Let $M \subset \mathbb{RP}^{n+1}$ a closed, connected hypersurface and let $\tilde{M} = \pi^{-1}(M) \subset \mathbb{S}^{n+1}$, where $\pi : \mathbb{S}^{n+1} \to \mathbb{RP}^{n+1}$ is the canonical projection. Assume that $\tilde{M} = M_1 \cup M_2$ is disconnected and that each connected component $M_i$ is homeomorphic to a sphere. I was asked to investigate whether $M$ itself is homeomorphic to a sphere.
In an attempt to show the affirmative answer, my idea was to prove that $\pi$ restricted to each $M_i$ is injective. The result would follow, since $\pi|_{M_i} : M_i \to M$ is a surjective local homeomorphism.
I defined
$$C = \{ p \in M : \pi^{-1}(p) = \{ x, -x\}, \text{ with } x \in M_1 \text{ and } -x \in M_2\}$$
Proving that $\pi|_{M_1}$ is injective is the same as proving that $C = M$. Obviously $C \neq \emptyset$. My question is: how does one prove that $C$ is open and closed in $M$?
You can do this:tThe restriction of $p:S^{n+1}\rightarrow \mathbb{R}P^{n+1}$ to each connected component $M_i, i=1,2$ is surjective. Since $p(M_i)$ is compact (closed) and open and $M$ is connected. Suppose that there exists distinct $x,y\in M_1$ with $p(x)=p(y)$, this implies that $y=-x$. There exists also $z\in M_2$ with $p(z)=p(x)$, this implies that $z=x$ or $z=-x$ contradiction since $M_1\cap M_2$ is empty. This implies that the restriction of $p$ to $M_i$ is an homeomorphism and $M$ is a sphere.