Proper map from continuous if it maps compact sets to compact sets

Question

Suppose that $X,Y$ are - say - normed vector spaces.

Let $f:X\to Y$ be such that for all sets $K\subset X$: $$f(K) \text{ compact}\iff K\text{ compact}.$$

Does it hold that $f$ is continuous?

Addendum: What if $X,Y$ are Hausdorff spaces?

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Intuitively it seems that this is wrong because there are both proper maps which are not continuous and maps that compact sets to compact sets which are also not continuous.

However, if $f$ is a compact injection then $f$ is continuous.

Normed vector spaces are Hausdorff. – Paul Frost Oct 17 '19 at 14:19 — Paul Frost, Oct 17 '19 at 14:19
@PaulFrost No shit, Sherlock. – amsmath Oct 17 '19 at 14:20 — amsmath, Oct 17 '19 at 14:20
@amsmath Thank you, Watson. – Paul Frost Oct 17 '19 at 14:32 — Paul Frost, Oct 17 '19 at 14:32

score 0 · Accepted Answer · answered Oct 18 '19 at 17:22

Yes, it is continuous. And it is enough that $X$ is Hausdorff compactly generated and $Y$ is Hausdorff.

If $X$ is compactly generated and we want to prove the continuity of $f$, then without loss of generality, $X$ is compact (as in the answer of the linked question), and so $f(X)$ is compact. Every closed $F ⊆ f(X)$ is compact, and so $f^{-1}(F)$ is compact since $f(f^{-1}(F)) = F$. Since $X$ is Hausdorff, $f^{-1}(F)$ is closed. That proves the continuity.

Moreover, we haven't used the full power of our assumption yet. For every $y ∈ Y$, the fiber $f^{-1}(y)$ is compact and discrete (every $A ⊆ f^{-1}(y)$ is compact and so closed since $f(A) ⊆ \{y\}$ is compact). Hence it is finite, and $f$ is finite-to-one.

Proper map from continuous if it maps compact sets to compact sets

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