Intersection of the image of a manifold and a line should be finite.

Question

The problem is the following:

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Let $f: M^n \rightarrow \mathbb{R}^{2n+1}$ be a smooth map such that $0 \notin f(M)$. Show that there exists a line in $\mathbb{R}^{2n+1}$ such that $f(M) \cap L$ is a finite number of point.

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First attempt: We can consider the map $g: TM \rightarrow \mathbb{R}^{2n+1}$ defined by $$ g(x,v) = f(x) + T_xf(v).$$ We know that the image of $g$ is a measure-zero set in $\mathbb{R}^{2n+1}$ (because $TM$ has dimension $2n$). Moreover, the map $g$ is transverse to a generic line $L$ of $\mathbb{R}^{2n+1}$ (isn't it ?), in this case the intersection $g(TM) \cap L$ is a zero dimensional manifold of $\mathbb{R}^{2n+1}$. Clearly, $f(M) \subset g(TM)$ hence $f(M) \cap L \subset g(TM) \cap L$.

Now if $M$ is compact we are done, but I wonder if this hypotesis of compacity is really needed. Should we use a different argument ?

Answer 1

Here is a sketch showing that there is a line through the origin which does not intersect $f(M)$ at all. I am not entirely convinced by it because it would allow you to change the hypotheses to something like $f:M^n \to \mathbb{R}^{n+2}$, but I will leave it here in case it turns out useful or delete it in case there is a mistake.

Since $0 \not\in f(M)$, we can compose $f$ with the projection $p:\mathbb{R}^{2n+1} \setminus \{0\} \to \mathbb{R}P^{2n}$ and obtain a smooth map $p \circ f:M \to \mathbb{R}P^{2n}$. By Sard's Theorem, there is a line $L \in \mathbb{R}P^{2n}$ which is a critical value of $p \circ g$, but considering the dimensions in question, this can only happen if $L \not\in (p\circ g)(M)$. Now we just need to notice that $f(x) \in L$ if and only if $p(f(x)) = L$, so $f(M) \cap L = \varnothing$.