Edit: Below, spurred by Jack Lee's comment, I am going to change the notation to more accurately reflect my question. I believe this can be done without creating a new question.
Let $M \subset R^{n+k}, \; n > 2,\;k > 1$ be a smooth, embedded submanifold of dimension $n-1$ and let $\rho : M \rightarrow R^k$ be the orthogonal projection onto the last $k$ coordinates. Further, suppose the map $\phi : R^l \times M \rightarrow M := \phi(t,x) = x + t$, maps each fiber $\rho^{-1}(y_1,\ldots,y_k)$ into itself (here $l < n$ and I am identifying $R^l$ with the subset of $R^{n+k}$ consisting of all vectors with last $n+k-l$ coordinates all equal to zero). In other words, $R^l$ together with $M$ satisfy the hypothesis of the Quotient Manifold Theorem (as presented in Lee, Introduction to Smooth Manifolds, Theorem 21.10). I would like to prove that $\rho(M)$ is a submanifold of $R^k$ (embedded, preferably, but even immersed will do).
Edit, an added assumption: I also know that $\rho$ is of constant rank $r < k$.
Edit, another property of $M$: I know that if two points of $M$ belong to the same "leaf" (with respect to the action) then their corresponding tangent spaces are exactly the same. That is, the tangent space of a point of $M$ depends only on the last $k$ coordinates.
Edit: In my mind, I see this as an undulating sheet of paper (Letter or A4, depending on where you live) standing on edge on a table top, and the projection being the "curve" where the paper meets the table. The fibers are just lines perpendicular to the table with $l = 1$ and the action just translation by a real number. I understand $n,l,k$ are not quite right (here $n=3,k=2$ but only $l = 1$), but that is the picture that made me think the projection might be a manifold.
From the cited Theorem I know there exists a smooth structure on $M\:/\:R^l$ such that the quotient map $\pi: M \rightarrow M\: /\: R^l$ is a smooth submersion. I also know (Theorem 4.30, also from Lee) there is a unique (should be injective) smooth map $\tilde{F} : M \: / \:R^l \rightarrow R^k$ such that $\rho = \tilde{F} \circ \pi$. I have two questions: First, am I correct in saying, because of the topologies involved and the fact that projections are open maps, that $\tilde{F}$ is a topological embedding onto its image ($\rho(M))$? Second, if the answer to the first question is no, can anyone give me a hint as to where I've gone wrong? If the answer to the first question is yes, is it then true that in order to conclude $\rho(M)$ is a submanifold (embedded) of $R^k$ I need to prove that $\tilde{F}$ is an immersion? I believe that with my added assumption on the rank of $\rho$ that this last is "obvious" and that $\rho(M)$ is indeed an embedded submanifold of $R^k$ of dimension $r$, but I am wary of my thought processes.
Thank you in advance for any comments and information.
This really is more of a comment, but it's a bit too long.
The setup still doesn't quite make sense. First of all, in order for the map $\phi$ to take its values in $M$, you need to assume that $M$ is invariant under all translations in the $(x^1,\dots,x^l)$ coordinates, and from that assumption it follows that $M$ is just a product manifold $\mathbb R^l\times M_0$, where $$ M_0 = \{ z\in \mathbb R^{n+k-l}: (0,z)\in M\}. $$ Moreover, the assumption that $\phi$ maps each fiber $\rho^{-1}(y_1,\dots,y_k)$ into itself is superfluous, because each map $x \mapsto \phi(t,x)$ doesn't change the last $k$ coordinates at all.
In any case, there's no reason why $\rho(M)$ should be a submanifold. Here's a counterexample.
Let $n=3$, $k=2$, and $l=2$, and let $M$ be the image of the following embedding $\alpha\colon \mathbb R^3\to \mathbb R^5$: \begin{equation*} \alpha(u,v,w) = (u,v,w,\sin 2w, \sin w). \end{equation*} Then $M$ is an embedded $3$-manifold, invariant under translations in the first two variables. But the projection onto $\mathbb R^2$ is the lemniscate traced out by the curve $\gamma(t) = (\sin 2t, \sin t)$, which is not a manifold. (See my Introduction to Smooth Manifolds, Example 4.19.)