# $\Gamma(U,\cdot)$ is a left-exact functor $\mathfrak{Ab}(X)\to\mathfrak{Ab}$

Exercise 1.8 in Hartshorne, Chapter II:

Let $0\to\mathcal{F}'\xrightarrow{\phi}\mathcal{F}\xrightarrow{\psi}\mathcal{F}''$ be a left-exact sequence. Then for any open set $U\subseteq X$, the sequence $0\to\Gamma(U,\mathcal{F}')\to\Gamma(U,\mathcal{F})\to\Gamma(U,\mathcal{F}'')$ is left-exact.

Here is the way I tried to prove this: Let $U\subseteq X$ be open. Since $\phi$ is injective, $\phi_U:\Gamma(U,\mathcal{F}')\to\Gamma(U,\mathcal{F})$ is injective. So it remains to show that $\ker(\psi_U)=\operatorname{im}(\phi_U)$. Then I looked at the diagram

$$\begin{array}{cccccc} 0 & \rightarrow & \Gamma(U,\mathcal{F}^') & \xrightarrow{\phi_U} & \Gamma(U,\mathcal{F}) & \xrightarrow{\psi_U} & \Gamma(U,\mathcal{F}^{''})\\ & & \downarrow & & \downarrow & & \downarrow\\ 0 & \rightarrow & \mathcal{F}^'_p & \xrightarrow{\phi_p} & \mathcal{F}_p & \xrightarrow{\psi_p} & \mathcal{F}^{''}_p \end{array}$$

which commutes for every $p\in U$ and with bottom row exact. Now if $s\in\Gamma(U,\mathcal{F}^')$, we have $\psi_U(\phi_U(s))_p=\psi_p(\phi_p(s_p))=0$ for every $p\in U$, hence $\psi_U(\phi_U(s))=0$. The inclusion left to show is $\ker(\psi_U)\subseteq\operatorname{im}(\phi_U)$.

For this, take $t\in\ker(\psi_U)$. $\psi_U(t)=0$ implies $\psi_p(t_p)=\psi_U(t)_p=0$ for every $p$. Since $\ker(\psi_p)=\operatorname{im}(\phi_p)$, there exist $s^'_p\in\mathcal{F}^'_p$ such that $t_p=\phi_p(s^'_p)$, hence also a section $s\in\Gamma(U,\mathcal{F}^')$ with $s_p=s^'_p$ (bad notation there, I didn't know how to name those germs). We have $\phi_U(s)_p=\phi_p(s_p)=t_p$ for every $p\in U$, so $\phi_U(s)=t\in\operatorname{im}(\phi_U)$.

Now my question is: is this proof correct? I usually look up some solutions in the internet after doing such an exercise, and for this one I found two solutions which both took a different approach from mine, which made me a bit insecure about my 'solution', since I took the 'direct approach' from my point of view.

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Since you're working with abelian sheaves, there's a shortcut: to show that an additive functor is left exact, it is enough to show that it preserves monomorphisms. So the only thing to do is to check that the monomorphisms in the category of abelian sheaves are what you think they are... –  Zhen Lin Feb 26 '12 at 10:42
Hello @Zhen Lin, that is very interesting, I did not know this. I'll read up on it, thank you! –  Rand al'Thor Feb 26 '12 at 11:49
In my opinion, you need to reason why the $s'_P$ do glue to some $s'\in\mathcal{F}'(U)$. However, this is not that tough. We pick neighborhoods $V_P\subseteq U$ for each point $P$ such that $s'_P=(f_P,V_P)$ with $f_P\in\mathcal{F}'(V_P)$. For $W:=V_P\cap V_Q\ne 0$, we have $\phi_W(f_P|_W)=t|_W=\phi_W(f_Q|_W)$ and we know that $\phi_W$ is injective.