ScalingStacks

Proof. Assume first $f$ is a non-singular cover of $Z^{\prime}$.

Let $\zeta^{\prime}\in L(\theta^{\prime})$. There are $i^{\prime}_{1}\neq i^{\prime}_{2}\in f(I)$ such that $\zeta^{\prime}\in I(\theta^{\prime}_{i^{\prime}_{i}},\theta^{\prime}_{i^{\prime}_{2}})$. By Lemma 7.3.24, there are elements $\zeta_{r}\in f^{-1}(\theta^{\prime}_{i_{r}})$ and $\zeta\in I(\zeta_{1},\zeta_{2})$ such $\zeta^{\prime}=f(\zeta)$. We define $\theta\in f^{-1}(\theta^{\prime})$ by setting $\theta_{\zeta_{r}(0)}=\zeta_{r}$ and by setting $\theta_{i}$ to be any lift of $\theta^{\prime}_{f(i)}$ for all $f(i)\notin\{i^{\prime}_{1},i^{\prime}_{2}\}$. This shows the surjectivity part of the first statement of the lemma.

Consider now $\theta$ and $\hat{\theta}$ maps in ${\mathcal{P}}_{f}^{\bullet}(Z)$ such that $f(\theta)=f(\hat{\theta})=\theta^{\prime}$. Let $\zeta\in L(\theta)$ and $\hat{\zeta}\in L(\hat{\theta})$ such that $f(\zeta)=f(\hat{\zeta})=\zeta^{\prime}$. There are $i^{\prime}_{1}\neq i^{\prime}_{2}\in f(I)$ such that $\zeta^{\prime}\in I(\theta^{\prime}_{i^{\prime}_{1}},\theta^{\prime}_{i^{\prime}_{2}})$. We have $\hat{\theta}_{\hat{\zeta}(t)},\theta_{\zeta(t)}\in f^{-1}(\theta^{\prime}_{i^{\prime}_{r}})$ for $t\in\{0,1\}$. It follows from Lemma 7.3.24 that $\zeta=\hat{\zeta}$. So, the first statement of the lemma holds.

Assume now $f$ is an open embedding. The injectivity of the first map of the lemma is clear, while the surjectivity follows from Lemma 7.3.25.

We deduce the first part of the lemma for $Z$ and $Z^{\prime}$ non-singular and the general case follows now by taking non-singular covers of $Z$ and $Z^{\prime}$ and the lift of $f$.

Let us prove now the second statement of the lemma about $D(\theta)$.

Consider $\theta\in f^{-1}(\theta^{\prime})$ and $\zeta\in L(\theta)$. It is clear that if $f(\zeta)\in D(\theta^{\prime})$, then $\zeta\in D(\theta)$.

Assume now $\zeta\in D(\theta)$. Fix $i_{1}\neq i_{2}\in I$ so that $\zeta\in I(\theta_{i_{1}},\theta_{i_{2}})$.

Let $\zeta^{\prime},\zeta^{\prime\prime}\in L(\theta^{\prime})$ such that $f(\zeta)=\zeta^{\prime}\circ\zeta^{\prime\prime}$. Let $z=\zeta^{\prime}(0)=\zeta^{\prime\prime}(1)$. If $|f^{-1}(\theta^{\prime}_{z})|>1$, then $z\in Z_{o}$ and $\theta^{\prime}_{z}=\operatorname{id}\nolimits$, hence $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations by Lemma 7.4.15. Assume now $\theta^{\prime}_{z}$ has a unique lift. Let $\hat{\zeta}^{\prime}$ and $\hat{\zeta}^{\prime\prime}$ be the unique lifts of $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ (first part of the lemma). By unicity of lifts, we have $\zeta=\hat{\zeta}^{\prime}\circ\hat{\zeta}^{\prime\prime}$. We have $\hat{\zeta}^{\prime},\hat{\zeta}^{\prime\prime}\in L(\theta)$, hence $\hat{\zeta}^{\prime}$ and $\hat{\zeta}^{\prime\prime}$ have opposite orientations. It follows that $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations as well.

Consider now $\zeta^{\prime}:f(i_{1})\to f(i_{2})$ a smooth homotopy class of paths such that $f(\zeta)\circ\zeta^{\prime-1}$ and $\zeta^{\prime-1}\circ f(\zeta)$ are smooth and have the same orientation as $f(\zeta)$ and $\zeta^{\prime}$. Let $\hat{\zeta}^{\prime}$ be the unique lift of $\zeta^{\prime}$. Since $f(\zeta)\circ\zeta^{\prime-1}$ is smooth, it follows that $\hat{\zeta}^{\prime}(0)=i_{1}$ and $\zeta\circ\hat{\zeta}^{\prime-1}$ is smooth and has the same orientation as $\zeta$. Similarly, $\hat{\zeta}^{\prime}(1)=i_{2}$ and $\hat{\zeta}^{\prime-1}\circ\zeta$ is smooth and has the same orientation as $\zeta$. A similar statement holds for $\zeta$ replaced by $\bar{\zeta}$. We deduce that $f(\zeta)\in D(\theta^{\prime})$. ∎