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Vague question: Is there a criterion to deduce that a morphism between algebraic stacks is finite based on the local deformation functors?

For sure this is not enough, so let me be more specific.

Suppose that $f: X \to Y$ is a morphism of algebraic stacks. Let us assume in addition that the diagonal $\Delta_f : X \to X \times_Y X$ is finite and unramified.

Now assume that for each point $x \in |X|$ of finite type over $y \in |Y|$ the morphism between the (pro-representing rings of the) deformation functors is finite.

Specific question: Under these conditions, can we assume that $f \colon X \to Y$ is finite?

Edit: This too is not enough as Piotr demonstrates in the comments. The key observation is that finiteness is not local on the source and even an open immersion is a counterexample to the question above.

Let me then modify the question like so:

Modified question: What global criteria, such as properness, on the morphism $f:X \to Y$ need to be assumed to make finiteness (formal) local on the source?

Improve this question
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  • $\begingroup$ If $Y$ is an abelian variety and $X$ is the blow-up of $Y$ at one point, then the blow-up morphism $\pi \colon X \to Y$ is not finite, but it induces an isomorphism of deformation functors $$\mathrm{Def}(X) \simeq \mathrm{Def}(Y),$$ see mathoverflow.net/questions/198049/deformations-of-a-blowup $\endgroup$
    – Francesco Polizzi
    Commented Dec 21, 2016 at 23:25
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    $\begingroup$ @FrancescoPolizzi I'm not sure this is what the question was asking. It seems that you have confused deforming $X$ with deforming a point $x\in X$. If $y\in Y$ is the point you blow up and $x\in X$ is a point over it, then $\widehat{\mathcal{O}}_{Y, y} \to \widehat{\mathcal{O}}_{X,x}$ is not finite. Or am I missing the point of your comment? $\endgroup$
    – Piotr Achinger
    Commented Dec 22, 2016 at 12:38
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    $\begingroup$ In any case, the answer to the question seems to be negative already for schemes in the case of an open immersion. Indeed, the diagonal of an open immersion $f:X\to Y$ is an isomorphism, so finite and unramified. However, if $x$ is a point over $y$, then the morphism on completed local rings is an isomorphism. The problem is that being finite is local on the target rather than on the source. $\endgroup$
    – Piotr Achinger
    Commented Dec 22, 2016 at 12:42
  • $\begingroup$ @PiotrAchinger Ah! Indeed a good point about finiteness being local on the target. Can you think of some global standard hypothesis (e.g. properness etc.) which combined with this source local criteria implies finiteness? $\endgroup$
    – Emre
    Commented Dec 22, 2016 at 14:01
  • $\begingroup$ @FrancescoPolizzi Thanks for your edits. Though indeed I was asking about the deformation of points, e.g. the complete local rings at those points. $\endgroup$
    – Emre
    Commented Dec 22, 2016 at 14:04

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