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

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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.

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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 $S^1$.

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  • $\begingroup$ 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

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