Is the hypersurface satisfying $\langle x-x_0,\nu\rangle>0$ diffeomorphic to sphere? 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.
 A: Let me suppose that $M$ is in addition compact, and for simplicity that $x_0=0$. Let $\phi : M \to S^n$ be the radial projection, which is a smooth map. 
If $\phi$ is not an immersion, there is a tangent vector $v \in T_x M$ such that $Dp(v)$ is parallel to the vector $p(x) \in \mathbb{R}^{n+1}$. But
$$\langle v, \nu(x)\rangle=0$$
as the tangent and normal spaces are orthogonal, so $\langle p(x), \nu(x)\rangle=0$, a contradiction. Thus $\phi$ is an immersion.
But $\phi$ is also proper, and a proper codimension 0 immersion is a covering map. As any compact covering space of $S^n$ is a finite union of copies of $S^n$ (this is phrased strangely so that it still holds for $n=1$), it follows that $M$ is diffeomorphic to a finite union of copies of $S^n$.
One clearly can't insist that it is a single copy of $S^n$ in general, just by taking e.g. two spheres of different radii. So assume $M$ is connected if this is what you want.
A: Assuming that $M$ is connected compact without boundary the answer is YES. 


*

*Consider a map $\pi: p \mapsto \frac{x(p)-x_0}{||x(p)-x_0||}$  from $M$ to $S^n$. 

*$\pi$ is an immersion since $\langle x(p)-x_0,\nu(p)\rangle >0$. 

*$\pi$ is a covering since $\pi$ is an immersion, $M$ is a closed $n-$manifold, and $S^n$ is connected. 

*$\pi$ is a homeomorphism since it is a covering over the simply connected $S^n$ . So $M$ is diffeomorphic to $S^n$.
