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.