Given an infinite set $X$ with the finite-complement topology, find a necessary and sufficient condition for a map $f:X\to X$ to be continuous.
I came up with the condition that $\lvert f^{-1}(\{x\})\rvert<\infty$ for all $x\in X$.
Edit: I forgot about the necessity of $f$ being surjective (which makes it a funny coincidence that I mentioned that in my example below).
Edit2: surjectivity is too strong; We need a condition like this to be able to argue as follows:
Assume $f$ is continuous, and that $f^{-1}(\{x\})$ is infinite. Then we consider $f^{-1}(X-\{x\})=X-f^{-1}(\{x\})$, and we need to conclude that this is now not-open (to derive a contradiction).
For this all we need is that $f^{-1}(\{x\})\not=X$, so this condition is enough (and much weaker that surjectivity).
However, I always find it annoying about these type of questions that there is always a trivial answer (in this case the condition that $f$ be continuous), so that I find it hard to judge if the answer I've given is not also similarly trivial.
So my question is if there are in a sense 'less trivial' conditions possible here. I realise that this is somewhat subjective, but I hope that people understand what I mean here.
For example an answer that seemingly has less direct connection to the finite-complement topology would be less trivial in my eyes, i.e. something like the condition that $f$ be surjective. This is 'less trivial' since it is at first sight 'less inspired' by the space $X$. To be clear: I don't mean to say that the surjectivety of $f$ is a correct answer, it was just an example.