Proof formalization help: Given a vector $u$ of Euclidean length $1$ in $\mathbb{R}^3$ and a compact 2 dimensional manifold $m$, $u$ is normal to the $m$ at at least 2 points.
I've thought about the question and I think I have some ideas, but I'm not sure how to formally write them.
Because $m$ is compact and a manifold, I think $m$ has to be homotopic to the sphere in $\mathbb{R}^3$ (as pointed out - this is not correct, but I think the idea still holds). What I mean to say is that it's a shape that has some space inside of it and is closed in the sense that if you were to fill it with water none of it would leak.
The fact that $m$ has some space inside of it stems from that fact that $m$ is two dimensional and that a manifold is locally a graph of a function. The fact that $m$ has no "holes" is because a manifold is a union of charts, which are inherently functions from open sets in $\mathbb{R}^2$, so any holes it would have would contradict the compactness of $m$.
If I have those two lemmas, I can assume WLOG that the point $(0,0,0)$ is inside $m$, and then of course I can extend $u$ to infinity from both ends and get that because $m$ has no holes, $u$ has to intersect it at least twice, (from the top and from the bottom), and we're done.
My explanations aren't formal, so I'd appreciate any help in getting them clearer.
Thanks in advance.
$M$ can easily look nothing like a sphere. Think of a torus, for example.
Consider the map $M \rightarrow \mathbb{R}$, $v \mapsto u \cdot v.$ Since $M$ is compact, this will take a minimum and maximum value. They are distinct ($M$ can't lie in a plane). Any points attaining these are your two points. You can prove this by locally writing $M$ as a level set $\{v : \, F(v) = 0\}.$