# Image sheaf is the sheafification of the image presheaf

This is an exercise in Vakil's notes on foundations of algebraic geometry.

Suppose $\Phi:\mathscr{F}\to\mathscr{G}$ is a morphism of sheaves of abelian groups, show that the image sheaf Im $\Phi$ is the sheafification of the image presheaf.

My solution

We use the key fact that in an abelian category, there is a natural isomorphism (preserving the arrow $\mathscr{F}\to \cdot$) between $$\text{Im }\Phi := \ker\text{coker}\,\Phi \cong \text{coker}\ker \Phi$$ Therefore we have $$\text{Im }\Phi \cong \text{coker}\ker \Phi = (\text{coker}_{pre} \ker \Phi)^{sh} = (\text{coker}_{pre} \ker_{pre} \Phi)^{sh} \cong (\text{Im}_{pre}\Phi)^{sh} .$$

Is this solution correct? To be rigorous, I must define a map between Im$_{pre} \Phi\to$Im $\Phi$.

• Well, have you proved that you can switch those two things in the category of Abelian sheaves? – Hoot Jan 25 '15 at 22:51
• Yes, in the sense that the natural map between them is an isomorphism. This holds in any abelian category, and in fact characterizes abelian categories (from additive categories). – mez Jan 25 '15 at 23:46
• Would this argument work? – Houndoom Mar 19 '16 at 4:59
• $$\text{Im}\Phi\cong\ker\text{coker}\Phi=\ker(\text{coker}_{pre}\Phi)^{sh} = (\ker_{pre}\text{coker}_{pre}\Phi)^{sh} \cong \text{Im}_{pre}\Phi^{sh}$$ – Houndoom Mar 19 '16 at 4:59

We know that the kernel and cokernel can be described in terms of compatible collections of germs. Using this characterization, the categorical image ($=\ker \operatorname{cok} \Phi$) should be the elements of $$\prod_{x \in X}\Phi_x(\mathscr{F}_x)$$ which are $\mathscr{G}$-compatible.
Let $\mathscr{P}$ be the image presheaf.
Lemma: Let $\mu \in \prod_{x \in X}\Phi_x(\mathscr{F}_x)$ be $\mathscr{G}$-compatible. Then $\mu$ is $\mathscr{P}$-compatible. (Note of course any $\mathscr{P}$-compatible collection is $\mathscr{G}$-compatible.)
Proof: Let $g$ be a section as in the definition of a compatible collection of germs, i.e. $g \in \mathscr{G}(U)$ be such that $g_y = \mu_y$ for all $y \in U$. Let $x \in U$. Since $\mu_x$ is in the image of $\Phi_x$, there is a neighborhood $V \subset U$ and $f \in \mathscr{F}(U)$ so that $\Phi_U(f) = g'$, and $g'_x = \mu_x$. Since $g'$ and $g$ have the same germ at $x$, after perhaps shrinking $V$, we have $g=g' \in \mathscr{G}(V)$. Taking such a representative for each $x$, we have shown the collection to be $\mathscr{P}$-compatible. $\blacksquare$
Hence the categorical image can be described as $\mathscr{P}$-compatible elements of $\prod_{x \in X}\Phi_x(\mathscr{F}_x)$. Now recall that the stalks of $\mathscr{P}$ are isomorphic to the images of the stalk maps $\Phi(\mathscr{F}_x)$. Therefore the categorical image is just $\mathscr{P}$-compatible elements of $\prod_{x \in X} \mathscr{P}_x$, which is just another way of saying the sheafification of $\mathscr{P}$.