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Let $M \subset \mathbb{R}^3$ be a regularly embedded surface. How can I show the Gauss map $n:M \to \mathbf{S}^2$ is bijective if and only if $M$ is convex?

In the reverse direction I planned to use the definition of convexity with tangent plane: $\forall p \in M$, the whole of $M$ lies on one side of the tangent plane $T_pM$. Thus the curvature at each point must be unique. Otherwise, there would be 3 planes tangent to $M$ that are parallel. Contradiction.

I am not sure how to prove the forwards direction.

mxnoqwerty
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Let $n$ be a bijection and take some point $p\in M$. Assume that $n$ is the outward unit normal. Choose a plane $P$ orthogonal to $n(p)$ such that $P\cap M $ is empty, and such that $P$ lies on that side of $M$ such that $n(p)$ points into the direction of $P$ (i.e. $\gamma(t) := p + tn(p)$ will meet $P$ for some positive $t$)

Now translate $P$ into direction of $M$ along $\gamma$. Denote by $P_t$ that translate which meets $\gamma$ in $t$

For continuity reasons there will be a first $t_0>0$ such that $P_{t_0}$ meets $M$. For that $t_0$, $P_{t_0}$ will be tangent to $M$ in every point of intersection. But since $n$ is a bijection this means that $P_{t_0}$ meets $M$ in $p$ (only). So $M$ lies entirely on one side of a plane tangent to $M$ in $p$. Since $p$ was arbitrary, $M$ is convex. (Assuming you know this kind of characterization of convex bodies).

Thomas
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