Let $p:M\to \mathbb{R}^{n+1}$ be the closed immersed hypersurface. Is the following thing right? If there exists a point $x\_0$ in $\mathbb{R}^{n+1}$ such that $\langle x(p)-x\_0,\nu(p)\rangle>0$ for all $p\in M$, then $M$ is diffeomorphic to a sphere. Here $\nu(p)$ is the outward normal vector.



Is the hypersurface satisfying $\langle x-x_0,\nu\rangle>0$ diffeomorphic to sphere?
Assuming that $M$ is connected compact without boundary the answer is YES. 


1. Consider a map $\pi: p \mapsto \frac{x(p)-x\_0}{||x(p)-x\_0||}$ from $M$ to $S^n$.
2. $\pi$ is an immersion since $\langle x(p)-x\_0,\nu(p)\rangle >0$.
3. $\pi$ is a covering since $\pi$ is an immersion, $M$ is a closed $n-$manifold, and $S^n$ is connected.
4. $\pi$ is a homeomorphism since it is a covering over the simply connected $S^n$ . So $M$ is diffeomorphic to $S^n$.


