Locally constant sheaf on a simply connected space I was reading an article and at some point the writer claims that 
1)A locally constant sheaf on a simply connected topological space is a constant sheaf.
2) $H^{i}(U,\mathcal{F})=0 \hspace{0.1cm}\forall i>1$ where U is a homotopically trivial open set and $\mathcal{F}$ a locally constant sheaf.
How could I prove that?
Thank you for your time.
 A: A more abstract way of reformulating the result is that the category of local systems ( = locally constant sheaves) on $X$ with stalk $M$ ($M$ is a $k$-vector space or a module)  is equivalent to the category of representations $\rho : \pi_1(X) \to GL(M)$. In particular, if $X$ is simply connected then every locally constant sheaf is constant.
This also shows that one could compute everything related to the local system from the representation, as an example if $\mathscr L$ is a local system on $D^*$ (punctured disk) then a local system is equivalent to an element $T \in GL(M)$. The cohomology of $\mathscr L$ is the cohomology of the complex $M \overset{d}{\to} M$ with $d = \text{id} - T$. I am not aware of a formula in the general case but in theory this should be possible, using Cech complex.
I think a good reference for this Galois groups and fundamental groups by Tamas Szamuely. 
A: I want to add some details to Berci's comment. Let's just concentrate on the case when $X=[0,1]$, and when the locally constant sheaf $\mathcal{F}$ has stalks isomorphic to a $k$-vector space $M$. Also, denote the corresponding constant sheaf by $\mathcal{M}$. We shall show $\mathcal{F}$ is actually isomorphic to $\mathcal{M}$.
By definition, $[0,1]$ has an open cover $\{U_i\}$ such that $\mathcal{F}|_{U_i}$ is isomorphic to the constant sheaf $\mathcal{M}|_{U_i}$, via natural transformation $\eta_i$. What this really means is that for any open subsets $U$ of $U_i$, we have a (iso)morphism $\eta_i(U) \to \mathcal{M}(U)=M$ such that the following diagram commutes (for any open subset $V\subset U$):
\begin{array}
A\mathcal{F}(U) & \stackrel{\eta_i(U)}{\longrightarrow} & M \\
\downarrow{res^U_V} &  & \downarrow{=} \\
\mathcal{F}(V) & \stackrel{\eta_i(V)}{\longrightarrow} & M  
\end{array}
so that we have $\eta_i(U)=\eta_i(V) \circ res^U_V$ for any open subsets $V \subset U \subset U_i$.
For simplicity, suppose we have $U_1=[0,2/3)$ and $U_2=(1/3,1]$. Now for any open subset $U$ of the overlap $(1/3,2/3)$, we have two isomorphisms :
\begin{array}
A\mathcal{F}(U) & \stackrel{\eta_1(U)}{\longrightarrow} & M \\
\downarrow{=} & &  \\
\mathcal{F}(U) & \stackrel{\eta_2(U)}{\longrightarrow} & M  
\end{array}
they differ by the transition function $\eta_1(U) \circ \eta_2(U)^{-1}$, denote it by $\eta_{12}(U) \in GL(M)$.
Now, the point is that these transition functions $\eta_{12}(U)$ is actually independent of the choice of $U \subset (1/3,2/3)$, since $\eta_i((1/3,2/3))=\eta_i(U) \circ res^{(1/3,2/3)}_U$ tells us $\eta_{12}(U)=\eta_{12}((1/3,2/3))$. So we may denote it by $\eta_{12}$.
Then we can replace the natural transformation $\eta_2: \mathcal{F}|_{U_2} \to \mathcal{M}|_{U_2}$ by $\eta_2'=\eta_{12} \circ \eta_2$. Now $\eta_2'$ agree with $\eta_1$ on the overlap and they together defines an isomorphism (natural equivalence) from $\mathcal{F}$ to $\mathcal{M}$. So we win!
Finally, a few comments:
(1) What we learn from this is that if some sheaf is isomorphic to a constant sheaf of $S$ in a category $\mathcal{C}$, then two such isomphism can only differ by some elements in $Aut_\mathcal{C}(S)$. So that a locally constant sheaf is really the data of a "$S$-bundle", i.e., how you twist $S$ along the base.
(2) Where do you need simply-connectedness? Well, think about the case when $X=S^1$ be a circle. Now if $\mathcal{F}$ is a locally constant sheaf on $X$, you may as well use the above method to modify $\eta_{i+1}$ to make it agree with $\eta_i$. However, when you go around and back to starting point, $\eta_n$ may not agree with $\eta_1$. P. S. What you need to make it work is called the cocycle condition.
