I am going through the proof of Borsul Ulam theorem in the book Algebraic Topology by Hatcher. Here is the snippet.

enter image description here

enter image description here

What I am unable to understand is the line in the proof where it said that "since $q$ is odd", $h$ is not nullhomotopic. Say in $\pi(S^1)$, we have $( cos(2\pi s), sin(2\pi s))$ as the generator element. Then if $q$ is odd, then as per the claim in the proof, $[ cos(6\pi s), sin(6\pi s)]=h (s)$ assuming $q=3$ is not nullhomotopic. But we can define a nullhomotopy on this circle as defined below ($x_0$ is the base point):

$$f_t(s) = (1-t)h(s) + x_0t$$

So I can't understand on why if $q$ is odd, $h$ is not nullhomotopic , because as described above, I can define a nullhomotopy on this circle.

Also why $h$ is NOT nullhomotopic when $q$ is odd, but nullhomotopic when $q$ is even.

What I am getting wrong here. Support would be greatly appreciated.


Your map $f_t(s)$ does not take values on the circle: it is a nullhomotopy for maps from $S^1$ to $\Bbb R^2$, not a nullhomotopy (as requires) for maps from $S^1$ to $S^1$.

  • $\begingroup$ Can you please give an example of a nullhomotopy when $q$ is even? I think in case $q$ is even, the condition: $h(s+1/2)=-h(s)$ won't be satisfied. $\endgroup$ – user3001408 Jul 1 '17 at 11:33
  • $\begingroup$ I understand that my argument is wrong, but why in odd case it is NOT nullhomotopic? $\endgroup$ – user3001408 Jul 1 '17 at 12:53
  • $\begingroup$ $\pi^1(S^1)$ is infinite cyclic. The point is that odd numbers are automatically nonzero. Your element corresponds to $q$ times a generator of $\pi^1(S^1)$ so cannot be nullhomotopic. $\endgroup$ – Lord Shark the Unknown Jul 1 '17 at 15:30

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.