On hypersurfaces of the projective space Let $\pi : \mathbb{S}^{n+1}  \to \mathbb{R}P^{n+1}$ the projection from the round sphere os radius $1$ to the projective space, and endow $\mathbb{R}P^{n+1}$ with a riemannian metric so that $\pi$ is a local isometry. Let $M^n \subset \mathbb{R}P^{n+1}$ be a hypersurface (orientable or not) and consider $\tilde{M} = \pi^{-1}(M)$. Since $\pi$ is a submersion, it is transverse to $M$, so that $\tilde{M}$ is indeed a hypersurface of $\mathbb{S}^{n+1}$. Is it always orientable? And connected?
I read that, whenever $n$ is odd, $\mathbb{S}^{n+1}$ is the oriented double cover of $\mathbb{R}P^{n+1}$ via $\pi$, and since the sphere is connected, it follows that the even-dimensional projective spaces are non-orientable. I wonder if the same holds for $\pi|_{\tilde{M}} : \tilde{M} \to M$: is it true that $M$ is non-orientable iff $\tilde{M}$ is connected?
Another question, now related to distances. The projective space has diameter of $\pi/2$, half of the diameter of the sphere. Suppose that the smallest (geodesic) ball of $\mathbb{S}^{n+1}$ that contains $\tilde{M}$ is $r$. Is it true, by analogy, that the radius of the smallest geodesic ball of $\mathbb{R}P^{n+1}$ that contains $M$ is $r/2$?
 A: I will give a partial answer. Assume that $\pi :S^{n+1}\rightarrow \mathbb{R}P^{n+1}$ is a
quotient map by $\mathbb{Z}_2$-free
isometric action.
If $M$ is a closed $n$-dimensional manifold, then let $\widetilde{M}:=\pi^{-1} (M)$. Since $\pi$ is
   locally isometric, so $\widetilde{M}$ is a closed $n$-manifold.  
Assume that $M_1$ is a connected component in $\widetilde{M}$. 
(1) $M_1$ is in some open hemisphere iff $M$ is in some
$B(x,r)$ where $r<\frac{\pi}{2}$
In this case $\widetilde{M}$ is disconnected and $M_1,\ M$ are
  orientable (cf. Proof of (2.2)). 
(2) By (1) we have that there is a point $\widetilde{x}\in
M_1$ with $-\widetilde{x} \in M_1$ iff $M$ is not in any geodesic
ball $B(x,r)$ where $r<\frac{\pi}{2}$.
(2.1) In this case, prove that $\widetilde{M}$ is connected. 
Proof : If not, then let $M_2$ to be another connected component.
Then $\widetilde{x},\ -\widetilde{x}\in
M_1$ and $\widetilde{y},\ -\widetilde{y}\in M_2$.
Note that $\pi\ (\widetilde{x})$ is not in $\pi\ (M_2)=M$. It is a
contradiction. 
(2.2) In this case, consider orientability of $M,\
\widetilde{M}$
Proof : $\widetilde{M}$ is in $S^{n+1}-\{ \widetilde{x}\}$
for some point $\widetilde{x}$ so that $\widetilde{M}$ is
orientable. 
If $M=\mathbb{R}P^n$, then it is orientable where $n$ is odd. And it
is non-orientable where $n$ is even. 
