5

I've already proved that set of points $z\in D^2$ such that $D^2-z$ is simply connected is precisely $S^1$. Now from this, I'm supposed to conclude that if $f:D^2\to D^2$ is a homeomorphism, then $f(S^1)=S^1$, but I'm not sure how to proceed. Can someone please help

user124910
  • 3,241
  • 4
    The restriction of $f$ to $D^2 \setminus {z}$ is still a homeomorphism. What can the image possibly be? – Alex G. May 07 '14 at 01:49
  • 4
    Well isn't $D^2-z$ homeomorphic to $D^2-f(z)$? If that's true, then $f$ induces a homomorphism $f_*:\pi (D^2-z)\to\pi (D^2-f(z))$, and so if $z\in S^1$, $\pi (D^2-z)=0$ which implies $\pi (D^2-f(z))=0$ which implies $f(z)\in S^1$. Is this right? – user124910 May 07 '14 at 02:03
  • 4
    Yes, exactly. This shows $f(S^1) \subseteq S^1$. The reverse inclusion is shown similarly. – Alex G. May 07 '14 at 02:06

0 Answers0