Let $\mathcal C$ be the class of morphisms $f\colon U\to V$ of schemes such that for every proper map $g\colon X\to Y$ between schemes and every commutative solid square

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there exists a lift $h$ making the diagram commute.

Is there a nice description of elements of $\mathcal C$? Or at least of some subclass of $\mathcal C$ beyond the morphisms used in the valuative criterion, where $V$ is the spectrum of a valuation ring and $U$ is the spectrum of its fraction field? I am particularly interested in the case where $V=\mathbb A^1\times S$ and $U=(\mathbb A^1\setminus 0)\times S$, for any scheme $S$. For $S$ the spectrum of a field, this follows from 0BX7 in the Stacks Project, but what about general $S$?

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    $\begingroup$ The lifting criterion for $(\mathbf A^1\setminus \{0\}) \times S \subseteq \mathbf A^1 \times S$ clearly does not hold. For instance, take $g \colon X \to Y$ the blowup of $Y = \mathbf A^2$ in the origin and take $S = \mathbf A^1$ with the identity map $\mathbf A^1 \times S \to Y$. Then the lift $h$ does not exist since $g$ does not have a section (but it does have a section over $(\mathbf A^1 \setminus \{0\}) \times \mathbf A^1$ as $g$ is an isomorphism away from the origin). $\endgroup$ Dec 31, 2022 at 0:39
  • $\begingroup$ Thank you, @R.vanDobbendeBruyn! $\endgroup$ Jan 1, 2023 at 13:55
  • $\begingroup$ I wonder whether the proposition for special $S$ on Stacks Project depends on the fact that $g$ is of finite type? I mean, whether it is a formal consequence of valuation criterion (i.e. "partially proper") or it also depends on some extra finiteness? $\endgroup$
    – Z. M
    Jan 1, 2023 at 17:27

1 Answer 1


I don't have a description of the class of morphisms you are describing, but let me provide a counterexample in your case of interest that can perhaps limit your search for a general result. Consider the natural inclusion $f: U = (\mathbb{A}^1_\mathbb{C} \setminus 0) \times \mathbb{A}^1_\mathbb{C} \to \mathbb{A}^1_\mathbb{C} \times \mathbb{A}^1_\mathbb{C} = \mathbb{A}^2_\mathbb{C} = V$, and let $g: X = \mathrm{Bl}_0 \mathbb{A}^2_\mathbb{C} \to \mathbb{A}^2_\mathbb{C}$ be the blow-up of the plane at the origin. The vertical maps are the natural inclusion and the identity, respectively. In this setting there is no lift $h$ because each direction in the plane would need a different value of $h(0)$ for the map to even be continuous.

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    $\begingroup$ ...this is identical to the comment I left... $\endgroup$ Dec 31, 2022 at 16:29
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    $\begingroup$ @R. van Dobben de Bruyn Yes, I'm sorry. I opened the question, came back to answer it several hours later and didn't refresh the page. I can delete the answer if you prefer to post it yourself. I do think it deserves to be an answer, since it addresses one of the main points of the question. $\endgroup$ Jan 1, 2023 at 12:36
  • $\begingroup$ Thank you, @EduardodeLorenzo! $\endgroup$ Jan 1, 2023 at 13:55
  • $\begingroup$ Ah I see. I was just a bit puzzled, but that explains. $\endgroup$ Jan 1, 2023 at 16:07

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