ScalingStacks

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

$$I_{l}=\{e^{2i\pi d/n_{x^{\prime}}}|0\leq d<n_{x^{\prime}},\ x_{l}\in\overline{f^{-1}(a({\mathbf{R}}_{>0}e^{2i\pi d/n_{x^{\prime}}}))}\}$$

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

$$I_{0}=\{e^{2i\pi d/n_{x^{\prime}}}|0\leq d<n_{x^{\prime}},\ f^{-1}(a({\mathbf{R}}_{>0}e^{2i\pi d/n_{x^{\prime}}}))\neq\emptyset,\ f^{-1}(x^{\prime})\cap\overline{f^{-1}(a({\mathbf{R}}_{>0}e^{2i\pi d/n_{x^{\prime}}}))}=\emptyset\}.$$

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 $f_{|f^{-1}(U)}=a\circ g\circ b^{-1}$. ∎