= Constant rank theorem
{wiki}
The constant rank theorem separates the directions a <smooth map> preserves from the directions it forgets. If a <smooth map> from an <open subset> of $\mathbb R^m$ to $\mathbb R^n$ has <derivative> of constant <rank> $r$ on a neighbourhood of a point, there are local <coordinate transformations> in source and target in which it becomes
$$
(x^1,\ldots,x^m)\longmapsto(x^1,\ldots,x^r,0,\ldots,0).
$$
It keeps $r$ coordinates, discards the other $m-r$, and inserts $n-r$ zeros. This is stronger than simplifying the <Jacobian matrix> at one point: it straightens the actual nonlinear map on a neighbourhood. https://web.stanford.edu/~lindrew/math215B.pdf[Eleny Ionel, Differential Topology, notes by Andrew Lin (2023), Theorem 44]{ref}
For a worked example, take
$$
f(x,y)=(x+y,(x+y)^2),\qquad
Df(x,y)=
\begin{pmatrix}1&1\\2(x+y)&2(x+y)\end{pmatrix}.
$$
The two columns agree, and the first row is nonzero, so the <rank> is exactly one everywhere. The <image of a function> is the <parabola> $b=a^2$. Every <line> $x+y=a$ is a <fiber> mapping to the same point $(a,a^2)$.
Choose input coordinates $u=x+y$, $v=x-y$, and output coordinates $\alpha=a$, $\beta=b-a^2$. Both <coordinate transformations> are <diffeomorphisms>: their inverses are $x=(u+v)/2$, $y=(u-v)/2$ and $a=\alpha$, $b=\beta+\alpha^2$. In these coordinates,
$$
(u,v)\longmapsto(u,u^2)\longmapsto(u,0).
$$
The curved image has become a straight <line>, and the unused coordinate $v$ is explicit.
\Image[/constant-rank-straightening.png]
{title=A rank-one map collapses input fibres to a parabola, which a change of output coordinates straightens}
Constant <rank> on a neighbourhood is essential. The map $(x,y)\mapsto(x,y^2)$ has <rank> one when $y=0$ and two when $y\ne0$, so the constant rank theorem does not supply one rank-one form across that line. By contrast, the full-rank square case of the theorem forgets no coordinates and recovers the <inverse function theorem>. The usefulness is that curved <fibers> and images can often be studied through this simpler local form.
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