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