Let $X$ and $Y$ be topological spaces taken with the cofinite topology. Show that $f:X\to Y$ is continuous if and only if it is constant or $\forall y\in Y$, $f^{-1}(\{y\})$ is finite.
What I have tried:
If $f$ is constant, then $f$ is continuous so we are done. Suppose $\forall y\in Y$, $f^{-1}(\{y\})$ is finite. Let $V$ be an open set of $Y$, then
$$f^{-1}(V)=f^{-1}\left(\bigcup_{y\in V}\{y\}\right)=\bigcup_{y\in V}f^{-1}(\{y\})$$
so that
$$X-f^{-1}(V)=\bigcap_{y\in Y}(X-f^{-1}(\{y\}))$$
now $f^{-1}(\{y\})$ is finite but i'm not sure how to show that the term $\bigcap_{y\in Y}(X-f^{-1}(\{y\}))$ is finite. I'm not sure how to prove the other direction either.
Best Answer
It would be easier to observe that if $V$ is a non-empty open set in $Y$, then $Y\setminus V$ is finite, so $f^{-1}[Y\setminus V]$ is also finite: it’s the union of finitely many finite sets. Thus, $f^{-1}[V]=X\setminus f^{-1}[Y\setminus V]$ is cofinite and hence open in $X$.
Once you’ve finished this off, you still have to show that if $f$ is continuous, then $f$ either is constant or has finite fibres. (The fibres of a function are the inverse images of singletons.) I suggest that you assume that $f$ is not constant and has an infinite fibre and show that in that case $f$ is not continuous.