ScalingStacks

Proof. $\bullet\$ Assume $\zeta_{1}$ or $\zeta_{2}$ is an identity. In that case, (1) and (2) follow from Lemma 7.3.21 and (3) follows from Lemmas 7.1.24 and 7.3.21.

From now on, we assume that neither $\zeta_{1}$ nor $\zeta_{2}$ is an identity.

$\bullet\$ Let $f:Z\to Z^{\prime}$ be an injective morphism of curves and assume $f(\zeta_{1})$ and $f(\zeta_{2})$ satisfy (1) and (2). We have $i(f(\zeta_{1}),f(\zeta_{2}))\leq i(\zeta_{1},\zeta_{2})$. There are minimal admissible paths $\gamma^{\prime}_{i}$ in $f(\zeta_{i})$ for $i\in\{1,2\}$ such that $i(f(\zeta_{1}),f(\zeta_{2}))=i(\gamma^{\prime}_{1},\gamma^{\prime}_{2})$. There are admissible paths $\gamma_{i}$ of $Z$ such that $\gamma^{\prime}_{i}=f(\gamma_{i})$ for $i\in\{1,2\}$. It follows that $i(\zeta_{1},\zeta_{2})\geq i(\gamma^{\prime}_{1},\gamma^{\prime}_{2})=i(\gamma_{1},\gamma_{2})$, hence $i(f(\zeta_{1}),f(\zeta_{2}))=i(\zeta_{1},\zeta_{2})$. We deduce also that (1) and (2) hold for $\zeta_{1}$ and $\zeta_{2}$.

$\bullet\$ Assume $Z=S^{1}$ unoriented. The assertions (1) and (2) follow from Lemma 6.2.3.

$\bullet\$ Assume $Z$ is non-singular and connected. There is an injective map $f:Z\to S^{1}$. It follows that $Z$ satisfies (1) and (2). This shows that (1) and (2) hold for a general non-singular curve.

$\bullet\$ Let $Z^{\prime}$ be an arbitrary curve and let $f:Z\to Z^{\prime}$ be the non-singular cover of $Z^{\prime}$. Assume $f(\zeta_{1}(t))\neq f(\zeta_{2}(t))$ for $t\in\{0,1\}$. Since all admissible paths in $Z^{\prime}$ lift to $Z$, it follows that $i(\zeta_{1},\zeta_{2})\leq i(f(\zeta_{1}),f(\zeta_{2}))$.

Consider two minimal admissible paths $\gamma_{1}$ and $\gamma_{2}$ in $\zeta_{1}$ and $\zeta_{2}$ such that $i(\gamma_{1},\gamma_{2})=i(\zeta_{1},\zeta_{2})$. We assume that given $\rho_{1},\rho_{2}:[0,1]\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}[0,1]$ any two homeomorphisms fixing $0$ and $1$ and such that $i(\gamma_{1}\circ\rho_{1},\gamma_{2}\circ\rho_{2})=i(\zeta_{1},\zeta_{2})$, we have $i(f(\gamma_{1}),f(\gamma_{2}))\leq i(f(\gamma_{1}\circ\rho_{1}),f(\gamma_{2}\circ\rho_{2}))$. Let $t_{0}\in(0,1)$ such that $\gamma_{1}(t_{0})\neq\gamma_{2}(t_{0})$ but $f(\gamma_{1}(t_{0}))=f(\gamma_{2}(t_{0}))$. There is a small open neighbourhood $U$ of $z^{\prime}=f(\gamma_{1}(t_{0}))$ homeomorphic to $\operatorname{St}\nolimits(n_{z^{\prime}})$ and with $U\cap f(Z_{f})=\{z^{\prime}\}$ and there are $0\leq t_{1}<t_{0}<t_{2}\leq 1$ such that $f(\gamma_{1})([t_{1},t_{2}])\subset U$ and $f(\gamma_{2})([t_{1},t_{2}])\subset U$. The paths $(\gamma_{1})_{|[t_{1},t_{2}]}$ and $(\gamma_{2})_{|[t_{1},t_{2}]}$ are contained in disjoint connected components of $f^{-1}(U)$, hence $f(\gamma_{1})([t_{1},t_{2}])\cap f(\gamma_{2})([t_{1},t_{2}])=\{z^{\prime}\}$. So, by reparametrizing $f(\gamma_{1})$ and $f(\gamma_{2})$ in the interval $[t_{1},t_{2}]$, we can assume they do not have a common value in that interval. This contradicts the minimality of $i(f(\gamma_{1}),f(\gamma_{2}))$. It follows that

$$i(\zeta_{1},\zeta_{2})=i(\gamma_{1},\gamma_{2})=i(f(\gamma_{1}),f(\gamma_{2}))\geq i(f(\zeta_{1}),f(\zeta_{2})),$$

hence $i(\zeta_{1},\zeta_{2})=i(f(\zeta_{1}),f(\zeta_{2})).$ This shows that (1) and (2) hold for $f(\gamma_{1})$ and $f(\gamma_{2})$. We deduce that (1) and (2) hold in full generality. It follows also that (3) holds when $f$ is injective.

$\bullet\$ Consider now a morphism of curves $f:Z\to Z^{\prime}$. Consider the map $\hat{f}:\hat{Z}\to\hat{Z}^{\prime}$ between non-singular covers corresponding to $f$. Let $\hat{\zeta}_{i}$ be the lift of $\zeta_{i}$ to $\hat{Z}$. Since $\hat{f}$ is injective, it follows that $i(\hat{f}(\hat{\zeta}_{1}),\hat{f}(\hat{\zeta}_{2}))=i(\hat{\zeta}_{1},\hat{\zeta}_{2})$. The study above shows that $i(\hat{f}(\hat{\zeta}_{1}),\hat{f}(\hat{\zeta}_{2}))=i(f(\zeta_{1}),f(\zeta_{2}))$ and $i(\hat{\zeta}_{1},\hat{\zeta}_{2})=i(\zeta_{1},\zeta_{2})$. It follows that $i(f(\zeta_{1}),f(\zeta_{2}))=i(\zeta_{1},\zeta_{2})$. This completes the proof of the lemma. ∎