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The question when $(-) \times X$ preserves colimits in topological spaces is well-studied. Since it always preserves arbitrary coproducts (disjoint unions), one only has to show when it preserves coequalizers. It is well-known that if $X$ is locally compact and $Y \to Z$ is a quotient map, then $Y \times X \to Z \times X$ is quotient map as well. Thus, taking product with a locally compact space preserves all colimits. (Here, I define a space to be locally compact if every point has a basis of (quasi-)compact neighborhoods. No Hausdorffness is required. For example, the underlying topological space of every scheme is locally compact.) Actually, $(-)\times X$ preserves colimits if and only if $X$ is core-compact in the sense of this nlab page.

My question is:

Is there a condition (without assuming Hausdorffness!) on a map $X \to X'$ such that $(-) \times_{X'} X$ preserves colimits, or, presumably equivalently: preserves quotient maps?

If $X'$ is Hausdorff, $Z\times_{X'} X$ is a closed subset of $Z\times X$. If $X$ is core-compact and $Y\to Z$ is a quotient map, $Y\times_{X'} X \to Z\times_{X'} X$ is thus a quotient map again. But I would like to apply the potential criterion to something like $\operatorname{Spec} A \to \operatorname{Spec} B$, which is the reason why I don't want to assume Hausdorffness.

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Thanks to the comments, I found the paper Exponentiability for maps means fibrewise core-compactness by G. Richter, which gives a more concrete (albeit still complicated) characterization of exponentiable maps, i.e. maps $p\colon X \to T$ such that $(-)\times_T X$ preserves colimits. Richter characterizes them as fibrewise core-compact maps. Here, $p$ is fibrewise core-compact if for every $x\in X$ and every neighborhood $U$ of $x$, there are neighborhoods $V\subset U$ of $x$ and $p(U)\subset W$ of $p(x)$ such that: for all $t\in W$ and every open cover $\mathcal{U}$ of $U\cap p^{-1}(t)$, there exists a finite subset $\mathcal{E}\subset \mathcal{U}$ and an open neighborhood $W'\subset W$ of $t$ such that $V\cap p^{-1}(W') \subset \bigcup \mathcal{E}$.

This is quite a mouthful. Let's first consider the case that $T = \mathrm{pt}$. Then this says that for every neighborhood $U$ of $x$, there exists a neighborhood $V$ of $x$ such that every open cover of $U$ admits a finite subcover of $V$. This is the definition of core-compact. If $X$ is locally-compact, we can choose $V$ as a compact subneighborhood of $U$.

Another special case: Let $T$ be locally compact and let $p$ be locally proper: for every $x\in X$ and a neighborhood $U$ of $x$, there exists a neighborhood $V$ of $x$ and a neighborhood $W$ of $p(x)$ containing $p(V)$ such that $V\to W$ is proper, i.e. $p$ is closed and preimages of compacts are compact. I claim that then $p\colon X\to T$ is fibrewise core-compact. Indeed: Let $x\in X$ and $U$ a neighborhood of $x$. Choose $V$ and $W$ as above. Let $t\in W$ and $\mathcal{U}$ an open cover of $U\cap p^{-1}(t)$. Then $\bigcup \mathcal{U}$ is open and thus $p(V \setminus V\cap \bigcup \mathcal{U}) \subset W$ is closed and does not contain $t$. Choose a compact subneighborhood $W'$ of $t$ inside of its complement. Then $V\cap p^{-1}(W')$ is compact and a subset of $\bigcup \mathcal{U}$. Thus, there exists a finite $\mathcal{E}\subset \mathcal{U}$ covering $V\cap p^{-1}(W')$.

Unfortunately, $p$ being locally proper in the sense above is rarely fulfilled for a the underlying map of topological spaces of a morphism between schemes. It may still be the case that these are often fibrewise core-compact, but I don't know.

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    $\begingroup$ I believe all spectral maps between spectral spaces are exponentiable. Category-theoretic models of certain type theories constructed by Hyland and Pitts in "The theory of constructions: Categorical semantics and topos-theoretic models" (a pdf is available) are based on this fact. Here, spectral maps are the ones with inverse images of compact open sets compact. $\endgroup$ Apr 30, 2023 at 16:24
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    $\begingroup$ Sorry, what I wrote is actually not so easy to extract from that paper, I had to be more specific. First, there is 4.7 (page 182) claiming that for any Grothendieck topos $\mathsf E$, the toposes algebraic over $\mathsf E$ are exponentiable in toposes over $\mathsf E$. Next, 5.2 (page 192) describes toposes over $\mathsf E$ which are both localic and algebraic. Then 5.6 says that exponentials of algebraic toposes over $\mathsf E$ that are localic over $\mathsf E$ are themselves localic over $\mathsf E$. Finally one must take $\mathsf E$ itself to be the topos of sheaves on some spectral space. $\endgroup$ May 12, 2023 at 15:58
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    $\begingroup$ Unfortunately that's not all, since I did not say anything yet about spectral maps. These appear in the localic version of the Stone duality between spectral spaces and distributive lattices. By the localic version I mean replacing spectral spaces with their locales, i. e. complete Heyting algebras of their open sets. The link to algebraic locales is through this. While spectral maps appear because in this form of duality, the distributive lattice is recovered from the Heyting algebra as the sublattice of its compact elements. So, the spectral maps which by definition preserve compact elements $\endgroup$ May 12, 2023 at 16:09
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    $\begingroup$ (under preimages), recover the homomorphism of distributive lattices in the opposite direction. Thus in essence the last part in Hyland & Pitts' paper can be interpreted as saying that in the opposite of the category of distributive lattices every morphism is exponentiable. And this opposite category is equivalent to the category of spectral spaces and spectral maps. $\endgroup$ May 12, 2023 at 16:11
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    $\begingroup$ One more thing. I was looking into "Sketches of an Elephant" to figure out whether a more straightforward argument is available, and, as a result, came up with the following. Let $X$ be an exponentiable space, and $f:X\to Y$ be a map with locally closed graph - more precisely, such that $(\operatorname{identity},f):X\hookrightarrow X\times Y$ is a homeomorphism of $X$ onto a locally closed subspace of $X\times Y$. Then $f$ is exponentiable in spaces over $Y$. This is because (1) the projection $p:X\times Y\to Y$ is exponentiable in spaces over $Y$, $\endgroup$ May 12, 2023 at 16:35

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