Welcome to our community

Be a part of something great, join today!

Problem of the Week #240 - Jan 03, 2017

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
Happy New Year, MHB! Since the year has just started I figured I'd start with a light problem which I'm sure several of you can solve.

-----
Prove that a homeomorphism of the closed unit disk onto itself must map $S^1$ onto $S^1$.

-----


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
Honorable mention goes to vidyarth for a partially correct solution. You can read my solution below.


Let $\bar{D}^2$ be the closed unit disk in $\Bbb R^2$. Let $f : \bar{D}^2 \to \bar{D}^2$ be a homeomorphism. Suppose there exists a $z\in S^1$ such that $f(z)\notin S^1$. There is a homeomorphism $\bar{D}^2\setminus\{z\} \xrightarrow{~} \bar{D}^2\setminus \{f(z)\}$ induced by the restriction of $f$ to $\bar{D}^2\setminus\{z\}$. As $z\in S^1$, $\bar{D}^2\setminus\{z\}$ is contractible (since it is a convex subspace of the plane). Since $f(z)$ lies in the interior of $\bar{D}^2$, $\bar{D}^2\setminus\{f(z)\}$ deformation retracts to $S^1$. So the induced map on fundamental groups gives an isomorphism $0 \to \Bbb Z$, which is a contradiction.
 
Status
Not open for further replies.