# Constant Rank theorem for domain with nonempty boundary

Problem 4-3 in J.M. Lee's introductory text about smooth manifolds, asks to formulate and prove a version of the constant rank theorem for a map of constant rank whose domain is a smooth manifold with boundary. That is, show that,

If $F:M\rightarrow N$ is smooth, $N$ with empty boundary, $F$ of constant rank $r$, then, for every point $p$ in $M$, $F$ has a local representation of the form $\tilde{F}(x)=(x_1, ...,x_r,0,...,0)$

Lee gives a hint: After extending $F$ (the interesting case is when $p\in\partial M$), follow the proof of the regular constant rank theorem, until you have to make use of the constant rank hypothesis. The problem may be that the extension has higher rank. Lee's hint is to modify the map so that it has constant rank.

I don't see how to do this.

(If it's a silly question, I'm sorry, I haven't slept in over 24hs.)

• Have you considered the case where 0P=? – Bananas Apr 23 '13 at 3:49
• How 'bout composing the extension with the projection on $\mathbb R^r$? – Ted Shifrin Apr 24 '13 at 23:10
• @TedShifrin: I don't see how this could work. – Bill Apr 25 '13 at 2:21

First, to get a good result, you'll have to add the assumption that $\ker dF_p\not\subseteq T_p\partial M$. After choosing smooth coordinates, you can assume $M \subseteq \mathbb H^m$ and $N\subseteq\mathbb R^n$, and extend $F$ to a smooth function $\widetilde F$ on an open subset of $\mathbb R^m$.
Now, as Ted suggested, assuming that $F$ has constant rank $r$, show that there is a coordinate projection $\pi\colon\mathbb R^n\to\mathbb R^r$ such that $\pi\circ \widetilde F$ is a submersion, and apply the rank theorem to $\pi\circ \widetilde F$ to find new coordinates in which $\widetilde F$ has a coordinate representation of the form $(x,y) \mapsto (x,R(x,y))$. Then use the rank condition to show that $R|_M$ is independent of $y$.