Prove that for any $x,y \in$ Int $D^n$, $\exists$ homeomorphism $\phi: D^n \rightarrow D^n$ such that $\phi(a)=b$ for all $n \geq 1$, fixing boundary.

Question

I know a geometric proof, is there any algebraic proof? My approach, say for $n=1$, say for any $a,b \in (0,1)$,(say $a < b$)We need to find a homeomorphism $\phi$ that fixes the boundary, that is $\{0,1\}$ and $\phi(a)=b$. Geometrically, thinking $[0,1]$ as a rubber sheet, upto homeomorphism we can always stretch a till we reach $b$ and shrink $b$ accordingly. The idea is similar in the case for $n=2$, take any two point $a,b \in$ Int ($D^2$). Then apply the above procedure for the line joining $a$ and $b$. Now, similarly we can do it for all the line in any neighborhood of this line and we are done! Any Algebraic proof? Thanks for any help!

Answer 1

• take a homeomorphism $f(x)=x/(1-\|x\|)$ from the interior of the disc to $R^n$
• compose with a translation that takes $f(a)$ to $f(b)$
• compose with the inverse of $f$
• check that this extends as the identity in the boundary.