Can anybody help me understand this proof?
In Rotman's 'An introduction to homological algebra' in Proposition 6.75 (iii). Flabby sheaves $\mathcal{L}$ are acyclic (Page 381), in the proof it says
Let $\mathcal{L}$ be flabby. Since there are enough injectives, there is an exact sequence $0 \rightarrow \mathcal{L} \rightarrow \mathcal{E} \rightarrow \mathcal{Q} \rightarrow 0$ with $\mathcal{E}$ injective. Now $\mathcal{E}$ is flabby, by Corollary 6.74 (Corollary 6.74 says that every injective sheaf $\mathcal{E}$ over a space $X$ is flabby), and so $\mathcal{Q}$ is flabby, by part (ii) (Part (ii) says: Let 0 \rightarrow \mathcal{L}' \rightarrow \mathcal{L} \rightarrow \mathcal{Q} \rightarrow 0 be an exact sequence of sheaves. If \mathcal{L}' and $\mathcal{L}$ are flabby, then $\mathcal{Q}$ is flabby). We prove that $H^{q}(\mathcal{L}) = 0$ by induction on $q \geq 1$. If $q = 1$, the long exact cohomology sequence contains the fragment
$H^{0}(\mathcal{E}) \rightarrow H^{0}(\mathcal{Q}) \rightarrow H^{1}(\mathcal{L}) \rightarrow H^{1}(\mathcal{E})$.
Since $H^{1}(\mathcal{E}) = \left\{ 0\right\}$, we have $H^{1}(\mathcal{L}) = coker(\Gamma(\mathcal{E}) \rightarrow \Gamma(\mathcal{Q}))$. But this cokernel is $0$, by part (i), and so $H^{1}(\mathcal{L}) = \left\{ 0\right\}$...
Q: I don't get where the coker part came from? Why is $H^{1}(\mathcal{L}) = coker(\Gamma(\mathcal{E}) \rightarrow \Gamma(\mathcal{Q}))$?? I'm lost here