TITLE: Constant Rank Theorem for Manifolds with Boundary
QUESTION [13 upvotes]: I'm trying to answer problem 4-3 from Lee's Introduction to Smooth Manifolds, 2nd edition.  The problem says:

Formulate and prove a version of the rank theorem for a map of constant rank whose domain is a smooth manifold with boundary.

Lee himself gave a hint in another question. Here is what I have so far:
I'm assuming I have a smooth map of constant rank $F: \mathbb{H}^m\rightarrow\mathbb{R}^n$. Let's say $\mathrm{rank}(F)=k$. Then I extend $F$ to a smooth map $\tilde{F}:\mathbb{R}^m\rightarrow\mathbb{R}^n$.  I can shrink the domain to a small enough neighborhood of $\mathbf{0}$, $U$, such that there is a projection $\pi:\mathbb{R}^n\rightarrow\mathbb{R}^k$ with $\pi\tilde{F}_{|U}$ a submersion. The actual constant-rank theorem then says we have charts $(A,\alpha)$ and $(B,\beta)$ with
$$ \beta\circ\pi\tilde{F}\circ\alpha^{-1}:\alpha(A)\rightarrow\beta(B)$$
If I write the coordinates of $\mathbb{H}^m$ as $(x,y)$ and the coordinates of of $\alpha(A)$ as $(a,t)$, then this map looks like
$$ \beta\circ\pi\tilde{F}\circ\alpha^{-1}(a,t) = a$$
I can switch the order so the projection is the last map applied:
$$ \pi\circ(\beta\times\mathrm{id})\circ\tilde{F}\circ\alpha^{-1}\equiv\beta\circ\pi\tilde{F}\circ\alpha^{-1}$$
Which finally means
$$(\beta\times\mathrm{id})\circ\tilde{F}\circ\alpha^{-1}(a,t)=(a,S(a,t))$$
Where $S:\alpha(A)\rightarrow\mathbb{R}^{(n-k)}$.
This is pretty similar to how the constant-rank theorem is proved, but now I'm stuck.  The map above, restricted to $\alpha(A\cap\mathbb{H}^m)$, has rank $k$. But that does not mean $S$ is independent of $y$. That's because $\alpha$ is not necessarily a boundary chart on $A\cap\mathbb{H}^m$. I also don't see any way to make $\alpha$ a boundary chart (something similar is proved in Lee's book, but crucially uses that $F$ is an immersion).
Can someone give me a hint as to how to finish this? Ideally I'd get to a place where (with possibly different charts)
$$ \tilde{\beta}\circ F\circ\tilde{\alpha}^{-1}(a,t) = (a,0)$$
and $\tilde{\alpha}$ is a boundary chart.
Bonus question: where do I use Lee's assumption (from the hint) that $\ker dF_p\not\subseteq T_p\partial\mathbb{H}^m$?

REPLY [3 votes]: Lee's Introduction to Smooth Manifolds deals with the case of local immersions for manifolds with boundary in Theorem 4.15. So let us suppose that $F:\mathbb{H}^m \rightarrow \mathbb{R}^n$ has $\mathrm{rank}(F)=k<m$. For $\mathbb{H}^m=\{(x^1,\dots,x^m)\in \mathbb{R}^m, x^m\geq 0\}$, the assumption $\ker dF_p\not\subseteq T_p\partial\mathbb{H}^m$ in Lee's answer to the original question means
that $dF_p(\partial/\partial x^m+a_1\partial/\partial x^1+\dots+a_{m-1}\partial/\partial x^{m-1})=0$ for some numbers $a_i$. The search for $k$ linearly independent tangent vectors in the image can therefore be restricted to $dF_p(\partial/\partial x^i), i<m$. Let's suppose that $k$ is the rank of $(\partial F_i/\partial x^j)_p, 1\leq i, j \leq k$. The coordinate change $(x^1,\dots,x^m) \rightarrow (F_1,\dots,F_k,x^{k+1},\dots,x^m)$ produces another boundary chart $x^m\geq 0$ for which the rest of the proof works as in the case of manifolds without a boundary.