I am reading Raoul Bott and Loring W. Tu Differential Forms in Algebraic Topology, and in the page 146 there is the next theorem:
Let $\mathfrak{U}$ be a good cover on a connected topological space $X$ and $N(\mathfrak{U})$ its nerve. If $\pi_1(N(\mathfrak{U})) = 0$, then every locally constant presheaf on $\mathfrak{U}$ is constant.
Where in this case $\pi_1(N(\mathfrak{U}))$ is the edge path group of the nerve of the cover.
The problem is that I am not used to the notions of constant and locally constant presheaf, and I don't understand why the proof is valid. In the book, they construct for every open subset and isomorphism between $\mathfrak{F}(U_{\alpha_0...\alpha_r})$ and the group $G$. Why this implies the constant character of the presheaf?
I know that is not well seen to post an image, but in this case I will do it for clarify the question. To sum up, why by this proof is the presheaf constant?:
