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

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...

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.
• 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 :). – Anette Jul 9 at 15:35
• 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. – crystalline Jul 9 at 16:51