Homeomorphism between a manifold minus distinct points [duplicate]

Suppose $M$ is a connected Hausdorff manifold, then show that for any $x,y \in M$ there exist a homeomorphism $\phi:M \to M$ such that $\phi(x)=y$

Using the definitions, I concluded that there exists open neighborhood $U,V$ of $x,y$ respectively and open sets $R_1 \subset \Bbb R^m$ and $R_2 \subset \Bbb R^m$ such that $U$ and $R_1$ , $V$ and $R_2$ are homeomorphic.

But I am unable to proceed from here.

Solution 1:

Lemma: Let $x,y\in \text{int}\left(\Bbb D^n\right):=\{z\in\Bbb R^n:||z||< 1\}$. Then, there is a homeomorphism $\varphi:\Bbb D^n\to \Bbb D^n$ such that $\varphi(w)=w$ for $||w||=1$ and $\varphi(x)=y$.

Consider the homeomorphism $\psi:\text{int}(\Bbb D^n)\to \Bbb R^n$ given by $$\psi(z)=\frac{z}{1-||z||}.$$ Let $T:\Bbb R^n\to \Bbb R^n$ be the translation given by $T(z)=z-\psi(x)+\psi(y)$. Then, consider $\psi^{-1}\circ T\circ \psi:\text{int}(\Bbb D^n)\to \text{int}(\Bbb D^n)$ and extend it to whole $\Bbb D^n:=\{z\in\Bbb R^n:||z||\leq 1\}$.

Why such an extension is possible? Note that for $||z||<1$, write $z=rv$ for some $v\in\Bbb S^{n-1}$ and some $r\in[0,1)$. Then, $\psi(z)=\frac{r}{1-r}v$ and for any $R\in [0,\infty)$ and any $w\in \Bbb S^{n-1}$ we have $\psi^{-1}(Rw)=\frac{R}{1+R}w$. So, for any $r\in [0,1)$ and any $v\in \Bbb S^{n-1}$ we have $$\psi^{-1}\circ T\circ \psi(rv)$$$$=\frac{\left|\left|\frac{r}{1-r}v-\psi(x)+\psi(y)\right|\right|}{1+\left|\left|\frac{r}{1-r}v-\psi(x)+\psi(y)\right|\right|}\cdot\frac{\frac{r}{1-r}v-\psi(x)+\psi(y)}{\left|\left|\frac{r}{1-r}v-\psi(x)+\psi(y)\right|\right|}$$$$=\frac{\left|\left|rv-(1-r)\psi(x)+(1-r)\psi(y)\right|\right|}{(1-r)+\left|\left|rv-(1-r)\psi(x)+(1-r)\psi(y)\right|\right|}\cdot\frac{rv-(1-r)\psi(x)+(1-r)\psi(y)}{\left|\left|rv-(1-r)\psi(x)+(1-r)\psi(y)\right|\right|}.$$

Every connected topological manifold is homogeneous i.e. for a connected manifold $M$ and any two points $a,b\in M$ there is a homeomorphism $\Phi:M\to M$ such that $\Phi(a)=b$.

To prove this, consider the non-empty set $$S:=\{x\in M|\text{ there is a homeomorphism }f:M\to M\text{ with }f(x)=b\}.$$ Now, consider $y\in S$ with a homeomorphism $g:M\to M$ such that $g(y)=b$. Let $\psi:U\big(\subseteq_{\text{closed}}M\big)\to \Bbb D^n$ be a homeomorphism with $y\in \text{int}(U)$. Now, for any $x\in\text{int}(U)$, choose $\varphi:\Bbb D^n\to\Bbb D^n$ such that $\varphi\left(\psi(x)\right)=\psi(y)$ and $\varphi(w)=w$ for $||w||=1$. So, define a homeomorphism $f:M\to M$ as $$f(z)=\begin{cases}g\big(\psi^{-1}\circ \varphi\circ \psi(z)\big) &\text{ if }z\in\text{int}(U),\\g(z)&\text{ if }z\in M\backslash \text{int}(U). \end{cases}$$ In other words, $\text{int}(U)\subseteq S$. That is $S$ is open in $M$. Similarly, prove that $M\backslash S$ is open. Now, $M$ is connected implies the result.

Some extra fact: Every connected manifold $M$ is $2$-homogeneous, i.e. given $\{a_1,a_2\}\cup\{b_1,b_2\}\subseteq M$, we have a homeomorphism $\psi:M\to M$ such that $\psi(a_k)=b_k$ for each $k=1,2$.

Let $$T:=\big\{(x_1,x_2)\in M\times M\big|\text{there is a homeomorphism }$$$$f:M\to M\text{ with }f(x_1)=b_1,f(x_2)=b_2\big\}.$$ Now, consider $(y_1,y_2)\in T$ with a homeomorphism $g:M\to M$ such that $g(y_1)=b,g(y_2)=b_2$. Let $\psi_k:U_k\big(\subseteq_{\text{closed}}M\big)\to \Bbb D^n$ be a homeomorphism with $y_k\in \text{int}(U_k)$ for $k=1,2$ with $U_1\cap U_2=\emptyset$. Now, for any $x_k\in\text{int}(U_k)$, choose $\varphi_k:\Bbb D^n\to\Bbb D^n$ such that $\varphi_k\left(\psi_k(x_k)\right)=\psi_k(y_k)$ and $\varphi_k(w)=w$ for $||w||=1$ where $k=1,2$. So, define a homeomorphism $f:M\to M$ as $$f(z)=\begin{cases}g\big(\psi_1^{-1}\circ \varphi_1\circ \psi_1(z)\big) &\text{ if }z\in\text{int}(U_1),\\g\big(\psi_2^{-1}\circ \varphi_2\circ \psi_2(z)\big) &\text{ if }z\in\text{int}(U_2),\\g(z)&\text{ if }z\in M\backslash\big( \text{int}(U_1)\cup \text{int}(U_2)\big). \end{cases}$$ In other words, $\text{int}(U_1)\times \text{int}(U_2)\subseteq T$. That is $T$ is open in $M\times M$. Similarly, $\big(M\times M\big)\backslash T$ is open. Now, $M\times M$ is connected implies the result.

Similarly, one can show, every connected topological manifold is $k$-homogeneous for every $k\in\Bbb N$.