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...



Is the pushforward of a closed immersion of spectral Deligne-Mumford stacks conservative?
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.


