If f:$X\rightarrow Y$ is continuous and onto and $X$ is a metric space Y Hausdorff, then $Y$ is metrizable iff $X$ is compact

Question

I came across this question recently:

“Let $X$ be a metric space, $Y$ Hausdorff, and let $f: X \rightarrow Y$ be continuous and onto. Then $Y$ is metrizable iff $X$ is compact.”

I believe I know how to do show the reverse direction, here’s an outline of my idea:

By Urysohn’s Metrization Theorem, it’s sufficient to show Y is regular and second countable. X is compact and f is continuous, so Y is compact. Y is a compact Hausdorff space, so Y is regular. X is a compact metric space, so X is second countable. To show Y is second countable, let C be the set of all finite unions of elements from the countable base for X. Then X-c is closed for any c in C. This X-c is compact, so f(X-c) is compact, and thus closed in Y. Then it’s straightforward to show the set of all Y-f(X-c) such that c is in C is a countable base for Y.

For the other direction, I’m struggling to find a place to start. Could anyone give me a hint for this direction?

Answer 1

The other direction is false. Let $X$ be any non-compact metric space. Then the identify function $X \to X$ is continuous and onto, and $X$ is Hausdorff and metrisable but not compact.