ScalingStacks

Proof. Let $\gamma_{1}$, $\gamma_{2}$ be two minimal paths in $X$ with $\gamma_{1}(1)=\gamma_{2}(0)$. The path $\gamma_{2}\circ\gamma_{1}$ is minimal if and only if there are $t_{1},t_{2}\in(0,1)$ such that $\gamma_{1}((t_{1},1))\cap\gamma_{2}((0,t_{2}))=\emptyset$. If $\gamma_{2}\circ\gamma_{1}$ is not minimal, then there are unique elements $t_{1}\in[0,1)$ and $t_{2}\in(0,1]$ such that $(\gamma_{2})_{|[0,t_{2}]}\circ(\gamma_{1})_{|[t_{1},1]}$ is homotopic to a constant path and $(\gamma_{2})_{|[t_{2},1]}\circ(\gamma_{1})_{|[0,t_{1}]}$ is minimal (if $t_{2}\neq 1$ or $t_{1}\neq 0$).

We deduce by induction that a composition of minimal paths is homotopic to a minimal path or to a constant path.

Let $\gamma$ be a path in $X$. If $X$ is homeomorphic to an interval of ${\mathbf{R}}$, then $\gamma$ is homotopic to a minimal path or a constant path. In general there is a finite subset $E$ of $X$ such that given $U$ a connected component of $X\setminus E$, the space $\bar{U}$ is homeomorphic to an interval of ${\mathbf{R}}$. By Lemma 7.1.16 there is a path $\gamma^{\prime}$ homotopic to $\gamma$ and such that $\gamma^{\prime-1}(E)$ is finite. So, $\gamma^{\prime}$ is a composition of paths contained in subspaces of $X$ that are homeomorphic to intervals of ${\mathbf{R}}$. Consequently, $\gamma^{\prime}$ is a composition of minimal paths. It follows that $\gamma^{\prime}$, hence $\gamma$, is homotopic to a minimal or constant path.

Let $\gamma$ be a path homotopic to a constant path. The image $\bar{\gamma}$ of $\gamma$ in $\bar{X}=X/(X_{exc}\cup\{\gamma(0),\gamma(1)\})$ is homotopic to a constant path. Since $\bar{X}$ is homotopy equivalent to a wedge of circles, its fundamental group is free and $\bar{\gamma}$ cannot be a minimal path. It follows that $\gamma$ is not minimal.

Let $\gamma$ be a minimal path. Let $\{0=t_{0}<t_{1}<\ldots<t_{n}=1\}=\{0,1\}\cup\gamma^{-1}(X_{exc})$. Note that $\gamma((t_{i},t_{i+1}))$ is contained in a connected component $U_{i}$ of $X\setminus X_{exc}$ and it is a connected component if $\gamma(t_{i}),\ \gamma(t_{i+1})\in X_{exc}$. If $\bar{U}_{i}$ is homeomorphic to an interval of ${\mathbf{R}}$, then $U_{i}\neq U_{i+1}$ and $U_{i}\neq U_{i-1}$. Otherwise, $\bar{U}_{i}$ is homeomorphic to $S^{1}$ and if $U_{i}=U_{i+1}$, then the paths $\gamma_{|U_{i}}$ and $\gamma_{|U_{i+1}}$ have the same orientation.

Let $\gamma^{\prime}$ be a minimal path homotopic to $\gamma$. We will show the existence of $\phi$ as in the lemma by induction on $n$. Since $\gamma\circ\gamma^{\prime-1}$ is not minimal, there is $\varepsilon>0$ such that $\gamma^{\prime}([0,\varepsilon])\subset\bar{U}_{1}$. Consider $\varepsilon$ maximal with this property.

Assume $\gamma^{\prime}(\varepsilon){\not\in}X_{exc}$. We have $\varepsilon=1$. Let $\varepsilon^{\prime}\in(t_{0},t_{1}]$ such that $\gamma(\varepsilon^{\prime})=\gamma^{\prime}(\varepsilon)$. The path $\gamma_{|[\varepsilon^{\prime},1]}$ is homotopic to the identity, hence $n=1$, $\varepsilon^{\prime}=1$ and $\gamma(t_{1})=\gamma^{\prime}(\varepsilon)$.

If $\gamma^{\prime}(\varepsilon)\in X_{exc}$, then $\gamma^{\prime}(\varepsilon)=\gamma(t_{1})$ as well. In both cases, the paths $\gamma_{|[0,t_{1}]}$ and $\gamma^{\prime}_{|[0,\varepsilon]}$ are injective and have the same image. So, there is a homeomorphism $\psi:[0,\varepsilon]\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}[0,t_{1}]$ such that $\gamma^{\prime}(t)=\gamma(\psi(t))$ for $t\in[0,\varepsilon]$ and the existence of $\psi$ follows by induction.

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