Let $X$ be a space and we denote by $\mathbb{Z}_X$ the sheaf of local constant sections on $X$.
We are going to compare the sheaf cohomology $H^i(X,\mathbb{Z}_X)$ with singular cohomology $H^i(X,\mathbb{Z})$ with coefficients in $\mathbb{Z}$ (take into account that $X$ has an underlying structure of topological space).
My question is why $H^i(X,\mathbb{Z}_X)$ and $H^i(X,\mathbb{Z})$ coinside (up to isomorphism) if every $x \in X$ has a contractable neighbourhood base?
My attempts:
Denote $S_n(X):= \bigoplus_{s: \Delta_n \to X \text{ contin.}} \mathbb{Z}$ and $S^n(X):= Hom(S_n(X),\mathbb{Z})= \prod_{s: \Delta_n \to X \text{ contin.}} \mathbb{Z}$.
The differential maps are given by $$d_n: S^{n+1} \to S^n, s\mapsto \sum_{i=0} ^{n+1} (-1)^i s \vert _{\{(x_0,..., x_{n+1}) \vert x_i=0 \}}$$
this gives the complex
$$S^0(X) \to S^1(X) \to S^2(X) \to ...$$
It's $i$-th kohomology is exactly the $i-th$ singular cohomology $H^i(X,\mathbb{Z})$.
Otherwise we can define the presheaf via $V \mapsto S^n(V)$ for $V \subset X$. Denote by $\mathcal{I}^n_X$ it's sheafification.
We obtain the complex with induced differentials
$$\mathcal{I}^0 _X \to \mathcal{I}^1 _X \to \mathcal{I}^2 _X \to ...$$
One can verify that $\mathcal{I}^n_X$ are flabby and the canonical inclusion $\mathbb{Z}_X \to\mathcal{I}^0_X$ induce the complex
$$0 \to \mathbb{Z}_X \to \mathcal{I}^0 _X \to \mathcal{I}^1 _X \to ...$$
If we would know that this complex is exact we could apply the global section functor $\Gamma(-,X)$ to this flabby resolution to obtain the desired result.
So the obstructing problem is: Why is this complex EXACT?
How does our assumption that every $x \in X$ has a contractable neighbourhood base imply it?