Factoring constant rank maps into a submersion and an immersion Let $X$ and $Z$ be smooth manifolds and $\phi: X \to Z$ a smooth map so that the differential $D \phi$ is everywhere of rank $d$. Is there necessarily a $d$-fold $Y$ so that $\phi$ factors as a submersion $X \to Y$ followed by an immersion $Y \to Z$?
I have no real motivation, it just seemed like a natural question. Notice that we can't assume $Y \to Z$ is injective: Consider mapping $\mathbb{R}^2 \to \mathbb{R}^2$ by $(x,y) \mapsto ((x-1)^2 (x+1), (x-1)(x+1)^2)$, whose image is a nodal cubic.
 A: The point of this answer is to note that the quotient $X/\sim$ is locally Euclidean (though, as pointed out above, not Hausdorff). To recall notation, $\phi: X \to Z$ is the map of constant rank and $\sim$ is the equivalence relation on $X$ where $x_1 \sim x_2$ iff $\phi(x_1) = \phi(x_2) =: z$ and $x_1$ and $x_2$ are in the same connected component of $\phi^{-1}(z)$. This may have been obvious to everyone else, but I didn't see it for a while.
The constant rank theorem tells us that $X$ can be covered by open sets $U_i$ where we can locally write $\phi$ as $U_i \overset{\pi_i}{\longrightarrow} V_i \overset{\iota_i}{\longrightarrow} W_i$, where $U$, $V$ and $W$ are open sets in $\mathbb{R}^m$, $\mathbb{R}^d$ and $\mathbb{R}^n$, the map $\pi_i$ is projection onto the first $d$ coordinates and the map $\iota_i$ is inclusion of the first $d$-coordinates. Replacing $U_i$ and $V_i$ by smaller open sets, we may assume that $U_i \cong V_i \times D_i$ with $D_i$ an open ball, and $\pi_i$ the projection onto $V_i$. Thus, $U_i/\sim$ will be isomorphic to $V_i$. 
So, $X/\sim$ will be glued from the $V_i$, and the map from each $V_i \to X/\sim$ will be an inclusion. But it wasn't obvious to me at first that the image of $V_i$ in $X$ will open. Once we establish this, the $V_i$ will give Euclidean patches on $X/\sim$.
Let's fix two of these open sets, $V_a$ and $V_b$. We must show that the part of $V_b$ which is $\sim$-identified with points in $X$ will be open; this establishes that $V_a$ is open in $X/\sim$.
Now, the fibers of $\phi$ will be manifolds, so we can replace "connected" with "path connected" in the definition of $\sim$. This means that if $x_1 \in U_a$ is equivalent to $x_2 \in U_b$, there must be a path from $x_1$ to $x_2$ on which $\phi$ is constant. Said path will be compact, so we can cover it with finitely many $U_i$. For each sequence $(a, c_0, c_1, c_2, \ldots, c_r, b)$ of indices, let $U(a, c_1, \ldots, c_r, b)$ be the set of points $x$ of $U_b$ such that there is a path starting in $U_a$ and passing through $U_{c_1}$, ..., $U_{c_r}$, $U_b$, ending at $x$. Clearly, $U(a, c_1, \ldots, c_r, b)$ is a union of $\sim$ equivalence classes in $U_b$, write $V(a, c_1, \ldots, c_r, b)$ for the quotient in $V_b$. The part of $V_b$ which is identified with $V_a$ is $\bigcup_{c_1 \ldots, c_r} V(a, c_1, \ldots, c_r, b)$. So it is enough to show that $V(a, c_1, \ldots, c_r, b)$ is open.
This last is pretty straightforward but annoying to write out, so I think I'll skip it unless someone tells me they want it.
A: A smooth factorization does not exist in all cases.
Let $X$ be $\mathbb{R}^2$ minus the closed ray $R = \{ (x,0)\ |\ x\ge 0\}$, and define $\phi:X\to\mathbb{R}^2 = Z$ by the rule
$$
\phi(x,y) = \begin{cases}(x,0), & \text{when $x\le 0$,}\\ 
\bigl(x,\phantom{-}e^{-1/x^2}\bigr), & \text{when $x> 0$ and $y>0$.}\\
\bigl(x,-e^{-1/x^2}\bigr), & \text{when $x> 0$ and $y<0$.} \end{cases}
$$
Then $D\phi = \phi'$ has rank $1$ everywhere on $X$, but there is no smooth manifold $Y$ for which there is a smooth submersion $\sigma:X\to Y$ and a smooth immersion $\iota:Y\to Z$ such that $\phi = \iota\circ\sigma$.  
(Proof: Since $X$ is connected, one can assume that $Y$, if such existed, be connected.  Now, considering the map $\pi:Z\to\mathbb{R}$ given by $\pi(x,y)=x$, one sees that $\pi\circ\phi:X\to\mathbb{R}$ is a surjective submersion.  If the desired factorization existed, then since $\pi\circ\phi = (\pi\circ\iota)\circ\sigma$, the Chain Rule would imply that $\pi\circ\iota:Y\to\mathbb{R}$ be a (surjective) local diffeomorphism.  Since $Y$ is connected and $1$-dimensional, it would follow that $\pi\circ\iota:Y\to\mathbb{R}$ be a diffeomorphism, which is impossible since $\pi\circ\iota$ cannot be one-to-one.)
