Let $X$ and $Y$ be topological spaces, $X$ compact, $f : X \to Y$ continuous. Then the preimage of each compact subset of $Y$ is compact.
With the stipulation that $X$ and $Y$ are metric spaces, this is a theorem in Pugh's Real Mathematical Analysis. The proof uses sequential compactness. Is this theorem true in general (i.e. can it be proved with covering compactness alone)?
Best Answer
The claim is true if $Y$ is Hausdorff: if $C\subseteq Y$ is compact, then it is closed; therefore $f^{-1}(C)$ is closed in $X,$ hence compact.
For a counterexample, take $Y=X$ with the indiscrete topology and $f$ the identity map. Then every subset of $Y$ is compact, which can be easily arranged for $X$ not to.