$T: \mathbb{R}^n\to \mathbb{R}^m$, $n>m$, $\operatorname{rank}(dT)=m$, show $T$ maps open sets to open sets.

Question

Suppose $T: \mathbb{R}^n\to \mathbb{R}^m$, $n>m$, with $dT$ having rank $m$ at all points in an open set $D \subset \mathbb{R}^n$. What is a proof that $T$ maps $D$ into an open set in $\mathbb{R}^m$?

One thing I'm thinking is that $dT$ cannot have a zero row, so at each point $p\in T(D)$, no component is independent of the domain...for each component $y_j$ in the target, at a point $p\in T(D)$, there is some variable $x_i$ in the domain such that if we wiggle $x_i$, $y_j$ should wiggle too. This should allow us to prove that $T(D)$ is open, right (since by "wiggling" in each component of the target, we can get an open neighborhood)? In fact $T(S)$ is open for $S\subset D$ open.

Is this the standard proof, or is there a "cleaner" one?

Answer 1

For every $x\in \mathbb R^n$ the differential matrix $dT(x)$ has a nonzero $m\times m$ minor. Pick the columns of this minor (say, $i_1,\dots, i_m$) and restrict your attention to the affine subspace $x+P$ where $P$ is the subspace generated by basis vector $x_{i_1},\dots,x_{i_m}$. The restriction of $T$ to $x+P$ satisfies the assumptions of the inverse function theorem. Therefore, it restricts to a diffeomorphism of a neighborhood of $x $ in $x+P$ onto a neighborhood of $T(x)$ in $\mathbb R^m$. Consequently, the image of any open set containing $x$ contains $T(x)$ as an interior point. The claim follows.

Answer 2

Full rank of $dT$ implies that the map is a submersion at that point. In essence, it means that under a suitable change of coordinates (on both the domain and codomain), the map acts as a projection. This is a theorem called the Local Submersion Theorem. Here's how the argument proceeds:

Fix a point $x\in U$. For some choice of charts, you get two diffeomorphisms $\phi:\mathbb{R}^m\to \mathbb{R}^m$ and $\psi: \mathbb{R}^n \to \mathbb{R}^n$ and a projection $\pi:\mathbb{R}^n\to\mathbb{R}^m$ such that in a suitably restricted domain, $T= \phi\circ\pi\circ \psi^{-1}$. It would be easy to see if you draw this in a commutative diagram. Therefore, $T$ is an open map on this restriction (since diffeomorphisms and projections are open and compositions of open maps are open) and we can choose a small open ball $B_x\subset U$ around $x$. It follows that $T(B_x)$ is open and inside the image $T(U)$ by construction. The claim follows by observing that $$T(U)=\bigcup_{x\in U} T(B_x).$$