# Why is the Image sheaf a subsheaf?

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.

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

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.