While understanding prop 5.6, II, Hartshorne, I try to solve the additional exercise : Let $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F''} \to 0 $ be a short exact sequence(SES), then global section is left-exact functor . i.e, $0 \to \Gamma(X,\mathcal{F}') \to \Gamma(X,\mathcal{F}) \to \Gamma(X,\mathcal{F}'') $.
My sketch of proof is here : Suppose $0 \to \mathcal{F}' \overset{\phi} {\to} \mathcal{F} \overset{\psi} {\to} \mathcal{F''} \to 0 $ is SES, then claim is obviously, i) $\overset{\sim}{\phi}: \Gamma(X,\mathcal{F}') \to \Gamma(X,\mathcal{F})$ is injective, ii) $Ker ~\overset{\sim} {\psi}= Im ~\overset{\sim} {\phi} $ (Of course, $\overset{\sim} {\psi}: \Gamma(X,\mathcal{F}) \to \Gamma(X,\mathcal{F}'') $)[But, my intersting on this question is only i)].
( The following is also my thought) i) is equivalent to $ker~\overset{\sim} {\phi} = 0$. Thus, I assume $ker~\overset{\sim} {\phi} \neq 0$, i.e there is a nonzero function $s \in \Gamma(X,\mathcal{F}') $ s.t, $\overset{\sim} {\phi}(s)=0$. Let $\left \{ V_{i} \right \}$ be a open covering of $X$. Then, by (contrapostive) condition of sheaf, $s|_{V_i} \neq 0$ for some $i$, then since $s|_{V_i} \in \mathcal{F}'(V_i)$ and $\phi$ is injective by assumption, $\phi(s|_{V_i}) \in \mathcal{F}(V_i)$ is cannot be zero for some $i$, Hence $0 \neq \overset{\sim} {\phi}(s) \in \Gamma(X,\mathcal{F}) $ This contradicts to the assumption.
However, clearly, my proof seems to have many flaws. Personally, the reason I cannot prove the above completely is that I cannot surely understand what is the morphism of sheaves is exact sequence. The definition of morphism is sheaves is clear: If $U,V (V \subset U)$ are open subset of $X$, and ( $\mathcal{F}$ and $ \mathcal{F'}$ are obviously sheaves) Then, $\phi : \mathcal{F}' \to \mathcal{F} $ is a morphism of sheaves means diagram is commutative something like this. But what is such morphism is exact? My thought is that the following diagram holds... But is it the right idea?
