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.

  • $\begingroup$ Imo the only part that really addresses the question (and has content to it) is the last sentence about the cokernel, which due its lack of detail may as well be written in a comment. $\endgroup$ – user347489 Apr 24 '18 at 10:17

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