Each connected component lies in a hemisphere Let $\pi : \mathbb{S}^n \to \mathbb{R}P^n$ be the canonical projection. Let $M \subset \mathbb{R}P^n$  be a hypersurface and consider $\tilde
{M} = \pi^{-1}(M)$. Is it true that if $\tilde{M}$ has two connected components then each one of them lies in oposite hemispheres?
 A: The answer is no when $n \ge 2$. To obtain a counterexample, it suffices to construct a closed connected hypersurface $H \subset \mathbb{S}^n$ with the following properties:


*

*$H$ doesn't contain any pair of antipodal points;

*$H$ is not contained in any hemisphere.


Indeed, we can then take $M = \pi(H)$. Condition 1 implies that $M$ is a hypersurface in $\mathbb{R}P^n$, and condition 2 ensures that the connected components of $\tilde{M} = H \sqcup a(H)$ cannot lie in opposite hemispheres (here $a : \mathbb{S}^n \to \mathbb{S}^n$ is the antipodal map).
I will now explain how to construct such an $H$. View $\mathbb{S}^n$ as the unit sphere in $\mathbb{R}^{n+1}$ and let $\mathbb{S}^{n-1}$ denote the "equator" $\mathbb{S}^n \cap (\mathbb{R}^{n} \times \{0\})$. Let $S_-$, $S_+$ denote the two closed hemispheres with common boundary $S_- \cap S_+ = \mathbb{S}^{n-1}$.
Fix a point $p$ in the interior of $S_+$ and choose $n + 1$ points $q_1, \dots, q_{n+1} \in \mathbb{S}^{n-1}$ such that the interior of the convex hull of $\{q_1, \dots, q_{n+1}\}$ in $\mathbb{R}^n \times \{0\}$ contains the origin. The point of this condition is that it implies that $S_+$ and $S_-$ are the only hemispheres of $\mathbb{S}^n$ that contain all the $q_i$'s. It follows that $S_+$ is the only hemisphere that contains $p$ and $q_1,\dots,q_{n+1}$.
Let $T \subset S_+$ be a smoothly embedded graph with vertices $p, q_1, \dots, q_{n+1}$ and with one edge connecting $p$ to $q_i$ for each $i \in \{1,\dots,n+1\}$. We can assume without loss of generality that there is no pair of antipodal points in $\{q_1,\dots,q_{n+1}\}$ and hence that there is no such pair in $T$. Thus, any sufficiently small regular neighborhood $U \subset \mathbb{S}^n$ of $T$ will also not contain any pair of antipodal points.
Note that $U$ is a connected open subset of $\mathbb{S}^n$ (in fact it is diffeomorphic to an $n$-ball) and has nonempty intersection with $S_-$. It follows that there exists a closed connected hypersurface $H \subset U$ which contains $p,q_1,\dots,q_{n+1}$ and intersects the interior of $S_-$ nontrivially (in fact we can take $H$ to be diffeomorphic to $\mathbb{S}^{n-1}$). This concludes the proof: $H$ satisfies condition 1 because $U$ does and condition 2 by our choice of $p,q_1,\dots,q_{n+1}$.
