ScalingStacks

Proof. Let $(\tilde{Z},\sim)$ and $(\tilde{Z}^{\prime},\sim^{\prime})$ be two non-singular curves with finite relations and let $q:\tilde{Z}\to Z=\tilde{Z}/\!\!\sim$ and $q^{\prime}:\tilde{Z}^{\prime}\to Z^{\prime}=\tilde{Z}^{\prime}/\!\!\sim^{\prime}$ be the quotient maps.

A morphism of curves $f:\tilde{Z}\to\tilde{Z}^{\prime}$ such that $z_{1}\sim z_{2}$ implies $f(z_{1})\sim^{\prime}f(z_{2})$ induces a morphism of curves $Z\to Z^{\prime}$. So, the quotient functor induces indeed a functor as claimed. Consider $f^{\prime}:\tilde{Z}\to\tilde{Z}^{\prime}$ such that $z_{1}\sim z_{2}$ implies $f^{\prime}(z_{1})\sim^{\prime}f^{\prime}(z_{2})$. If $q^{\prime}\circ f=q^{\prime}\circ f^{\prime}$, then $f$ and $f^{\prime}$ coincide outside a finite set of points, hence $f=f^{\prime}$. So, the quotient functor is faithful.

Consider now a morphism of curves $g:Z\to Z^{\prime}$. Let $E^{\prime}$ be the finite subset of $\tilde{Z}^{\prime}$ of points that are not alone in their equivalence class and $E=q^{-1}(g^{-1}(q^{\prime}(E^{\prime})))$. Consider the composition of continuous maps

$$f:\tilde{Z}-E\xrightarrow{q}Z-q(E)\xrightarrow{g}Z^{\prime}-q^{\prime}(E^{\prime})\xrightarrow{(q^{\prime}_{|\tilde{Z}^{\prime}-E^{\prime}})^{-1}}\tilde{Z}^{\prime}-E^{\prime}.$$

Given $z\in E$, the $\iota$-equivariance of $C(g):C_{Z}(q(z))\to C_{Z^{\prime}}(g(q(z)))$ ensures that $f$ extends to a continuous map at $z$. So, $f$ extends (uniquely) to a continuous map $\tilde{Z}\to\tilde{Z}^{\prime}$, and that map is a morphism of $1$-dimensional spaces.

We have $\tilde{Z}_{u}\subset\tilde{Z}-E$ and $f(\tilde{Z}_{u})\subset\tilde{Z}^{\prime}_{u}$. Since $g_{|g^{-1}(\tilde{Z}^{\prime}_{o})-E}$ is orientation-preserving, it follows that $f_{|f^{-1}(\tilde{Z}^{\prime}_{o}-E^{\prime})}$ is orientation-preserving. So, $f:\tilde{Z}\to\tilde{Z}^{\prime}$ is a morphism of curves and it is compatible with the relations. This shows that the quotient functor is fully faithful.

Let now $Z$ be a curve. Let $z\in Z_{exc}$ and $U_{z}\subset Z_{o}$ be a small open neighbourhood of $z$. Fix an isomorphism of curves $f_{z}:U_{z}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\mathrm{St}(n_{z}),\ z\mapsto 0$. The equivalence relation on $\pi_{0}(U_{z}-\{z\})$ whose equivalence classes are the orbits of $\iota$ defines via $f_{z}$ the equivalence relation on $\{e^{i\pi r/2n_{z}}\}_{0\leq r<2n_{z}}$ given by $\zeta\sim\zeta^{\prime}$ if and only if $\zeta^{\prime}=\zeta^{\pm 1}$.

The proof of Lemma 7.1.14 provides us a non-singular curve $\hat{Z}$ with a finite relation. Indeed, with the notations of the proof of Lemma 7.1.14, we have $\hat{U}_{z}=\coprod_{0\leq r<n_{z}}{\mathbf{R}}e^{i\pi r/n_{z}}$. Note that $\hat{Z}_{o}$ is the subspace of $\hat{Z}$ obtained by adding to $Z_{o}-Z_{exc}$ the point $0$ of ${\mathbf{R}}e^{i\pi r/n_{z}}$ for each $r\in\{0,\ldots,n_{z}-1\}$ and each $z\in Z_{exc}$.

This gives $\hat{Z}$ a structure of non-singular curve. As in the proof of Lemma 7.1.14, we obtain a finite relation on $\hat{Z}$ and an isomorphism of curves $Z\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\hat{Z}/\!\sim$. This shows that the quotient functor is essentially surjective. ∎