Lemma 103.17.5. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. The category of coherent $\mathcal{O}_\mathcal {X}$-modules is abelian. If $\varphi : \mathcal{F} \to \mathcal{G}$ is a map of coherent $\mathcal{O}_\mathcal {X}$-modules, then

the cokernel $\mathop{\mathrm{Coker}}(\varphi )$ computed in $\textit{Mod}(\mathcal{O}_\mathcal {X})$ is a coherent $\mathcal{O}_\mathcal {X}$-module,

the image $\mathop{\mathrm{Im}}(\varphi )$ computed in $\textit{Mod}(\mathcal{O}_\mathcal {X})$ is a coherent $\mathcal{O}_\mathcal {X}$-module, and

the kernel $\mathop{\mathrm{Ker}}(\varphi )$ computed in $\textit{Mod}(\mathcal{O}_\mathcal {X})$ may not be coherent, but it is in $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$ and $Q(\mathop{\mathrm{Ker}}(\varphi ))$ is coherent and is the kernel of $\varphi $ in $\textit{Coh}(\mathcal{O}_\mathcal {X})$.

The inclusion functor $\textit{Coh}(\mathcal{O}_\mathcal {X}) \to \mathit{QCoh}(\mathcal{O}_\mathcal {X})$ is exact.

**Proof.**
The rules given for taking kernels, images, and cokernels in $\textit{Coh}(\mathcal{O}_\mathcal {X})$ agree with the prescription for quasi-coherent modules in Remark 103.10.5. Hence the lemma will follow if we can show that the quasi-coherent modules $\mathop{\mathrm{Coker}}(\varphi )$, $\mathop{\mathrm{Im}}(\varphi )$, and $Q(\mathop{\mathrm{Ker}}(\varphi ))$ are coherent. By Lemma 103.17.1 it suffices to prove this after restricting to $U_{\acute{e}tale}$ for some surjective smooth morphism $f : U \to \mathcal{X}$. The functor $\mathcal{F} \mapsto f^*\mathcal{F}|_{U_{\acute{e}tale}}$ is exact. Hence $f^*\mathop{\mathrm{Coker}}(\varphi )$ and $f^*\mathop{\mathrm{Im}}(\varphi )$ are the cokernel and image of a map between coherent $\mathcal{O}_ U$-modules hence coherent as desired. The functor $\mathcal{F} \mapsto f^*\mathcal{F}|_{U_{\acute{e}tale}}$ kills parasitic modules by Lemma 103.9.2. Hence $f^*Q(\mathop{\mathrm{Ker}}(\varphi ))|_{U_{\acute{e}tale}} = f^*\mathop{\mathrm{Ker}}(\varphi )|_{U_{\acute{e}tale}}$ by part (2) of Lemma 103.10.2. Thus we conclude that $Q(\mathop{\mathrm{Ker}}(\varphi ))$ is coherent in the same way.
$\square$

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