Use the fact that $\text{id}:S^n\to S^n$ is not homotopic to a constant to show that there is no retraction of $B^{n+1}$ onto $S^n$.

I tried to look up online but most of the solutions use the concept of fundamentally group, which we never mentioned.

  • $\begingroup$ @JimBelk I wrote down the definition of "retraction" and was trying to argue by contradiction. We didn't actually talk about "homotopic" so I really only know the definitions :( $\endgroup$ – Toujou Jun 5 '15 at 0:45
  • $\begingroup$ It's very strange that you would be assigned this problem in a course that hasn't yet covered fundamental group or homotopy. Is it possible that you're mistaken about what the course has covered? Alternatively, is it possible that this course assumes knowledge of fundamental group and homotopy as a prerequisite? $\endgroup$ – Jim Belk Jun 5 '15 at 0:50
  • $\begingroup$ @JimBelk We've never mentioned fundamental group. We talked about the definition of homotopy and it's an equivalent relation, and that's all. It's an undergrad introductory class. We also talked about Brouwer fixed point theorem, but not the proof. $\endgroup$ – Toujou Jun 5 '15 at 1:09

No knowledge about the fundamental group is necessary to solve this problem, although it leads to more "algebraic" (and even shorter) proof of the non-existence of such a retraction (at least for $n=1$). But the essential idea is the same.
Here is a hint: The inclusion map $i:S^n\hookrightarrow B^{n+1}$ is null-homotopic, i.e. there is a homotopy $H$ from $i$ to a constant map, for example $H(x,t)=tx$. If there were a retraction $r:B^{n+1}\to S^n$, what would that mean for the composite $rH:S^n\times I\to S^n$ ?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.