For a function $f: X \to Y$, if $Y-V$ is finite, when is $X - f^{-1}(V)$ finite? I apologize if this is a silly question but I just do not know enough set theory (i.e., sizes) to understand if it's even silly.
My question is

Let $f: X \to Y$ be a function of sets. Suppose $V \subset Y$, and that $Y - V$ is finite (countable). Is $X - f^{-1}(V)$ finite (countable)? What conditions do we need to place on $f$ to guarantee it will be finite (countable)?

So I kind of have two questions. But I will take either.
Why I care:
I'm asking this because I want to know the following. Recall that for a set $X$, we can endow $X$ with the finite complement topology where a set $U \subset X$ is open if $X - U$ is finite. Denote this topology on a set $X$ as $FC_X$.
Now suppose $f: X \to Y$ is a function. As $X$ and $Y$ are sets, we may ask:

Does the function $f: X \to Y$ extend to a continuous function $f: (X, FC_X) \to (Y, FC_Y)$?

For such a function to be continuous, we need that $f^{-1}(U)$ is open if $U$ is open. In this case, that means that if $Y - U$ is finite then $X - f^{-1}(U)$ must be finite. But I do not know when that last statement holds, and hence it's my question above. My guess is that if $f$ is injective, then it will be true, but I don't really know if that's true. I lack the set theory knowledge to really attack such a problem.
Finally, if the above answer is true, it tells me that I've got a functor $F: \textbf{Set} \to \textbf{Top}$ where $X$ is sent to the finite complement topology. I'm also interested in the countable case, since that would give me another different functor. But that's besides the question.
 A: As pointed out by @Cronus in the comments, it is sufficient that $f$ has the property that the preimage of every point is a finite set. I believe it is also a necessary condition.
Let's just consider surjective functions, just for convenience (it really doesn't change anything since any function $f:X \to Y$ can be 'made surjective' by replacing its codomain with its image.)
To see that the condition is sufficient, note that if some $f: X \to Y$ has the property, then for finite $Y \setminus V$,
$$ X \setminus f^{-1}(V) = f^{-1}(Y \setminus V) = \bigcup_{y \in Y \setminus V} f^{-1}(y).$$
In other words, $X \setminus f^{-1}(V)$ is the union of a finite number of finite sets.
I believe this condition is also necessary, and here is a proof that seems to be correct. Assume that $f$ doesn't have the property, and thus there exists some $y$ with $f^{-1}(y)$ not finite. Taking $V = Y \setminus \{ y \}$, one then concludes that
$$ X \setminus f^{-1}(V) = f^{-1}(y),$$
which is not finite. Importantly, we have that $f^{-1}(V)$ does not intersect $f^{-1}(y)$ since no $x \in X$ can be mapped both into $V$ and into $\{ y \}$. Thus by contradiction we have that $f$ must have the finite-to-one property.
A: Let $X$ be an infinite set, $Y$ a finite set and $V = \emptyset$. Then for any function $f : X \to Y$ we have that $Y\setminus V = Y$ is finite but $X \setminus f^{-1}(V) = X$ is infinite.
We get a similar conclusion for your other question if $X$ is an uncountable set and $Y$ is a countable set.
