I wish to prove that for $\pi : Y \to X$ a covering space (spaces involved are Hausdorff, semilocally simply connected, path connected, and locally path connected, among other things), the sheaf $$ \mathcal{S}(U) = \{\sigma: U \to Y \text{ continuous } \vert \pi \circ \sigma = \text{id}_U\}$$ is locally constant (i.e. for every $x \in X$ there is a neighbourhood on which the restriction of $\mathcal{S}$ is the constant sheaf). I know this is equivalent to the function being constant on connected components.
I have tried taking an evenly covered neighbourhood $U$ of $x$ and showing that $\mathcal{S}\vert_U$ is constant, but don't believe this is the correct approach: for $V \subset U$ an open subset, a locally constant function $\sigma: U \to Y$ (here $Y$ regarded as a set) does not satisfy $\pi \circ \sigma = \text{id}_V$.
But other than the above naive choice, it doesn't seem like there can be a restriction of $\mathcal{S}$ to any other open set for which locally constant functions coincide with functions for which $\pi \circ \sigma = \text{id}$. Somehow it seems the only way that this can work is if we require singletons are open in $X$ which can't always be true.
Otherwise due to the property that $\pi^{-1}(U) = \sqcup_\alpha U_\alpha$ and $U_\alpha \cong U$ for an evenly covered neighbourhood, the fact that $\pi \circ \sigma = \text{id}$ is something which seems to conflict with $\sigma$ being locally constant. I figure that I could also take $U$ to be simply connected too, but since $Y$ is any such covering space I don't think this gives any 'easier' restrictions on what my functions $\sigma$ should look like. Some resolution/hint towards proving this fact would be appreciated.