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Let $ X \hookrightarrow Y$ be a closed immersion of (connective) spectral Deligne-Mumford stacks, is $ i_* : Qcoh(X) \rightarrow Qcoh(Y)$ conservative? Somehow I couldn't find the statement in SAG...

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First, recall this is true when $X$ and $Y$ are spectral affine schemes. Indeed by [SAG, Proposition 2.5.1.1], $i_*$ is t-exact so it suffices to show this on the hearts, and in particular you can reduce to the case where $X$ and $Y$ are 0-truncated/ordinary affine schemes, which is obvious.

The general case follows by descent. Let $F \to G$ be a morphism in $Qcoh(X)$ such that $i_*(F) \to i_*(G)$ is an equivalence. We want to show that $F \to G$ is already an equivalence. Let $q : V \to Y$ be an etale atlas with $V$ an affine scheme, and $p : U = V \times_Y X \to X$ the induced etale atlas. Since closed immersions are affine, $U$ is also an affine scheme. By etale descent for $Qcoh$, $p^*$ is conservative, so it suffices to show that $p^*(F) \to p^*(G)$ is an equivalence in $Qcoh(U)$. By the first paragraph, push forward along the closed immersion $j : U \to V$ induces a conservative functor $j_*$ so it further suffices to show that $j_*p^*(F) \to j_*p^*(G)$ is an equivalence. By base change, this is the same as $q^*i_*(F) \to q^*i_*(G)$, which is an equivalence by assumption.

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  • $\begingroup$ I suppose the line "V is also affine scheme" should read "U is also an affine scheme"? Also, is it necessary to assume the stack $Y$ is quasi-compact? Anyway thanks for the answer :). $\endgroup$ – Anette Jul 9 at 15:35
  • $\begingroup$ Typo corrected thanks. I don't think $Y$ needs to be quasi-compact. For the base change formula you need $i$ to be quasi-compact quasi-separated, but this is true. $\endgroup$ – crystalline Jul 9 at 16:51

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