The cokernel of $H^0(\hat{X},\sigma^*L^k)\to H^0(E,\mathcal{O}_E)\otimes L^k(x)$ Let $X$ be a compact manifold. $\sigma:\hat{X}\to X$ is the blow up of $X$ of $x\in X$. Denote $\sigma^{-1}(x)$ by $E$. $L^k\to X$ is a very ample line bundle. $L^k(x)$ is a fiber. And by $L^k\to X$ and $\sigma:\hat{X}\to X$, we can pullback $L^k$ along $\sigma$ to get $\sigma^*L^k$.
Then we have the following claim:

I can't see why the sequence is exact and why the cokernel is contained in $H^1(\hat{X},\sigma^*L^k\otimes\mathcal{O}(-E))$ by the sequence.
 A: First you have the exact sequence 
$$
0 \to \mathcal{O}_{\hat X}(-E) \to \mathcal{O}_{\hat X} \to \mathcal{O}_{E} \to 0
$$
Since $\sigma^\ast L^k$ is locally free, you can take the tensor products
$$
0 \to \sigma^\ast L^k \otimes \mathcal{O}_{\hat X}(-E) \to \sigma^\ast L^k \to \sigma^\ast L^k \otimes \mathcal{O}_{E} \to 0
$$
Finally, $\sigma^\ast L^k \otimes \mathcal{O}_{E} = L^k(x)\otimes \mathcal{O}_{E} $ because $\sigma$ is the blowup at $x$.
Now, taking the long exact sequence of cohomology we have
$$
0 \to \mbox{H}^0(\hat X ,\sigma^\ast L^k \otimes \mathcal{O}_{\hat X}(-E)) \to \mbox{H}^0(\hat X, \sigma^\ast L^k)  \to \mbox{H}^0(E,\sigma^\ast L^k \otimes \mathcal{O}_{E}) = \mbox{H}^0(E,\mathcal{O}_{E})\otimes L^k(x) \to \mbox{H}^1(\hat X ,\sigma^\ast L^k \otimes \mathcal{O}_{\hat X}(-E)) \to \dots
$$
To conclude we just need to compute the cokernel. Consider an exact sequence of vector spaces
$$
A \xrightarrow{f} B \xrightarrow{g} C.
$$
Then the cockernel of $f$ is 
$$
\frac{B}{f(A)} = \frac{B}{\ker g} \simeq g(B) \subset C
$$
Applying this to $\mbox{H}^0(\hat X, \sigma^\ast L^k)  \to  \mbox{H}^0(E,\mathcal{O}_{E})\otimes L^k(x) \to \mbox{H}^1(\hat X ,\sigma^\ast L^k \otimes \mathcal{O}_{\hat X}(-E))$ we prove the claim.
