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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

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Unrelated: Did "coker" come from here? –  user21436 Feb 17 '12 at 16:55
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@Kannappan; No. It just means cokernel. –  Zhen Lin Feb 17 '12 at 17:28
    
Unrelateder: ...or from this ? –  Georges Elencwajg Feb 17 '12 at 18:14
    
Haha @GeorgesElencwajg Good one, that's a lovely dog, Swag –  Mario Carrasco Feb 20 '12 at 2:00
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up vote 4 down vote accepted

The exact sequence fragment $$H^{0}(\mathcal{E}) \rightarrow H^{0}(\mathcal{Q}) \rightarrow H^{1}(\mathcal{L}) \rightarrow H^{1}(\mathcal{E})$$ becomes, after the observation that the last term is $0$, $$H^{0}(\mathcal{E}) \rightarrow H^{0}(\mathcal{Q}) \rightarrow H^{1}(\mathcal{L}) \rightarrow 0.$$ Because this is exact, the map $H^0(\mathcal{Q})\to H^1(\mathcal{L})$ is onto. Therefore, by the isomorphism theorems, $$H^1(\mathcal{L}) \cong \frac{H^0(\mathcal{Q})}{\mathrm{ker}(H^0(\mathcal{Q})\to H^1(\mathcal{L}))}.$$ But since the sequence is exact, $\mathrm{ker}(H^0(\mathcal{Q})\to H^1(\mathcal{L})) = \mathrm{Im}(H^0(\mathcal{E})\to H^0(\mathcal{Q}))$. So: $$H^1(\mathcal{L}) \cong \frac{H^0(\mathcal{Q})}{\mathrm{ker}(H^0(\mathcal{Q})\to H^1(\mathcal{L}))} = \frac{H^0(\mathcal{Q})}{\mathrm{Im}(H^0(\mathcal{E})\to H^0(\mathcal{Q}))} = \mathrm{Coker}(H^0(\mathcal{E})\to H^0(\mathcal{Q})).$$

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This is awesome, thanks bro –  Mario Carrasco Feb 20 '12 at 1:57
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