# How to show that the hom sheaf is left exact

Let $(X,\mathcal{O}_X)$ be a ringed space, and let $\mathcal{F}$ and $\mathcal{G}$ be $\mathcal{O}_X$-modules. Then, we can define an $\mathcal{O}_X$-module by $$\mathcal{Hom}(\mathcal{F},\mathcal{G})\quad :\quad U \longmapsto \mathrm{Hom}_{\mathcal{O}_X}(\mathcal{F}|_U, \mathcal{G}|_U),$$ where $\mathrm{Hom}_{\mathcal{O}_X}$ is the set of $\mathcal{O}_X$-module homomorphisms. The $\mathcal{O}_X$-module is called the sheaf hom of $\mathcal{F}$ and $\mathcal{G}$.

If a sequence of $\mathcal{O}_X$-modules $$\mathcal{F}_1 \xrightarrow{f} \mathcal{F}_2 \xrightarrow{g} \mathcal{F}_3\to 0 \quad\quad(\natural)$$ is exact, then the sequence $$0\to\mathcal{Hom}(\mathcal{F}_3,\mathcal{G}) \xrightarrow{g^*} \mathcal{Hom}(\mathcal{F}_2,\mathcal{G}) \xrightarrow{f^*} \mathcal{Hom}(\mathcal{F}_1,\mathcal{G}) \quad\quad(\ast)$$ is exact as $\mathcal{O}_X$-modules.

Why does the euqality $\mathrm{Im}(g^{\ast})=\mathrm{Ker}(f^{\ast})$ in the sequence $(\ast)$ hold?

Since $g\circ f=0$, it is clear that the image of $g^{\ast}$ is contained in the kernel of $f^{\ast}$. I want to show the converse. By exactness of $(\natural)$, we have the exact sequence of stalks at any point $x\in X$ $$\mathcal{F}_{1,x} \xrightarrow{f_x} \mathcal{F}_{2,x} \xrightarrow{g_x} \mathcal{F}_{3,x}\to 0.$$ If $\phi_x\in \mathrm{Ker}(f^{\ast}_x)$, there uniquely exists $\psi:\mathcal{F}_{3,x}\to \mathcal{G}_x$ such that $\phi_x=\psi\circ g_x$ since $\phi_x\circ f_x=0$.

How do we show that $\psi\in\mathcal{Hom}(\mathcal{F_3,\mathcal{G}})_x$?

Thank you.

It is enough to show that the sequence $(\ast)$ is exact on sections, i.e. that for every open $U$ of $X$, the sequence $$0\to \operatorname{Hom}_{\mathcal{O}_X|_U}(\mathcal{F}_3|_U,\mathcal{G}|_U)\to \operatorname{Hom}_{\mathcal{O}_X|_U}(\mathcal{F}_2|_U,\mathcal{G}|_U)\to \operatorname{Hom}_{\mathcal{O}_X|_U}(\mathcal{F}_1|_U,\mathcal{G}|_U)$$ is exact. But since restriction to an open is exact, sequence $(\sharp)$ remains exact when restricted to $U$ and exactness of the above sequence is basically the universal property of the cokernel.