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I have encountered the following exercise in some lecture notes for an undergraduate course in topology.

Consider the real projective space $\mathbb{P}^n_{\mathbb{R}}$ (the $\mathbb{R}$ will be omitted in what follows), defined as the set of equivalence classes of the relation $\sim$ over $S^n$, where $x \sim y$ if there is a line passing through the origin containing both $x$ and $y$. In this way, $\mathbb{P}^n = \{ [x] = \{ x, -x\} \ : \ x \in S^n \}$.

Let $p: S^n \to \mathbb{P}^n$ be the standard projection, that is, $p(x) = [x]$.

The question is: why is there no continuous function $q: \mathbb{P}^n \to S^n$ such that $p \circ q$ is the identity map over $\mathbb{P}^n$?

boejiden
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    Do you know connectivity? compactness? – Arctic Char Aug 18 '21 at 16:14
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    Basic things about the fundamental group, functorial properties, things like that. We saw the fundamental group of $S^1$ and similar things, but not much more. Yes we also saw connectivity and compactness – boejiden Aug 18 '21 at 16:19
  • You might find it useful to visualize the special case $n=1$, where the projection winds the circle $\mathbb S^1$ twice around the circle $\mathbb P^1$. – Andreas Blass Aug 18 '21 at 18:52

2 Answers2

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Below is an elementary proof.

Assume that such a continuous $q: \mathbb {RP}^n \to \mathbb S^n$ exists. Then $p\circ q$ is the identity implies $f(p) \in \{ p, -p\}$. Let $A: \mathbb S^n \to \mathbb S^n$ be the antipodal map $A(p) = -p$. Then $q$, $A\circ q$ are both continuous and

$$ \operatorname{Im} (q) \cap \operatorname{Im} (A\circ q) = \emptyset, \ \ \ \operatorname{Im} (q) \cup \operatorname{Im} (A\circ q) =\mathbb S^n.$$

Also, both $\operatorname{Im} (q)$, $\operatorname{Im} (A\circ q)$ are compact and thus are closed, and are clearly non-empty. But that contradicts the connectivity of $\mathbb S^n$. Thus $q$ does not exists (The above argument of course assumed $n>0$: the case $n=0$ is simple though).

Arctic Char
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Arctic Char gives a very nice elementary proof. It is an incarnation of a more homotopical fact that if we have a group $G$ acting on a space $X$ such that $p:X \rightarrow X/G$ is a covering map, then a section $s:X/G \rightarrow X$ implies that our group action is homeomorphic to the obvious one one $X/G \times G$. The proof of this is very easy, simply map $x \in X$ to $(p(x),s(p(x)))$ which is a homeomorphism commuting with the group action.

So one way to phrase Arctic Char's answer is "No, a section of the $\mathbb{Z}/2$ action on $S^n$ cannot be continuous, since if it were $S^n \rightarrow \mathbb{R}P^n$ would be a trivial $\mathbb{Z}/2$ covering map, but this is impossible since $S^n$ is connected and $\mathbb{R}P^2 \times \mathbb{Z}/2$ is not."

Connor Malin
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