# Pullback of a locally constant sheaf by a function whose domain is simply connected

Let $\mathcal{A}$ be a locally constant sheaf on a topological space $X$ and let $\sigma:\Delta_p\to X$ denote a singular $p$-simplex. Writing the pullback of $\mathcal A$ by $\sigma$ as $\sigma^\ast(\mathcal A)$, Bredon's book on sheaf theory (page 26 in the second edition) says:

Since $\mathcal{A}$ is locally constant and $\Delta_p$ is simply connected, $\sigma^{\ast}(\mathcal{A})$ is a constant sheaf on $\Delta_p$.

(Emphasis mine).

I realize this is pretty basic, but I can't seem to figure out why the simply-connectedness of $\Delta_p$ enters into this. Is it in general true that the pullback of a locally constant sheaf by a continuous function whose domain is simply connected is a constant sheaf on said domain? Could someone hint at a proof for this?

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To get some intuition, you might try to find, say, a map $f:S^1 \rightarrow X$ and a locally constant sheaf $\mathscr{F}$ over $X$ such that $f^*\mathscr{F}$ isn't constant. –  Aaron Mazel-Gee May 16 '12 at 19:25
The espace étalé of a locally constant sheaf on a space $X$ is a covering space of $X$. If $X$ is simply connected, any covering space is trivial, and thus the sheaf you started with must be constant.