In Part II of General Topology, Munkres, Theorem 57.1 states that if $h:S^1\to S^1$ is continuous and antipode-preserving, then $h$ is not nulhomotopic.
There are some definitions I should clarify here: $S^1$ is the unit circle on the $xy$-plane, so we could parametrize $S^1$ as $(\cos \theta, \sin\theta)$, $0\leq \theta\leq 2\pi$. A map $f:S^1\to S^1$ is antipode-preserving if for $x\in S^1$ we have $f(-x)=-f(x)$, which means for every pair of antipodes $\{x, -x\}$, their images under $f$ are also antipodes $\{f(x), f(-x)\}=\{f(x),-f(x)\}$. Finally, $h$ is nulhomotopic means $h$ is path-homotopic to a constant map $e: S^1\to S^1$, by definition this means there is a continuous map $H:S^1\times [0,1]\to S^1$ such that $H(x, 0)=h(x), H(x, 1)=e(x)$. There are other details about this proof, but these should be enough for the question I would like to ask.
In the proof, by composition with some rotation map, we can further assume $h(b_0)=b_0$, where $b_0=(1,0)$. We define a map $q:S^1\to S^1$ by $q(x)=x^2$ (in the complex sense), hence
$$q(\cos \theta,\sin\theta)=(\cos2\theta, \sin2\theta).$$
It is easy to see $q$ is a quotient map, because it is a continuous closed surjective map. Note that $q$ maps each pair of antipode points to the same point, and the inverse of any point in $S^1$ under $q$ is a pair of antipode.
My question: The steps thereafter states that since $q$ is a quotient map, so $q\circ h$ induces a continuous map $k:S^1\to S^1$ such that $k\circ q=q\circ h$. I didn't quite get this point because I don't remember such result from the chapter of quotient map. My idea to get myself through this property is to let $k=q\circ h\circ q^{-1}$ and show that $k$ is well-defined. Indeed, for $x\in S^1$, $q^{-1}(x)$ is not a single point, but a pair of antipodes, then $h(q^{-1}(x))$ is also a pair of antipodes, then $q$ maps $h(q^{-1}(x))$ into a single point, thus $k$ is well-defined. I believe $k$ is continuous so I'm not troubling myself to show it, but is there a general way to see why $q$ being a quotient map will generate the continuous map $k$?
I have noted this question has been asked twice before, but the specific question is different from mine in this link and the answer to this link doesn't answer my question in more detail.
Best Answer
You have to apply Munkres' Theorem 22.2. Consider the map $g = p \circ h : S^1 \to S^1$. Each set $p^{-1}(z)$ consists of two antipodal points $w_1, w_2 = -w_1$ (these are the two complex square roots of $z$). But obviously $g$ is constant on all $p^{-1}(z)$ since $g(w_2) = h(w_2)^2 = h(-w_1)^2 = (-h(w_1))^2 = h(w_1)^2 = g(w_1)$.