Kernel of the tangent map If $\varphi:U\subset \mathbb{R}^n \to \mathbb{R}^m$ is $C^1$, let $\mathrm{T}\varphi:\mathrm{T}U \to \mathrm{T}R^m$ be its tangent map. The inverse function theorem tells us that if $\ker(\mathrm{T}\varphi(x))$ is zero, $\varphi$ is injective in some neighborhood of $x$. If the kernel is non-zero, what can we say about $\varphi$ near $x$ provided we know the kernel? In particular, can we say anything about curves through $x$ whose tangents belong to this kernel? 
 A: As Wikipedia says: 
"The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with locally constant rank near a point can be put in a particular normal form near that point." 
See http://en.wikipedia.org/wiki/Derivative_rule_for_inverses#Constant_rank_theorem
The Constant Rank Theorem is stated as Theorem (7.1) p. 47 of An Introduction to Differentiable Manifolds and Riemannian Geometry, Revised Second Edition, William M. Boothby, Academic Press. (This is the reference given by Wikipedia.)
Here is, for the reader's convenience, a statement of the Constant Rank Theorem.
Let $k,n$ and $r$ be positive integers, let $a$ be in $\mathbb R^n$, let $b$ be in $\mathbb R^k$, let $f$ be a smooth map from a neighborhood of $a$ to $\mathbb R^k$ sending $a$ to $b$, and let $\ell$ be the linear map from $\mathbb R^n$ to $\mathbb R^k$ sending $x$ to $(x_1,\dots,x_r,0,\dots,0)$. Assume that the rank of the tangent map to $f$ at $x$ is equal to $r$ for all $x$ in our neighborhood of $a$.
Then there is a diffeomorphism $g$ from a neighborhood of $a$ to a neighborhood of 0 in $\mathbb R^n$, and a diffeomorphism $h$ from a neighborhood of 0 in $\mathbb R^k$ to a neighborhood of $b$, such that the equality $f=h\circ\ell\circ g$ holds in some neighborhood of $a$. 
EDIT OF MARCH 19, 2011
Here is a statement and a proof of the Constant Rank Theorem. 

Theorem. Let $U$ be open in $\mathbb{R}^n$, let $a$ be a point in $U$, and let $f$ be $C^p$ map ($1\le p\le\infty$) of rank $r$ from $U$ to $\mathbb{R}^k$. Then there are open sets $U_1,U_2\subset\mathbb{R}^n$, $U_3\subset\mathbb{R}^k$ and $C^p$ diffeomorphisms $\varphi:U_1\to U_2$, $\psi:U_3\to U_3$ such that $a\in U_1$ and $(\psi\circ f\circ\varphi^{-1})(x)=(x_1,\dots,x_r,0,\dots,0)$ for all $x$ in $U_2$. 

Proof. For $$x\in\mathbb{R}^r,\quad y\in\mathbb{R}^{n-r},\quad(x,y)\in U$$ write 
$$f(x,y)=(f_1(x,y),f_2(x,y)),\quad f_1(x,y)\in\mathbb{R}^r,\quad f_2(x,y)\in\mathbb{R}^{k-r}.$$
We can assume that $\partial f_1(x,y)/\partial x$ is invertible for all $(x,y)\in U$. Define $$\varphi:U\to\mathbb{R}^n,\quad(x,y)\mapsto(f_1(x,y),y).$$ By the Inverse Function Theorem, there are open sets 
$$U_1\subset\mathbb{R}^n,\quad U_4\subset\mathbb{R}^r,\quad U_5\subset\mathbb{R}^{n-r}$$ such that $a\in U_1\subset U$, $\varphi$ is a $C^p$ diffeomorphism from $U_1$ onto $U_2:=U_4\times U_5$, and $U_5$ is connected. 
Then $f(\varphi^{-1}(x,y))=(x,g(x,y))$ for some $C^p$ map $g$ from $U_2$ to $\mathbb{R}^{k-r}$. As $\partial g/\partial y=0$, we can write $g(x)$ for $g(x,y)$, and it suffices to set $U_3:=U_4\times\mathbb{R}^{k-r}$ and $\psi(u,v):=(u,v-g(u))$ for $ u\in U_4$ and $v\in\mathbb{R}^{k-r}.$
