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Let $X$ be a projective variety and $Y$ an Artin stack. Suppose that $f:X\to Y$ is a morphism of Artin stacks. Is $f(X)$ necessarily a closed substack of $Y$?

This seems like it should be true and probably one can find it somewhere in the stacks project, but I cannot locate a good source.

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    $\begingroup$ This is false. Take for example $Y = [\mathbb{A}^n/\mathbb{G}_m]$ and $X = \mathbb{P}^{n-1}$. Then $X$ is projective and the natural inclusion $f : X \to Y$ is an open embedding with dense image but $f(X)$ is not closed. The issue is that $Y$ is not separated so $f : X \to Y$ is not proper. $\endgroup$
    – Dori Bejleri
    Commented Jun 16, 2021 at 3:36
  • $\begingroup$ No, $Y$ is the Artin stack $[\mathbb{A}^n/\mathbb{G}_m]$. If you unravel the definition of the quotient stack, a map $T \to Y$ is the same as a a tuple $(L, s_1, ...., s_n)$ where $L$ is a line bundle on $T$ and $s_i$ are sections. On the other hand, a map $T \to X$ is the same data but with the condition that the $s_i$ don't simultaneously vanish. Dropping this condition induces the embedding $f : X \to Y$ which you can equivalently think of as identifying $X$ with the quotient $[(\mathbb{A}^n\setminus 0)/\mathbb{G}_m]$ which is an Artin stack which happens to be isomorphic to a scheme. $\endgroup$
    – Dori Bejleri
    Commented Jun 16, 2021 at 3:59
  • $\begingroup$ Here I was implicitly assuming that the action of $\mathbb{G}_m$ is the usual one by scaling with all weights equal to $1$. $\endgroup$
    – Dori Bejleri
    Commented Jun 16, 2021 at 4:01
  • $\begingroup$ @DoriBejleri If we assume that $Y$ is a scheme, then is the answer to my question affirmative? $\endgroup$
    – LAGC
    Commented Jun 16, 2021 at 4:33

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To elaborate on Dori's comment, consider $[\mathbf{A}^1/\mathbf{G}_m]$ which consists of two points: The closed point corresponding to the origin and the orbit of $1$ (which is open). Take $\operatorname{Spec} k \to [\mathbf{A}^1/\mathbf{G}_m]$ corresponding to this open point. The image is not closed and certainly $[\mathbf{A}^1/\mathbf{G}_m]$ is not isomorphic to $\operatorname{Spec} \Gamma(\mathbf{A}^1, \mathcal{O}_{\mathbf{A}^1})^{\mathbf{G}_m} = \operatorname{Spec} k$, for instance because the stack has two points.

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  • $\begingroup$ Does assuming that $Y$ is a scheme in my original question change the answer? $\endgroup$
    – LAGC
    Commented Jun 16, 2021 at 4:41
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    $\begingroup$ @LAGC Not unless $Y$ is assumed separated. If, say, $x$ is a (nonisolated) closed point of $X$, with complement $U$, and $Y$ is the union of two copies of $X$ glued alog $U$, then the natural embeddings $X\hookrightarrow Y$ have nonclosed images. $\endgroup$
    – Laurent Moret-Bailly
    Commented Jun 16, 2021 at 7:10

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