My question is, I hope, simple. I have a smooth, properly embedded one dimensional submanifold $M \subset R^n, \; n > 2$ (i.e. a smooth curve, which is actually the level set of a smooth function $\phi : R^n \rightarrow R^{n-1}$). I know from separate considerations that the projection $\pi$ of $M$ onto its first two coordinates is a smooth injective immersion. Can I conclude that then $\pi (M)$ is an embedded submanifold of $R^2$? My sense is no, as I have tried without success (following results in Lee's ISM) to prove that $\pi$ is either a closed or open or proper map onto $\pi(M)$ (with the relative topology, of course), so that $\pi$ will actually be an embedding.
I would appreciate any pointers or comments.
Edit After reading the helpful comments I'm thinking the problem may be connected to a sign change of the tangent vector along the curve or perhaps that $dy / dx$ goes undefined. I will explore whether an assumption that the coordinates of the tangent vector never change sign would make a difference. If I come up with anything or have any questions in connection with that I'll post another question.
The answer is no. Suppose $M$ is the image of the embedding: $$f(t) =(\cos t,\cos t\sin t, t)$$ defined on $(-3\pi/2, \pi/2)$. This is an embedded submanifold of $\mathbf{R}^3$, but its projection onto the first two coordinates $\pi(M) = \{(\cos t,\cos t \sin t) : -3\pi/2 < t < \pi/2\}$ is a lemniscate in $\mathbf{R}^2$, but with the smooth structure of $\mathbf{R}$. In particular, it is not an embedded submanifold of $\mathbf{R}^2$.
While I am not sure if this is the specific example he uses, I owe this idea to Tu's book on smooth manifolds.