Suppose $M \subset R^n$ is a properly embedded submanifold of dimension $m$ and let $\pi : M \rightarrow R^2$ be the projection onto the first two coordinates. Further, suppose the rank of $\pi$ on $M$ is everywhere equal to one. My question is: What additional condition or conditions could I impose on $M$ to insure that the image $\pi(M)$ is a one-dimensional submanifold of $R^2$?
Now I know that I need an additional assumption since without one you could have self-intersections in $\pi(M)$; e.g. if $M$ is a toroidal knot. So I decided to assume, for each $(x,y) \in R^2$, that for every $p \in \pi^{-1}(x,y)$ the tangent space $T_p M$ was the same. I pictured $M$ as a sheet of paper on edge on a table top (could be flat or smoothly curved) and $\pi$ as the projection onto the table top. Extreme, I know, but I wanted to get an idea how restrictive I needed to be to have $\pi(M)$ be a manifold.
So the way I approached a proof was to choose some $p \in M$ and infer, from the Constant Rank Theorem, that there were neighborhoods $U$ of $p$ and $V$ of $\pi(p)$ such that $\pi(U)$ was a curve $C$ (i.e. one-dimensional manifold) containing $q = \pi(p)$. We should then have $T_q C = \pi (T_p M)$. The problem, of course, is that my choice of $p$ was arbitrary, but by my second assumption this equality should be true regardless of which element of $\pi^{-1}(q)$ I choose. That is, the tangent to $C$ at $q$ should be equal to $ \pi(T_{p^\prime} M) = \pi(T_pM)$ for every $p^\prime \in \pi^{-1} (q)$. I thought this might be enough but there is no way I can exclude the possibility, no matter how much I shrink the neighborhood $V$, that there aren't points $r \in M, r \neq p$ such that $\pi(r) \in V \setminus C$.
Perhaps what I'd like cannot be achieved regardless of the assumptions I place on $M$, I'm hoping someone can help me with some info or pointers to a reference that discusses this rather unique situation.
Edit: Thanks to Moishe Kohan, who has kindly pointed out that injective immersions that have images that are not manifolds are an obvious counter-example to my conjecture. I guess I'll have to look elsewhere.
