ScalingStacks

Proof. A minimal path is a locally injective path. Since every point of $X$ has an open neighbourhood on which $f$ is injective (cf Lemma 7.1.8), the image by $f$ of a minimal path is a minimal path.

Consider the set $\Omega=\{t\in[0,1]\ |\ \gamma(t)\neq\gamma^{\prime}(t)\}$, an open subset of $[0,1]$. Let $I$ be a connected component of $\Omega$. If $I=[0,1]$, then $\gamma$ and $\gamma^{\prime}$ are constant paths at distinct points of $X$ with the same image under $f$. Otherwise, let $s\in\overline{I}-I$. There is an open neighbourhood $U$ of $\gamma(s)=\gamma^{\prime}(s)$ such that $f_{|U}$ is injective. There is $t\in I$ such that $\gamma(t)$ and $\gamma^{\prime}(t)$ are in $U$, hence $\gamma(t)=\gamma^{\prime}(t)$, a contradiction. This shows the second assertion of the lemma.

Assume $\gamma$ and $\gamma^{\prime}$ are minimal. Since $f(\gamma)$ and $f(\gamma^{\prime})$ are minimal and homotopic, it follows from Lemma 7.1.18 that there is $\phi:[0,1]\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}[0,1]$ with $\phi(0)=0$ and $\phi(1)=1$ such that $f(\gamma^{\prime})=f(\gamma)\circ\phi=f(\gamma\circ\phi)$. It follows from the previous assertion of the lemma that $\gamma^{\prime}=\gamma\circ\phi$.

Assume now $\gamma$ is minimal. Since $f(\gamma)$ is minimal, it follows that $[f(\gamma^{\prime})]$ is not the identity, hence $[\gamma^{\prime}]$ is not the identity. We deduce that the third assertion of the lemma holds when $[\gamma]$ and $[\gamma^{\prime}]$ are not both identities. The case where they are both identities is clear. ∎