ScalingStacks

Proof. Fix, for every $x\in X_{exc}$, a small open neighbourhood $U_{x}$ of $x$ and a homeomorphism $f_{x}:U_{x}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\mathrm{St}(E_{x})$, where $E_{x}$ is a finite subset of $S^{1}$. We choose now an equivalence relation on $E_{x}$ whose classes have cardinality at most $2$. Note that $f_{x}$ induces a bijection between $C(x)$ and $E_{x}$, hence the equivalence relation can be viewed on $C(x)$.

Define $\hat{U}_{x}=\coprod_{E^{\prime}\in E_{x}/\!\sim}\operatorname{St}\nolimits(E^{\prime})$. The map $f_{x}$ provides an open embedding

$$U_{x}-\{x\}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\mathrm{St}^{\circ}(E_{x})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\coprod_{E^{\prime}\in E_{x}/\!\sim}\operatorname{St}\nolimits^{\circ}(E^{\prime})\hookrightarrow\hat{U}_{x}.$$

We put

$$\hat{X}=(X-X_{exc})\coprod_{(\coprod_{x\in X_{exc}}(U_{x}-\{x\}))}\bigl(\coprod_{x\in X_{exc}}\hat{U}_{x}\bigr).$$

Note that $\hat{X}$ is a $1$-dimensional manifold. Let $q:\hat{X}\to X$ be the canonical map: it identifies $X$ with the quotient of $\hat{X}$ by the equivalence relation given by $\hat{x}_{1}\sim\hat{x}_{2}$ if $q(\hat{x}_{1})=q(\hat{x}_{2})$. Up to isomorphism, $\hat{X}$ depends only on the choice of an equivalence relation on $C(x)$ for $x\in X_{exc}$. ∎