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Suppose I have a smooth (overlap maps are $C^\infty$) embedded submanifold $S \subset R^n$ of dimension $d > 2$. Let $\pi : S \rightarrow R^2$ be the projection of $S$ onto its first two coordinates and assume that the rank of $\pi$ (i.e. the rank of its differential) is, everywhere on $S$, equal to one. Thus, from the constant rank theorem, or by using the implicit function theorem, I can infer, locally, that the image of $\pi$ is a one-dimensional manifold. Additional info is that $\pi$ may not be injective.

I am having a slight disagreement with someone regarding what one can say about $\pi(S)$: He claims that if this rank one condition is true everywhere on $S$ that $\pi(S)$ must be a one-dimensional manifold, I say that the most you can conclude is that $\pi(S)$ is locally a curve in the sense that for each $p \in S$ there exists a neighborhood $V$ of $p$ such that $\pi(V)$ is a curve. The reason I say this is based on the fact that $\pi: S \rightarrow R^2$ may not be injective.

I would appreciate any information that would help us resolve this issue (I really like the guy).

Edit: Ted Shifrin pointed out the title was confusing (components could easily refer to connected components) so I changed it.

Moishe Kohan alerted me to the fact that I had typed "$\pi(S)$ might not be injective", I promise it was a typo, since corrected.

user167131
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    The title is very confusing, as some of us think of connected components, rather than of a coordinate plane. – Ted Shifrin Oct 14 '23 at 16:34
  • @TedShifrin: Sorry, I see that now, no matter how careful I try to be to respect the audience here I seem to mess up. I don't know if I can change the title but I will try. Thanks for the heads up. – user167131 Oct 14 '23 at 16:44
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    Start by drawing picture of a knot in $R^3$ and it's projection to a coordinate plane. Also, one talks about injectivity of maps, not injectivity of their images. – Moishe Kohan Oct 14 '23 at 16:44
  • @MoisheKohan: Yes, that's exactly the type of example I wanted to show my friend but couldn't come up with an equation for the knot (he wants an equation). Essentially I explained that if two points on the submanifold project to the same point $q \in R^2$ then the local curves can be different, they could easily intersect at $q$. But he's a stubborn fella'. Thanks for the comment. – user167131 Oct 14 '23 at 16:49
  • Why do you need an equation for this purpose? Just use a torus knot to make things concrete. – Moishe Kohan Oct 14 '23 at 16:52
  • @MoisheKohan: Ahh, yes, you're right of course, I shouldn't, but..... I will try again. Actually $S$ is a level set and I will have a tough time convincing him that a knot is applicable. I didn't mention that fact (i.e. that $S$ is a level set) in the question because I didn't think it relevant (I could be wrong though). – user167131 Oct 14 '23 at 16:59
  • Just take a direct product of a knot with a linear subspace of large dimension as your $S$. Or, do you want $S$ to have a small codimension? – Moishe Kohan Oct 14 '23 at 17:02
  • Makes no difference, in some cases its large, in some small. Let me think about my approach in light of your comments, thanks. I just need to be more assertive..... – user167131 Oct 14 '23 at 17:12

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