I don't understand the implication in the boxed line or why $q$ is odd. In the mentioned proof of the fundamental group of $S^1$ the covering map was $p(s)=(\cos2 \pi s, \cos2 \pi s )$. i don't see how we can use the same covering map here. Any ideas what is the map $p$ here and how it gives the implication? thanks
(This is from Algebraic Topology by Hatcher, page 33)
We have $p(s) = e^{2\pi i s}$. Therefore $h(t) = p \tilde{h}(t) = e^{2\pi i \tilde{h}(t)}$ and $$e^{2\pi i\tilde{h}(s+1/2)} = h(s+1/2) = -h(s) = e^{\pi i}e^{2\pi i\tilde{h}(s)} = e^{2\pi i\tilde{h}(s) + \pi i} .$$ This equation implies that $$2\pi i\tilde{h}(s+1/2) = 2\pi i\tilde{h}(s) + \pi i + 2k\pi i$$ for some $k \in \mathbb Z$. Hence $$\tilde{h}(s+1/2) = \tilde{h}(s) + 1/2 + k = \tilde{h}(s) + q/2$$ with $q = 2k+1$.
Edited on request:
Hatcher shows that $h$ represents $q$-times the standard generator of $\pi_1(S^1)$. Since $q$ is odd, we have $q \ne 0$, thus $h$ represents a non-zero element of $\pi_1(S^1)$. We conclude that $h$ is not nullhomotopic.