Homeomorphism that maps given points to given points Let $\{x_{1},x_{2},...,x_{m}\}$ be a set of $m$ distinct points in $\mathbb{R}^{n}$, $\{y_{1},y_{2},...,y_{m}\}$ be another set of $m$ distinct points in $\mathbb{R}^{n}$.
Is there a homeomorphism $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$, which maps $x_{i}$ to $y_{i}$ for all $i$?
Well obviously it's not true for $n=1$, so let's assume $n\ge 2$
 A: 
Lemma 1. Let $n\in\mathbb{N}$, $D\subseteq\mathbb{R}^n$ be a closed disk and let $v,w\in \mbox{int}(D)$. Then there exists a homeomorphism $f:D\to D$ such that $f(u)=w$ and $f(x)=x$ for $x\in\partial D$.

Proof. It is a simple generalization of this: A homeomorphism of B^n fixing the boundary?
$\Box$

Lemma 2. Let $n>1$, $A=\{v_1,\ldots,v_m\}$ be a finite subset of $\mathbb{R}^n$ and $v,w\in\mathbb{R}^n$ are such that $v,w\not\in A$. Then there exists a homeomorphism $f:\mathbb{R}^n\to\mathbb{R}^n$ such that
$$f(v_i)=v_i$$
$$f(v)=w$$

Proof. Since $A$ is finite then there exists a point $z\in\mathbb{R}^n$ such that line segments $[v,z]$ and $[z,w]$ do not contain any of $A$ points. Define $P:=[v,z]\cup [z,w]$. Both $P$ and $A$ are closed and thus there is an $\epsilon$-neighbourhood $U$ of $P$ such that $U\cap A=\emptyset$. Since $\overline{U}$ is homeomorphic to a closed ball (not an obvious statement at all, also another non obvious statement is that the homeomorphism has to preserve boundaries) then lemma 1 applies and there is a homeomorphism $f:\overline{U}\to\overline{U}$ such that $f(v)=w$ and $f(x)=x$ for $x\in\partial U$. This homeomorphism can be easily extended to whole $\mathbb{R}^n$ by putting $F(x)=x$ for $x\not\in U$. This is what we are looking for. $\Box$

Lemma 3. Let $n>1$, $A=\{x_1,\ldots, x_m\}$ and $B=\{y_1,\ldots, y_m\}$ be finite subsets of $\mathbb{R}^n$. Then there exists a homeomorphism $f:\mathbb{R}^n\to\mathbb{R}^n$ such that $f(x_i)=y_i$.

Proof. By induction on $m$:
1) $m=1$ is trivial, you just take translation $f(x)=(y_1-x_1)+x$
2) Let $m>1$ be given. Take $A'=\{x_1,\ldots, x_{m-1}\}$ and $B'=\{y_1,\ldots, y_{m-1}\}$. By induction hypothesis there is a homeomorphism $f:\mathbb{R}^n\to\mathbb{R}^n$ such that $f(x_i)=y_i$ for $i=1,\ldots,m-1$.
Now by Lemma 2 there is a homeomorphism $g:\mathbb{R}^n\to\mathbb{R}^n$ such that
$$g(y_i)=y_i\mbox{ for all }i=1,\ldots,m-1$$
$$g(f^{-1}(x_m))=y_m$$
Put $F(x)=g(f(x))$. This is the homeomoprhism we are looking for. $\Box$
