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This post states that we can examine the deck transformations of the covering space $S^2 \rightarrow \mathbb{R}P^2 \rightarrow X$ to see that only the identity and antipodal maps are deck transformations. I am not sure how to show this is true. The homeomorphism is clear to me once I prove this.

If $q: S^2 \rightarrow \mathbb{R}P^2$ is the quotient covering map, and $p: \mathbb{R}P^2 \rightarrow X$ is any covering map, then a deck transformation $\phi: S^2 \rightarrow S^2$ for $p \circ q$ satisfies $$p \circ q = (p \circ q) \circ \phi.$$ For any $x \in S^2$, we have $p([x]) = p([\phi(x)])$ where $[x] = q(x)$ denotes an equivalence class in $\mathbb{R}P^2$. We don't have any further assumptions on $p$, so I don't know how to proceed from here. I can see that deck transformations with respect to $q$ are just the identity or antipodal maps, but composing with $p$ messes me up.

Dalop
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  • Have you read the answers in the linked post? What do you find unclear in the proofs given in these answers? They have nothing to do with the "proof" that you are attempting to write. – Moishe Kohan Mar 10 '24 at 16:13
  • @MoisheKohan The answers in the linked post are irrelevant to what I am asking. I was just interested in the initial comment that the original poster made and wanted to work out the details. – Dalop Mar 10 '24 at 16:20
  • Well, what poster wrote as an idea of the proof turned out to be useless. Is it what you wanted to know? – Moishe Kohan Mar 10 '24 at 16:22
  • @MoisheKohan Please elaborate? The poster first states that one can solve the problem by examining deck transformations. Then they ask if the problem can be solved by showing the covering is one-sheeted. They later make an edit requesting for alternative solutions if the one-sheeted idea doesn't work. Is the idea to examine deck transformations the same as the one-sheeted idea? – Dalop Mar 10 '24 at 16:29
  • Actually, on the 2nd thought, the idea of examining deck transformations can be used to give an alternative proof, but it is harder than the written answers. It uses Lefschetz fixed point theorem. – Moishe Kohan Mar 10 '24 at 16:49

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Here is a proof based on examining covering transformations of $p\circ q$. The group $G$ of covering transformations of $p\circ q$ contains the antipodal map. In order to show that $p$ is 1-1, it suffices to prove that $G$ has order 2. For each continuous map $f: S^2\to S^2$ define the induced map of homology group: $$ f_{*,i}: H_i(S^2; {\mathbb Q})\to H_i(S^2; {\mathbb Q}). $$

Let $G_0< G$ denote the index 2 subgroup consisting of covering transformations preserving orientation of $S^2$, i.e. inducting the identity map $f_{*,2}$.

For an element $f\in G_0$ compute its Lefschetz number $\Lambda_f$, $$ \Lambda_f= Tr(f_{*,0}) - Tr(f_{*,1}) + Tr(f_{*,2})= 1 -0+1=2. $$ Hence, by the Lefschetz theorem, $f$ has a fixed point in $S^2$. Since $f$ is a covering transformation, it follows that such $f$ is the identity map, i.e. $G_0=1$, i.e. $G$ has order $2$, i.e. $p$ is 1-1.

Moishe Kohan
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