ScalingStacks

Proof. The implication $(1)\Rightarrow(2)$ follows from the fact that $X_{f}\subset f^{-1}(f(E_{1}))$. For the implication $(2)\Rightarrow(3)$, take $E_{2}=X_{f}$. For $(3)\Rightarrow(4)$, take $E_{x}=(X-\{x\})\cap(f^{-1}(f(x))\cup E_{2})$. The implication $(4)\Rightarrow(5)$ is immediate.

Let us show that $(5)\Rightarrow(1)$. Note first that an injective continuous map ${\mathbf{R}}\to{\mathbf{R}}$ is open and a homeomorphism onto its image. It follows that the implication holds when $X$ and $X^{\prime}$ are homeomorphic to ${\mathbf{R}}$ and $E_{3}=\emptyset$.

Consider now the general case. There is a finite subset $E_{1}$ of $X$ containing $E_{3}$ such that $X-E_{1}$ and $X^{\prime}-f(E_{1})$ are homeomorphic to a finite disjoint union of copies of ${\mathbf{R}}$. By the discussion above, the restriction of $f$ to a connected component of $X-E_{1}$ is open and a homeomorphism onto its image, so the same holds for $f_{|X-E_{1}}$.

∎

Proof. Let $E$ be a finite subset of $X$ such that $f^{-1}(f(E))=E$, $f(X-E)$ is open in $X^{\prime}$ and $f_{|X-E}:X-E\to f(X-E)$ is a homeomorphism. Let $U$ be a small open neighbourhood of $x^{\prime}$ such that $U-\{x^{\prime}\}\subset X^{\prime}-f(E)$. Note that $f(X)\cap(U-\{x^{\prime}\})$ is open in $X^{\prime}$ and $f_{|f^{-1}(U-\{x^{\prime}\})}:f^{-1}(U-\{x^{\prime}\})\to f(X)\cap(U-\{x^{\prime}\})$ is a homeomorphism.

Let $L$ be a connected component of $U-\{x^{\prime}\}$. Note that $f(f^{-1}(L))$ is an open $1$-dimensional subspace of $L$ and $L$ is homeomorphic to ${\mathbf{R}}$. By shrinking $U$, we can assume that $f^{-1}(L)=\emptyset$ or $f(f^{-1}(L))=L$. So, we can assume that given $L$ a connected component of $U-\{x^{\prime}\}$ with $f^{-1}(L)\neq\emptyset$, the map $f_{|f^{-1}(L)}:f^{-1}(L)\to L$ is a homeomorphism.

Since $U$ is small, there is a homeomorphism $a:\operatorname{St}\nolimits(n_{x^{\prime}})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}U,\ 0\mapsto x^{\prime}$. Let $\{x_{1},\ldots,x_{r}\}=f^{-1}(x^{\prime})$ and define

for $l\in\{1,\ldots,r\}$. Define

Note that $a$ restricts to a homeomorphism $\operatorname{St}\nolimits(\bigcup_{0\leq l\leq r}I_{r})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}f(f^{-1}(U))$.

The composition $a\circ g$ takes values in $f(f^{-1}(U))$. Its restriction to $\operatorname{St}\nolimits^{\circ}(I_{0})$ defines a homeomorphism $\operatorname{St}\nolimits^{\circ}(I_{0})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}a(\operatorname{St}\nolimits^{\circ}(I_{0}))$. Since $f_{|f^{-1}(a(\operatorname{St}\nolimits^{\circ}(I_{0})))}:f^{-1}(a(\operatorname{St}\nolimits^{\circ}(I_{0})))\to a(\operatorname{St}\nolimits^{\circ}(I_{0}))$ is a homeomorphism, we have a homeomorphism $b_{0}=(f_{|f^{-1}(a(\operatorname{St}\nolimits^{\circ}(I_{0})))})^{-1}\circ(a\circ g)_{|\operatorname{St}\nolimits^{\circ}(I_{0})}:\operatorname{St}\nolimits^{\circ}(I_{0})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}f^{-1}(a(\operatorname{St}\nolimits^{\circ}(I_{0})))$.

Consider now $l\in\{1,\ldots,r\}$. We construct as above a homeomorphism $b^{\prime}_{l}:\operatorname{St}\nolimits^{\circ}(I_{l})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}f^{-1}(a(\operatorname{St}\nolimits^{\circ}(I_{l})))$ such that $(a\circ g)_{|\operatorname{St}\nolimits^{\circ}(I_{l})}=f\circ b^{\prime}_{l}$. The homeomorphism $b^{\prime}_{l}$ extends uniquely to a homeomorphism $b_{l}:\operatorname{St}\nolimits(I_{l})\to f^{-1}(a(\operatorname{St}\nolimits(I_{l})))$. We define $b=b_{0}\sqcup b_{1}\sqcup\cdots\sqcup b_{r}$. We have $f_{|f^{-1}(U)}=a\circ g\circ b^{-1}$. ∎

Proof. Fix, for every $x\in X_{exc}$, a small open neighbourhood $U_{x}$ of $x$ and a homeomorphism $f_{x}:U_{x}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\mathrm{St}(E_{x})$, where $E_{x}$ is a finite subset of $S^{1}$. We choose now an equivalence relation on $E_{x}$ whose classes have cardinality at most $2$. Note that $f_{x}$ induces a bijection between $C(x)$ and $E_{x}$, hence the equivalence relation can be viewed on $C(x)$.

