Let $q: \mathbb{R} \rightarrow S^1$ be the map $\theta \rightarrow e^{i\theta}$. Given any continuous function $f: S^1 \rightarrow S^1$ with $f(1,0)=(1,0)$, show that there is a continuous map $\tilde{f}: \mathbb{R} \rightarrow \mathbb{R}$ with $\tilde{f}(0)=0$ and $q\tilde{f}(\theta)=fq(\theta)$
So at first I did this problem by saying since $f$ is continuous from $S^1 \rightarrow S^1$ then it must be that $f(\theta)=r\theta$ for some $r \in \mathbb{R}$, but then I realized that is not true.
So now i'm trying to define something like $\tilde{f}(\theta)=q^{-1}(f(\theta))$... Unfortunately the preimage of $q$ has more than one value in $\mathbb{R}$ so this map is not well defined!! Would it be possible to like, select one of the values in the set $q^{-1}f(\theta)$ in order to define $\tilde{f}$ or am I not on the correct track at all here?
Usually when I see questions like this I want to use a lifting map somewhere, but as $f_*(\pi(S^1,x_0))$ is not contained in $q_*(\pi(\mathbb{R},r))$, the lifting criteria isn't satisfied so I couldn't do something like that.
Advice appreciated!! Thanks!
You can do something which involves the lifting criteria, and it's the simplest way to solve this. But the map that you want to lift is not $f$, it is $f\circ q$. You have that the following condition holds $$ (f\circ q)_*(\pi_1(\Bbb R,0))\subset q_*(\pi_1(\Bbb R,0))$$ simply because $(f\circ q)_*(\pi_1(\Bbb R,0))$ is trivial. Hence you can lift $f\circ q:\Bbb R\to S^1$ to a unique continuous map $\tilde{f}:\Bbb R\to \Bbb R$ such that $\tilde{f}(0)=0$, which therefore satisfies $q\circ \tilde{f}=f\circ q$.