I have been trying to prove the following result as an exercise, and I would like some help.
First, I don't know if the proof I have given is correct, or even if the initial statement is true. I just thought the statement sounded reasonable. In particular I do not actually fully understand how to establish results about the fundamental group of the Cech nerve of a space equipped with an open cover from the fundamental group of the original space, or conversely, by appealing to the nerve theorem or results about the geometric realization and I would like guidance where I attempt to use these results.
Second, if the proof is correct, I don't actually know what the necessary hypotheses are in order to make the result goes through, so it's possible some of my assumptions are irrelevant, and I would like to know in what generality the result holds.
"Theorem:" Let $M$ be a smooth two-dimensional manifold, and suppose $M$ is simply connected. Then if $\mathcal{F}$ is the locally constant sheaf of continuous functions from $M$ into a discrete Abelian group $A$, then the first Cech cohomology group $\check{H}^1(M;\mathcal{F})$ is zero.
Here is my attempt. Any open cover can be refined to to a "good cover" in the sense of (for example) Bott and Tu in ``Differential Forms in Algebraic Topology'', i.e. an open cover $\left\{ U_i \right\}_{i\in I}$ where each nonempty finite intersection $U_{i_0}\cap \dots U_{i_n}$ is diffeomorphic to $\mathbb{R}^2$.
Proof: Since $M$ is connected, the Cech nerve of this cover is $1$-connected, as if $U_i$ and $U_j$ are any two open sets, we can draw a path from a point in $U_i$ to a point in $U_j$, and this path can be covered with finitely many open sets from the cover. It follows that the 1-truncation of the nerve, as a connected graph, has a spanning tree. Choose such a spanning tree; we can consider this as a well-founded partial ordering where $i<j$ if $j$ is a child of $i$; the root node is the minimal element.
Now, let $\left\{ \sigma_{ij} \right\}_{i,j\in I}$ be a \v{C}ech 1-cocycle with respect to this cover, i.e. a family of sections of the sheaf over the intersections $U_i\cap U_j$ satisfying $\sigma_{ij}+\sigma_{jk}=\sigma_{ik}$ on the triple intersection $U_i\cap U_j\cap U_k$. Note that by the good-cover assumption, a $1$-cochain is equivalent to giving an element of $\prod_{i,j\in I}A$, as every section over $U_i\cap U_j$ is constant.
We will demonstrate that the $1$-cocycle admits a lift along the coboundary map by giving a $0$-cocycle $\left\{ \sigma_i \right\}_{i\in I}$ by constructing one by induction on the well-ordered set $I$. Let $\sigma_0 \in \mathcal{F}(U_0)$ be the constant zero section. Now we induct on the height of the well-founded tree mentioned earlier; if $j$ is a child of $i$, we define $\sigma_j$ as $\sigma_{ij}+\sigma_i$. (Here we interpret $\sigma_{ij}$ and $\sigma_j$ as ranging over elements of $A$ rather than as sheaf sections per se so that this addition is well-defined.)
We must now prove that this is actually a lift, so we show that for all $i,j$ with $U_i\cap U_j=\emptyset$ we have $\sigma_j-\sigma_i=\sigma_{ij}$. So let $i,j$ be arbitrary with $U_i\cap U_j\neq \emptyset$. In the spanning tree mentioned earlier, there is a unique path from $i$ to $j$ through the tree, whose nodes $i=i_0,i_1 ,\dots, i_{n-1},i_n=j$ define a sequence of open sets $U_{i_0}, U_{i_1},\dots, U_{i_n}$ where adjacent pairs meet nontrivially, and where by definition we have $\sigma_{i_{k+1}}-\sigma_{i_k}=\sigma_{i_k,i_{k+1}}$ for any $0\leq k < n$. It is clear that $\sigma_j-\sigma_i = \sum_{0\leq k<n}(\sigma_{i_{k+1}}-\sigma_{i_k})=\sum_{0\leq k<n}\sigma_{i_k,i_{k+1}}$; so it must be shown that \begin{equation} \label{eq:2} \sum_{0\leq k<n}\sigma_{i_k,i_{k+1}} = \sigma_{i,j} \end{equation} by some appeal to the cocycle condition.
For a change of perspective let us examine this problem from the perspective of the Cech nerve of the cover $\left\{ U_i \right\}_{i\in I}$ regarded as a simplicial complex. Each Cech $1$-cochain determines a map from the edges of the $1$-truncation of the Cech nerve into the group $A$. The cocycle condition $\sigma_{ij}+\sigma_{jk}=\sigma_{ik}$ can be viewed as a functoriality condition; if $\sigma$ is a $1$-cocycle then there is an induced functor from the edge-path groupoid of $N(\left\{ U_i \right\}_{i\in I})$ into $A$ (regarded as a category with one object), sending the edge $(i,j)$ associated to the intersection $U_i\cap U_j$ to $\sigma_{i,j}$, and the equation $\sigma_{ij}+\sigma_{jk}=\sigma_{ik}$ is a guarantee of compositionality, respecting the equivalence relation of homotopy between paths in the edge-path groupoid. To verify that the above equation holds is to verify that two paths, $i_0\to i_1\to\dots \to i_n$ and $i_0\to i_n$ in the edge-path groupoid, both map to the same morphism in $G$. To do this it suffices to guarantee that in fact the two paths $i_0\to i_1\to \dots \to i_n$ and $i_0\to i_n$ are equal in the edge-path groupoid; that is, they are homotopic in the Cech nerve. Because $M$ is simply connected and $\left\{ U_i \right\}$ is a good cover, the Cech nerve $N(\left\{ U_i \right\})$ is also simply connected (is this a corollary of the Nerve theorem???), and so there is exactly one isomorphism in the edge-path groupoid between any two open sets in the cover. This gives us the desired ``transitivity'' condition sought above, and so we are done.