# Most general lifting property for proper morphisms

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

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 $$S$$?

• 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). Dec 31, 2022 at 0:39
• Thank you, @R.vanDobbendeBruyn! Jan 1, 2023 at 13:55
• 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?
– Z. M
Jan 1, 2023 at 17:27

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.