Fiber from $S^1$ to $\mathbb{R} P^1$ that does not admit a section.

Question

Consider the map $q:S^1 \to \mathbb{R} P^1$ given by $v \mapsto \mathbb{R} v$. I want to show that this map does not admit sections. I've previously shown that $S^1$ is diffeomorphic to $\mathbb{R} P^1$ but this doesnt seem to be of much help. I've also shown that $TS^1$ is trivial, however I don't see how to use this. My only idea is that such a section wouldn't be well defined since it could go to either antipidal points when composing with $q$. Any hint or help with this would be amazing!

Answer 1

You certainly know that $p : S^1 \to \mathbb RP^1$ is a quotient map identifying pairs of antipodal points $z, -z$ in $S^1$ to a single point of $\mathbb RP^1$. Consider the map $2 : S^1 \to S^1, 2(z) = z^2$, where we regard $S^1$ as the unit circle in $\mathbb C$. This map also identifies pairs of antipodal points $z, -z$ in $S^1$ to a single point of $S^1$. Since $S^1$ is compact Hausdorff, $2$ is a closed map, hence a quotient map. Since $h = q \circ 2^{-1} : S^1 \to \mathbb RP^1$ is single-valued, it is a well-defined function such that $h \circ 2 = q$. By the universal property of the quotient, $h$ is continuous. It is clearly a bijection, thus a homeomorphism because domain and range are compact Hausdorff.

This reduces your question to showing that $2$ does not have a section $s : S^1 \to S^1$. Assume that a section $s$ exists. Then $2 \circ s = id$ which implies that $s$ must be injective. Moreover $s$ must be surjective. Assume it were not. Then there is $z_0 \notin s(S^1)$ and $s$ factors through the contractible space $S = S^1 \setminus \{z_0\}$. Therefore also $id = 2 \circ s$ factors through $S$ which implies that $id$ is null-homotopic. This is a contradiction. Thus $s$ is a continous bijection, hence a homeomorphism. We conclude that $2 = s^{-1}$, i.e. also $2$ is homeomorphism which is a contradiction.

Another way to see that no section exists is to observe that $p$ is a coverimg map with two sheets. Then apply If a covering map has a section, is it a $1$-fold cover?