Can a homeomorphism of the closed unit disk map $S^1$ onto $S^1$?

[Euge]

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 http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

Honorable mention goes to vidyarth for a partially correct solution. You can read my solution below.

Spoiler

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.