Functor of section over U is left-exact I am trying to prove $\Gamma(U,\cdot)$ is a left-exact functor $\mathfrak{Ab}(X)\to\mathfrak{Ab}$. This is Exercise 1.8 in Hartshorne, Chapter II or exercise 2.5.F of Vakil's notes (Nov 28,2015 version).
Since monomorphism of sheaves implies injection on open sets we have the exactness at the left place. For the middle place I really do not have idea. Actually I have a very silly question, it seems to me that if kernel of morphism of sheaves equals to the image of a morphism, then the corresponding kernel and image on some U should also be the same. But in this way the functor should be exact. So why this is not true?
 A: Method 1: Use Exercise II.1.4b to relate the image presheaf and its sheafification. Then by exactness of the original sequence, the isomorphism follows on sections of $U$. 
Method 2: Proceed just as in proof of Proposition II.1.1.
Let $\varphi: \mathscr{F} \to \mathscr{F''}$ and $\psi : \mathscr{F'} \to \mathscr{F}$. We want to show that the kernel and images are equal after applying $\Gamma (U, \cdot)$. This can be checked on the stalks. 
In one direction, we have that 
$$(\varphi _U \circ \psi _U (s))_P = \phi _P \circ \psi _ P (s_P) = 0$$
By the sheaf condition, this shows that $\phi _U \circ \psi _ U = 0$. 
Conversely, let's suppose $t \in \textrm{ker } \varphi _U$, i.e. $\varphi _U (t) = 0$. Again we know that the stalks is an exact sequence so for each $P \in U$, there is a $s_P$ such that $\psi _P (s_P) = t_P$. Let's represent each $s_P = (V_P , s(P))$ where $s(P) \in \mathscr{F'} (V_P)$. Now, $\psi (s(P))$ and $t \mid _{V_P}$ are elements of $\mathscr{F} (V_P)$ whose stalks at $P$ are the same. Thus, WLOG, assume $\psi (s(P)) = t \mid _{V_P}$ in $\mathscr{F} (V_P)$. $U$ is covered by the $V_P$ and there is a corresponding $s(P)$ on each $V_P$ which, on intersections, are both sent by $\psi$ to the corresponding $t$ on the intersection. Here, we apply injection (exactness at left place, which you showed in your OP) which allows us to glue via sheaf condition to a section $s \in \mathscr{F'} (U)$ such that $s \mid _ {V_P} = s(P)$ for each $P$. Verify that $\psi (s) = t$ and we're done by applying the sheaf property and the construction to $\psi (s) - t$. 
