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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