## 101.40 Valuative criterion for second diagonal

The converse statement has already been proved in Lemma 101.39.2. The criterion itself is the following.

Lemma 101.40.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\Delta _ f$ is quasi-separated and if for every diagram (101.39.1.1) and choice of $\gamma$ as in Definition 101.39.1 the category of dotted arrows is a setoid, then $\Delta _ f$ is separated.

Proof. We are going to write out a detailed proof, but we strongly urge the reader to find their own proof, inspired by reading the argument given in the proof of Lemma 101.39.2.

Assume $\Delta _ f$ is quasi-separated and for every diagram (101.39.1.1) and choice of $\gamma$ as in Definition 101.39.1 the category of dotted arrows is a setoid. By Lemma 101.6.1 it suffices to show that $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a closed immersion. By Lemma 101.6.4 it in fact suffices to show that $e = \Delta _{f, 2}$ is universally closed. Either of these lemmas tells us that $e = \Delta _{f, 2}$ is quasi-compact by our assumption that $\Delta _ f$ is quasi-separated.

In this paragraph we will show that $e$ satisfies the existence part of the valuative criterion. Consider a $2$-commutative solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_ x \ar[d]_ j & \mathcal{X} \ar[d]^ e \\ \mathop{\mathrm{Spec}}(A) \ar[r]^{(a, \theta )} & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} }$

and let $\alpha : (a, \theta ) \circ j \to e \circ x$ be any $2$-morphism witnessing the $2$-commutativity of the diagram (we use $\alpha$ instead of the letter $\gamma$ used in Definition 101.39.1). Note that $f \circ \theta = \text{id}$; we will use this below. Observe that $e \circ x = (x, \text{id}_ x)$ and $(a, \theta ) \circ j = (a \circ j, \theta \star \text{id}_ j)$. Thus we see that $\alpha$ is a $2$-arrow $\alpha : a \circ j \to x$ compatible with $\theta \star \text{id}_ j$ and $\text{id}_ x$. Set $y = f \circ x$ and $\beta = \text{id}_{f \circ a}$. Reading the arguments given in the proof of Lemma 101.39.2 backwards, we see that $\theta$ is an automorphism of the dotted arrow $(a, \alpha , \beta )$ with

$\gamma : y \circ j \to f \circ x \quad \text{equal to}\quad \text{id}_ f \star \alpha : f \circ a \circ j \to f \circ x$

On the other hand, $\text{id}_ a$ is an automorphism too, hence we conclude $\theta = \text{id}_ a$ from the assumption on $f$. Then we can take as dotted arrow for the displayed diagram above the morphism $a : \mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ with $2$-morphisms $(a, \text{id}_ a) \circ j \to (x, \text{id}_ x)$ given by $\alpha$ and $(a, \theta ) \to e \circ a$ given by $\text{id}_ a$.

By Lemma 101.39.11 any base change of $e$ satisfies the existence part of the valuative criterion. Since $e$ is representable by algebraic spaces, it suffices to show that $e$ is universally closed after a base change by a morphism $I \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ which is surjective and smooth and with $I$ an algebraic space (see Properties of Stacks, Section 100.3). This base change $e' : X' \to I'$ is a quasi-compact morphism of algebraic spaces which satisfies the existence part of the valuative criterion and hence is universally closed by Morphisms of Spaces, Lemma 67.42.1. $\square$

Comment #2100 by Matthew Emerton on

Does always a setoid'' meanfor any choice of $\gamma$''? If so, it would be good to make this explicit, since the fact that we have to allow this for every choice of $\gamma$ rather than just our favourite choice seems to be a slightly subtle point.

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