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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?

William
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Someone
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  • Where was the exercise from? – Oliver Jones Jan 15 '20 at 23:57
  • An old exam from my topology class – Someone Jan 15 '20 at 23:58
  • And there were no conditions on $f$ and $g$? – Oliver Jones Jan 16 '20 at 00:01
  • Ah yeah forgot to mention they are continuous – Someone Jan 16 '20 at 00:01
  • Okay. This might be true but I need to think about it for a bit. – Oliver Jones Jan 16 '20 at 00:03
  • You've made a mistake though. $f\circ g$ is a map ${\Bbb R}P^2\rightarrow {\Bbb R}P^2$. – Oliver Jones Jan 16 '20 at 00:08
  • Yeah i switched the functions , im going to edit it – Someone Jan 16 '20 at 00:10
  • Something looks wrong here. Take $f:{\Bbb R}P^2\rightarrow S^1$ to be any constant map and $g:S^1\rightarrow {\Bbb R}P^2$ to be any continuous map. Then $f\circ g$ is constant. – Oliver Jones Jan 16 '20 at 00:35
  • I already answered, but I realized I'm not sure if I'm parsing the question correctly. Is the question "Are there a pair of continuous functions such that $f\circ g \sim id_{S^1}$?", or is it asking to show "If $f$ and $g$ are continuous show that $f\circ g \sim id_{S^1}$"? If it's the first case then my answer is valid, but in the second case the problem is obviously false by Oliver's easy counter-example, and more generally because of my answer. – William Jan 16 '20 at 01:43
  • It's possible "the identity on $S^1$" is a typo for "the constant map on $S^1$" in the problem statement. This is rendered plausible by the fact that $1\in S^1\subset\mathbb C$ is a multiplicative identity. – Cheerful Parsnip Jan 16 '20 at 01:56
  • @CheerfulParsnip I think you're correct. Any continuous map $f:{\Bbb R}P^2\rightarrow S^1$ is null-homotopic. Hence $f\circ g$ is also null-homotopic for any $g$. – Oliver Jones Jan 16 '20 at 02:57
  • Is it straightforward to show $[\mathbb{R}P^2 , S^1] = 0$? I can't come up with something simple off the top of my head. Is it easier than using functoriality of $\pi_1$ to turn it into an algebraic problem? – William Jan 16 '20 at 03:10
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    @William Here's a link to a proof: https://math.stackexchange.com/a/143112/55622 – Oliver Jones Jan 16 '20 at 03:50

2 Answers2

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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 $\pi_1$.

William
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  • Isn't the answer much simpler? In my comment above, I show it's easy to construct $f$ and $g$ so that $f\circ g$ is the constant map. – Oliver Jones Jan 16 '20 at 02:32
  • @OliverJones I'm still a bit confused about what the question specifically is (see my comment), I'd like more clarification before I decide what to do with my answer – William Jan 16 '20 at 02:42
  • As suggested by Cheerful in his comment above, there's probably a typo in the question It should read $f\circ g$ is null-homotopic. See my comment for a proof. – Oliver Jones Jan 16 '20 at 03:00
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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 $\operatorname{id}_{S^1}$

Oiler
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