Given a short exact sequence of sheaves on $X$:
$0 \rightarrow \mathcal{A} \rightarrow \mathcal{B} \rightarrow \mathcal{C} \rightarrow 0$
Then there is an induced exact sequence on the left on global sections:
$0 \rightarrow \mathcal{A}(X) \rightarrow \mathcal{B}(X) \rightarrow \mathcal{C}(X)$
I'm studying on Bredon's book, and there the proof of this statement is said to be trivial because "look at stalks". But to me is not so trivial, because with stalks one loses all the global information about a sheaf.
So, why is this true?
This is usually the last part of the sequence which is not exact but the left part is exact and this is not so hard since it essentially involve injectivity.
For example, assume that $\phi : F \to G$ is a morphism of sheaf, then $\phi_U$ is injective for all $U$ if and only if $\phi_p$ is injective for all $p$.
One implication is clear, and if $s \in F(U)$ and $\phi_U(s) = 0$, then $\phi_p(s_p) = 0 = \phi_U(s)_p$ so it follows by injectivity of $\phi_p$ that $\phi_U(s)$ has all stalk zero, i.e that it's the zero section and $\phi_U$ is indeed injective.
Same kind of argument will give you the exactitude of this sequence. Again the only tricky part is the surjectivity.