Define $\hat{U}_{x}=\coprod_{E^{\prime}\in E_{x}/\!\sim}\operatorname{St}\nolimits(E^{\prime})$. The map $f_{x}$ provides an open embedding

We put

Note that $\hat{X}$ is a $1$-dimensional manifold. Let $q:\hat{X}\to X$ be the canonical map: it identifies $X$ with the quotient of $\hat{X}$ by the equivalence relation given by $\hat{x}_{1}\sim\hat{x}_{2}$ if $q(\hat{x}_{1})=q(\hat{x}_{2})$. Up to isomorphism, $\hat{X}$ depends only on the choice of an equivalence relation on $C(x)$ for $x\in X_{exc}$. ∎

Proof. Given $e\in E$, let $U_{e}$ be a connected and simply connected open neighborhood of $e$. Choose $U_{e}$ small enough so that $U_{e}\cap U_{e^{\prime}}=\emptyset$ for $e\neq e^{\prime}$. Let $U=\bigcup_{e\in E}U_{e}$. Let $V$ be an open subset of $X\setminus E$ containing $X\setminus U$.

Let $C$ be the set of connected components $I$ of $[0,1]\setminus\gamma^{-1}(E)$ such that $\bar{I}$ is not contained in $\gamma^{-1}(U)$ nor in $\gamma^{-1}(V)$. By Lebesgue’s number Lemma, that set is finite. Since the restriction of $\gamma$ to $\bar{I}$ is nullhomotopic for $I\in C$, it follows that $\gamma$ is homotopic to a path $\gamma^{\prime}$ that is constant on $\bar{I}$ for $I\in C$ and that coincides with $\gamma$ on $[0,1]-\bigcup_{I\in C}I$. Let $I^{\prime}$ be a connected component of $[0,1]\setminus\gamma^{-1}(E)$ with $I^{\prime}{\not\in}C$. We have $\bar{I}\cap\gamma^{-1}(E)\neq\emptyset$, hence $\bar{I}\subset\gamma^{-1}(U)$. We deduce that $\gamma^{\prime}([0,1])\subset U$, hence $\gamma^{\prime}$ is nullhomotopic. ∎

Proof. Let ${\mathcal{U}}$ be an open covering of $X$ by connected and simply connected subsets, each of which contain at most one element of $E$. By Lebesgue’s number Lemma, there are only finitely many $I\in\pi_{0}([0,1]\setminus\gamma^{-1}(E))$ such that $\bar{I}$ is not contained in an element of $\gamma^{-1}({\mathcal{U}})$. So, $B$ is finite.

We can write $\gamma$ as a finite composition of its restrictions to $\bar{I}$ for $I\in B$ interlaced with finitely many paths that satisfy the assumptions of Lemma 7.1.15. Thanks to that lemma, we obtain a path $\gamma^{\prime}$ satisfying the requirements of the lemma. By shrinking the intervals on which $\gamma^{\prime}$ is constant to points, we obtain a path $\gamma^{\prime\prime}$ as desired. ∎

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.

∎

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. ∎

Proof. The first statement is clear. Let us now prove the second statement. That statement is clear if $\gamma((0,1))\cap(X_{exc}\cup\{x\})=\emptyset$.

The left side of the equality is additive under compositions of paths, and so is the right side by Lemma 7.1.22 below.

Assume now $\gamma^{-1}(X_{exc}\cup\{x\})$ is finite. The path $\gamma$ is a (finite) composition of paths mapping $(0,1)$ into the complement of $X_{exc}\cup\{x\}$, hence the statement holds for $\gamma$.

Consider now the general case. The proof of Lemma 7.1.16 for $E=X_{exc}\cup\{x\}$ produces a path $\gamma^{\prime}$ homotopic to $\gamma$ such that $\gamma^{\prime-1}(E)$ is finite and such that $|I_{c}^{+}(\gamma)|-|I_{c}^{-}(\gamma)|=|I_{c}^{+}(\gamma^{\prime})|-|I_{c}^{-}(\gamma^{\prime})|$. Since the statement holds for $\gamma^{\prime}$, it follows that it holds for $\gamma$. ∎