V \in B^*$.

Question: Can we extend $F^*$ to a function $F : B \to A$ that commutes with arbitrary unions and finite intersections? Is this extension unique?

Stated from a different perspective, is there a continuous function $f: X \to Y$ such that $f^{-1}(U) = F^*(U)$ for each $U$ in $B^*$? If it exists, how do we define it on individual points of $X$? Is $f$ unique?

What I see so far: I can think of at least a partial answer to my questions.

First note that by definition, $F^*$ preserves finite intersections from $B^*.\ $ For $U_i$ an arbitrary family of open sets in $B^*$, define $F(\cup U_i) = \cup F^*(U_i)$. This extends $F^*$ to all of $B$, and obviously commutes over arbitrary unions. To see it commutes over finite intersections let $U_i, V_j$ be arbitrary families of open sets in $B^*$ and consider

$$F((\cup U_i)\cap(\cup V_j)) = F(\bigcup_i \bigcup_j U_i \cap V_j) = \bigcup_i \bigcup_j F^*(U_i)\cap F^*(V_j)$$ $$=(\cup F^*(U_i)) \cap (\cup F^*(V_j))$$ $$=F(\cup U_i)\cap F(\cup V_j)$$

How can we decide whether $F$ is unique? And can we take F at large an use it to construct $f$'s action on individual points of $X$?

Answer 1

The answer to the general question

is there a continuous function $f:X\to Y$ such that $f^{-1}(U)=F^*(U)$ for each $U\in B^*$

is "No". Consider the following example:

Let $X=Y=\mathbb{N}$ and let $L = \{S \subseteq \mathbb{N}: 1\in S \textrm{ and }\mathbb{N}\setminus S \textrm{ is finite}\}$.

Put $A = B^* = L \cup \{\emptyset\} \cup \{\{1\}\}$. It is easy to verify that $A= B^*$ is a topology (so in particular, $B^*$ is a base for the topology $B^*$). Let $F^*:B^*\to A$ be defined by

• $F^*(S) = S$ for all $S\in L$;

• $\{1\} \mapsto \emptyset$ and $\emptyset \mapsto \emptyset$.

Then $F^*$ obeys the conditions 1. to 3. in the question, but note that $F^*$ does not commute with arbitrary intersections, because $F^*(\bigcap L) = F^*(\{1\}) = \emptyset \neq \{1\} = \bigcap F^*(L)$. Now, for any $f:X\to Y$ the map $f^{-1}(\cdot)$ commutes with arbitrary intersections. So there cannot be $f:\mathbb{N}\to\mathbb{N}$ such that $f^{-1}(U) = F^*(U)$ for all $U\in B^*$.