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I am doing some problems on Algebraic Topology, this one I came across and currently have no idea how to solve for it, I am glad to have some hints to get me started with (don't give me the full answer, please)

if $f : B^2 \to B^2$ is a continuous map so that the restriction $f|_{S^1}$ is a homeomorphism from $S^1\to S^1$, then $f$ is surjective.

I should say I just finished my reading on the first chapter of Algebraic Topology from Munkers's Topology, Second Edition

user63279
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Suppose that $x_0\in B^2 \setminus S^1$ is not in the image of $f$ then you can find an homeomorphism $\phi\colon B^2\to B^2$ which is the identity on $S^1$ and such that $\phi(x_0)=0$. Then define $$ g(x) = \frac{f(\phi(x))}{|f(\phi(x))|} $$ which would be a retraction of $B^2$ onto its boundary. This is a contradiction because the fundamental group of $B^2$ is trivial while that of $S^1$ is not.

Emanuele Paolini
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  • I see, thank you very much, but I did not find "second homotopy group" on my book, is it the same as the "Fundamental group?" – user63279 Feb 21 '13 at 07:48
  • another question, how's do I get the composition of $\pi$ and $f$ to be a retraction? The restriction of $f$ on $S^1$ is a homeomorphism, but for retraction don't we need an the composition to be identity on the boundary? – user63279 Feb 21 '13 at 07:51
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    @user63279: The second homotopy group is different from the fundamental group. I think Emanuele means to use the fundamental group, since the second homotopy group of both spaces are trivial. Also, you're right that this map is not the identity on the boundary, but you can fix that by composing it with $f|_{S^1}^{-1}$. – Jason DeVito - on hiatus Feb 21 '13 at 14:17
  • @Jason DeVito good point, thank you, I think I understand it now. – user63279 Feb 21 '13 at 14:56
  • you are right... I made some corrections – Emanuele Paolini Feb 21 '13 at 18:10