question related to perfect maps preserving compactness

Question

A perfect map $\text{f}$ is a closed continuous surjective function such that the preimage of every point is compact. One property of perfect maps is that if $f \, \colon \, X \to Y$ is perfect, and $\text{Y}$ is compact, then $\text{X}$ is compact too.

My question (rephrased): if $\text{f}$ is a continuous surjective function such that the preimage of every point is compact, and $\text{Y}$ is compact, does it follow that $\text{X}$ is compact?

Answer 1

No.

Let $Y$ be the closed unit interval with the usual Euclidean topology, and let $X$ be the closed unit interval with the discrete topology. Let $f:X\to Y$ be the identity map. Clearly $f$ is continuous and surjective and has compact fibres, and $Y$ is compact, but $X$ is not. Note that $X$ and $Y$ are even metrizable, so they’re about as nice as one could hope for.