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Any idea about this problem concerning the Constant Rank Theorem:

Let $f:U\subset\mathbb{R}^m \longrightarrow U,U $ open connected $ ,f\in C^1(\mathbb{R}^m)$ such that: $f\circ f=f$

Prove that $f$ has constant rank in a neighborhood $V$ of $f[U] ,(Rank(f,a)=constant,\forall a \in V)$

$Rank(f,a) = $ dim $ Im[f'(a)]$

Any hints would be appreciated.

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An application of the chain rule, perhaps? –  Shawn Henry Oct 25 '12 at 17:59
    
Sure $f$ has constant rank on $Im(f)$, which is closed and connected, but what about a neighborhood? –  scineram Mar 28 '13 at 18:03

1 Answer 1

up vote 1 down vote accepted

The rank ($T\mapsto \dim Im T$) is a lower semicontinuous function on linear operators, and $f'$ is continuous, so $x\mapsto Rank(f,x)$ is lower semicontinuous as well. Thus $\{x\in U\mid Rank(f,x)>k\}$ is an open set for any $k\in\mathbb{R}$. Let $n:=\max\limits_{x\in U} Rank(f,x)$. For any $a\in U$ $$f'(a)=(f\circ f)'(a)=f'(f(a))\circ f'(a),$$ thus $Rank(f,a)\le Rank(f,f(a))$. If $a\in f[U]$ the above equality reduces to $f'(a)^2=f'(a)$. Since for projections the rank and trace are equal, and the trace is continuous, $x\mapsto Rank(f,x)$ is continuous on $f[U]$, which is connected, thus $$f[U]\subset\{x\in U\mid Rank(f,x)=n\}=\{x\in U\mid Rank(f,x)>n-1/2\}=:V.\quad\square$$

I would like to add that this implies that $f[U]$ is an $n$-dimensional submanifold: by the constant rank theorem every $a\in V$ has an open neighborhood $W\subseteq V$ such that $f[W]$ is an $n$-dimensional submanifold. But then so is $f[W]\cap W$, and we have $$f[U]\cap W=f[f[U]\cap W]=f[W]\cap W.$$

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