Homotopy of functions of $S^1$ and $\mathbb{R}P^2$

Question

So I was doing an exercise that asked me that if there exist two functions $g \colon S^1 \rightarrow \mathbb{R}P^2$ and $f\colon\mathbb{R}P^2 \rightarrow S^1 $ then $f \circ g$ is homotopic to the identity in $S^1$. I think this is false and my argument is as follows. Lets suppose that it is true, then we know that the homomorphism of fundamental groups induced by $f\circ g$ is the trivial one since we are going to have in one part a homomorphism from $\mathbb{Z}_2$ to $\mathbb{Z}$. So if we have a loop $\gamma$ in the fundamental group of $S^1$, $f\circ g\circ \gamma \sim e$ where $e$ is the identity. Since we are assuming that $f\circ g $ is homotopic to the identity function, then we can use this homotopy to show that $\gamma \sim f\circ g \circ \gamma$ but by what we showed before this would imply that $\gamma \sim e$ for all the elements of the fundamental group of the circle, which would be a contradiction. So my question is, is this argument correct?

Answer 1

I think your argument is correct, but it can be expressed more succinctly using the functoriality of $\pi_1$.

You can show that any continuous composition $S^1 \stackrel{g}{\to} \mathbb{R}P^2 \stackrel{f}{\to} S^1$ is null-homotopic by considering the composition $\pi_1(S^1) \stackrel{\pi_1(g)}{\to} \pi_1(\mathbb{R}P^2) \stackrel{\pi_1(f)}{\to} \pi_1(S^1)$, which must be $0$ since there are no non-zero homomorphisms $\mathbb{Z}/2 \to \mathbb{Z}$. Your arguments about homotopies are already encapsulated in the definition and basic properties of $\pi_1$.

Answer 2

Another solution/approach:

Let $p : \mathbb{R} \rightarrow S^1$ be the universal covering $t \mapsto e^{i t}$. The induced map on fundamental groups $f_{\ast} : \pi_{1}(\mathbb{RP}^2) \rightarrow \pi_{1}(S^1)$ must be trivial. Hence $f_{\ast}(\pi_{1}(\mathbb{RP}^2)) \subseteq p_{\ast}(\pi_{1}(\mathbb{R}))$, which implies that there exists a lift $\tilde{f} : \mathbb{RP}^2 \rightarrow \mathbb{R}$ such that $f = p \circ \tilde{f}$. Now $\tilde{f}$ is homotopic to the constant map that is identically zero via the homotopy $\tilde{f}_{t}(x) : = (1 - t)\tilde{f}(x)$, which in turns yields a homotopy $f_{t} : = p \circ \tilde{f}_{t}$ between $f_{0} = f$ and $f_{1} = 1$. Since $f$ is homotopic to a constant map, so too is $f \circ g$ and this is NOT homotopic to $\operatorname{id}_{S^1}$