ScalingStacks

Proof. Let $Z^{\prime}$ be a non-singular curve. We have a map $h:\operatorname{Hom}\nolimits(Z^{\prime},\hat{Z})\to\operatorname{Hom}\nolimits(Z^{\prime},Z),\ g\mapsto q\circ g$. Since $Z_{q}$ is finite, it follows that $h$ is injective.

Consider now a morphism of curves $f:Z^{\prime}\to Z$. We factor $f$ as $Z^{\prime}\xrightarrow{f_{1}}Z^{\prime}/\!\sim\ \xrightarrow{f_{2}}Z$ as in Lemma 7.2.7. By Proposition 7.2.8, there is a morphism $\hat{f}:Z^{\prime}\to\hat{Z}$ such that $q\circ\hat{f}=f$, hence $h(\hat{f})=f$. So $h$ is surjective. ∎