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Suppose $X$ is a topological space. We have the following criterion for compactness:

Theorem. $X$ is compact if and only if for every space $Y$, the second projection $\pi_2: X\times Y \to Y$ is a closed map.

This property is known as being universally closed, and also plays an important role in algebraic geometry (in the definition of a proper morphism). The proof of the result above can be found on this MSE thread.

My question is whether we can strengthen this result by only asking that the graphs of continuous functions have closed image in $Y$. From now on, we will consider only Hausdorff topological spaces.

Question. Is it true that a Hausdorff space $X$ is compact if and only if for every continuous map $f: X\to Y$ with $Y$ Hausdorff has a closed image $f(X)$ in $Y$.

More context for the question: let $X$ and $Y$ be any two Hausdorff spaces. For any continuous function $f: X\to Y$ we can consider its graph $\Gamma(f) = \{(x, y)\in X\times Y: y=f(x)\}$. Note that $\Gamma(f)\subset X\times Y$ is a closed subset of $Y$ (see this MSE thread for proof). If $X$ were compact, then we know that the image of $\Gamma(f)$ under the second projection map $X\times Y\to Y$ would be closed. Note that the image of $\Gamma(f)$ is precisely the image $f(X)=\{f(x)\in Y: x\in X\}$, and so $f(X)$ would be closed in $Y$. This shows that the forward implication is true (one can show this implication in a more direct away). It makes sense to ask whether the converse also holds.

Prism
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    This is related: https://en.wikipedia.org/wiki/H-closed_space – Cronus Jul 05 '20 at 20:08
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    Moreover, if your space is $T_{3.5}$, you can embed it in a compactification; since its image is by assumption closed, it means your space really is compact. I imagine there is a counterexample for non T3.5 spaces. – Cronus Jul 05 '20 at 20:10
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    The Wikipedia article on Stone–Čech compactification mentions the existence of a Hausdorff space on which any continuous real-valued function is constant, although it does not demonstrate such one and only cites a book as a reference. – Sangchul Lee Jul 05 '20 at 20:12
  • @SangchulLee There are descriptions of such a space on MSE if you look for them.. – Henno Brandsma Jul 05 '20 at 21:24
  • @Henno Could you provide a reference for one of those MSE threads? I couldn't locate them myself. Thank you in advance! – Prism Jul 07 '20 at 03:13
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    @SangchulLee this therad has some references and descriptions. So we can take $Y=\Bbb R$ and we get and $X$ (regular $T_1$) that only has constant maps into $\Bbb R$. Nothing special about the reals. – Henno Brandsma Jul 07 '20 at 04:52

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A Hausdorff space $X$ is H-closed if, for every Hausdorff space $Y$ and a topological embedding $f:X\to Y$, the image $f(X)$ is closed.

Lemma. A Hausdorff space $X$ is H-closed if and only if for every Hausdorff space $Y$ and a continuous map $f:X\to Y$, the image $f(X)$ is closed.

Lemma. The topological space $[0,1]$, with the smallest topology containing the both the standard one and the set $\Bbb{Q}\cap [0,1]$, is H-closed.

The latter space is not compact, since it contains $[0,1]\setminus\Bbb{Q}$ (with the standard topology) as a closed subset, while $[0,1]\setminus\Bbb{Q}$ is not compact (with the standard topology).

Cronus
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    Why does H-closed imply that every continuous function to a Hausdorff space has closed image? –  Jul 05 '20 at 20:39
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    there's a proof here: https://mathoverflow.net/a/237974. In fact it is more or less the same question... – Cronus Jul 05 '20 at 21:05
  • Ok. That is neat. –  Jul 05 '20 at 21:31
  • Yeah, it is! Although perhaps I should have added that this is impossible for T3.5 spaces... Not sure about T3, though. – Cronus Jul 06 '20 at 05:23