ScalingStacks

Proof. Given $e\in E$, let $U_{e}$ be a connected and simply connected open neighborhood of $e$. Choose $U_{e}$ small enough so that $U_{e}\cap U_{e^{\prime}}=\emptyset$ for $e\neq e^{\prime}$. Let $U=\bigcup_{e\in E}U_{e}$. Let $V$ be an open subset of $X\setminus E$ containing $X\setminus U$.

Let $C$ be the set of connected components $I$ of $[0,1]\setminus\gamma^{-1}(E)$ such that $\bar{I}$ is not contained in $\gamma^{-1}(U)$ nor in $\gamma^{-1}(V)$. By Lebesgue’s number Lemma, that set is finite. Since the restriction of $\gamma$ to $\bar{I}$ is nullhomotopic for $I\in C$, it follows that $\gamma$ is homotopic to a path $\gamma^{\prime}$ that is constant on $\bar{I}$ for $I\in C$ and that coincides with $\gamma$ on $[0,1]-\bigcup_{I\in C}I$. Let $I^{\prime}$ be a connected component of $[0,1]\setminus\gamma^{-1}(E)$ with $I^{\prime}{\not\in}C$. We have $\bar{I}\cap\gamma^{-1}(E)\neq\emptyset$, hence $\bar{I}\subset\gamma^{-1}(U)$. We deduce that $\gamma^{\prime}([0,1])\subset U$, hence $\gamma^{\prime}$ is nullhomotopic. ∎