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.