ScalingStacks

Proof. The implications $(1)\Rightarrow(2)$, $(1)\Rightarrow(3)\Rightarrow(4)$ are clear. We can assume that $\gamma^{\prime}$ is not constant, for otherwise the other implications are trivial.

Assume $(2)$. Let $\hat{f}:\hat{Z}\to\hat{Z}^{\prime}$ be the map between non-singular covers corresponding to $f$. Since $\gamma^{\prime}$ is smooth, it lifts uniquely to a path $\hat{\gamma}^{\prime}$ on $\hat{Z}^{\prime}$ and $\hat{\gamma}^{\prime}([0,1])\subset\hat{f}(\hat{Z})$. Since $\hat{f}$ is an open embedding, it follows that $\hat{\gamma}^{\prime}$ is the image of a path of $\hat{Z}$. Its image in $Z$ is a smooth path that lifts $\gamma^{\prime}$, hence $(1)$ holds.

Assume $(4)$. Let $\gamma^{\prime}_{0}$ be a minimal smooth path homotopic to $\gamma^{\prime}$ (cf Properties 7.3.5(4)). We have $\gamma^{\prime}_{0}([0,1])=\operatorname{supp}\nolimits([\gamma^{\prime}])\subset f(Z)$, hence $\gamma^{\prime}_{0}$ lifts to a smooth path in $Z$. So $(3)$ holds.

Assume $(3)$ and $f$ is strict. Note that $\gamma^{\prime}([0,1])\cap Z^{\prime}_{o}=\operatorname{supp}\nolimits([\gamma^{\prime}])\cap Z^{\prime}_{o}$ and $\gamma^{\prime}([0,1])\cap Z^{\prime}_{u}$ is contained in the union of the connected components of $Z^{\prime}_{u}$ that have a non-empty intersection with $\operatorname{supp}\nolimits([\gamma^{\prime}])$ (Properties 7.3.2(1)). Since $f(Z_{u})$ is open and closed in $Z^{\prime}_{u}$, it follows that $\gamma^{\prime}([0,1])\subset f(Z)$, so $(2)$ holds.

Assume $\gamma^{\prime}$ is admissible and lifts to $Z$. Since $f$ is strict, it follows that the lift is oriented. ∎