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Given a morphism of sheaves $\psi:\mathcal{F} \rightarrow \mathcal{G}$, we define the image sheaf Im $\psi$ to be the sheaf associated to the image presheaf $\text{Im}^{\text{pre}} \psi$.

By Hartshorne's construction, the sheaf $\mathcal{H}^+(U)$ associated to a presheaf $\mathcal{H}(U)$ is the set of functions $\{s:U \rightarrow \bigcup \mathcal{H}_p\}$ such that $\forall p \in U, s(p) \in \mathcal{H}_p$ and $\exists V \subset U$ containing p and $t \in \mathcal{H}$ such that $\forall q \in \mathcal{H}(V), s(q) = t_q$.

So my question is that why is the image subsheaf Im $\psi$ a subsheaf of $\mathcal{G}$? Clearly $\text{Im}^{\text{pre }} \psi (U)$ will always be a subgroup/subring of $\mathcal{G}(U)$ by definition, but I don't see why this sheafification of the image presheaf on an open $U$, i.e. $\text{Im}^{\text{pre }} \psi (U)$, needs to be a subgroup/subring of $\mathcal{G}(U)$

Edit: I don't know if Hartshorne mentions this but it is mentioned in Liu's Algebraic geometry and arithmetic curves in Lemma 2.16.

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  • $\begingroup$ Could you provide us with a problem number within Hartshorne? $\endgroup$ Commented Feb 5, 2021 at 4:07
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    $\begingroup$ I don't know if Hartsehorne mentions it but it is mentioned in Liu's Algebraic geometry and arithmetic curves in Lemma 2.16 $\endgroup$
    – User20354
    Commented Feb 5, 2021 at 4:15
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    $\begingroup$ I'll make an edit about that $\endgroup$
    – User20354
    Commented Feb 5, 2021 at 4:15

1 Answer 1

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You should look at Proposition-Definition 1.2 of Hartshorne. He defines the sheafification of the presheaf $\mathcal{F}$ to be the sheaf $\mathcal{F}^+$ equipped with a morphism of presheaves $\theta:\mathcal{F}\to \mathcal{F}^+$ satisfying the following universal property. Given a morphism of presheaves $\varphi:\mathcal{F}\to \mathcal{G}$ where $\mathcal{G}$ is a sheaf, there exists a unique morphism of sheaves $\psi:\mathcal{F}^+\to \mathcal{G}$ such that $\varphi=\psi\circ \theta$.

In this case, associated to a morphism of sheaves $\varphi:\mathcal{F}\to \mathcal{G}$ there the image presheaf $\varphi(\mathcal{F})$ which is as you say is a subsheaf of $\mathcal{G}$. We define $\operatorname{im}\varphi$ to be the sheaf associated to the presheaf $\varphi(\mathcal{F})$. As there is an inclusion morphism of presheaves $i:\varphi(\mathcal{F})\to \mathcal{G}$, we get a morphism of sheaves $j:\operatorname{im}\varphi\to \mathcal{G}$ by the universal property.

Lemma: Given an injective map of presheaves $\alpha:\mathcal{E}\to \mathcal{E}'$ the induced map on the associated sheaves $\mathcal{E}^+\to \mathcal{E}'^+$ is injective also.

Hence, as $i:\varphi(\cal{F})\to \mathcal{G}$ is injective, the associated map $j:\operatorname{im}\varphi\to \mathcal{G}$ is also, so that $\operatorname{im}\varphi$ can be identified with a subsheaf of $\mathcal{G}$.

Proof of the Lemma: Examine $\varphi_P^+:\mathcal{F}_P^+\to \mathcal{G}_P^+$ noting that $\mathcal{F}_P^+=\mathcal{F}_P$ and $\mathcal{G}_P^+=\mathcal{G}_P$. Recall, also that $\mathcal{F}\hookrightarrow{} \mathcal{F}_P^+$ as the maps $s:U\to\coprod_{p\in U} \mathcal{F}_P$ given by $s(P)=s_P$. We can check injectivity of a morphism of sheaves at the stalks. If $\varphi_P^+(t,U)=0$, then there exists $W\subseteq U$ containing $P$ such that $\varphi_W^+(t|_W)=0$. But, take an open set $W'\subseteq W$ containing $P$ on which $t|_{W'}=\tau \in \mathcal{F}(W')$. Then $\varphi^+_{W'}|_{\mathcal{F}(W')}$ agrees with $\varphi_{W'}$, and hence is injective. So, $\varphi_{W'}^+(\tau)=0$, and this implies $\tau =0$. So, the germ $(t,U)$ is $0$ at $P$. Therefore, we have injectivity at the stalks, and our map of sheaves is injective, too.

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    $\begingroup$ I see. This makes sense. Thanks so much! $\endgroup$
    – User20354
    Commented Feb 5, 2021 at 4:42

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