rank of function on connected manifold

Question

Let $X$ be a connected $n$-dimensional manifold and $f:X\to X$ a differentiable function satisfying: $f\circ f =f$. Prove that for all $p\in X$ that $rk_pf\leq rk_{f(p)}f$ and subsequently that $rk \;f$ is constant along $f[X]$. Can anyone give me a hint? I have no idea where to begin

Answer 1

John Hughes has already provided a wonderful hint for the fact that $rk_p f \leq rk_{f(p)} f$, so I'll focus on the other part of the statement.

First, a counterexample if $X$ is not connected is provided by letting $X$ denote two disjoint copies of $\mathbb{R}^2$ and letting $f$, on the first copy, project to the $x$ axis, while $f$, on the second copy, it is just the identity.

Now, some hints for the case where $X$ is connected. First, since $f$ is continuous, $f(X)$ is also connected.

Now, since the rank is an integer bounded above by $\dim X$, it follows that there is some $p\in X$ achieving the maximum rank. The fact that $rk_p f \leq rk_{f(p)} f$ now implies that $q = f(p)$ also achieves maximal rank.

Set $Z = \{s\in f(X): rk_s f = rk_q f\}$. Now, $q\in Z$ so $Z$ is non-empty. Prove that $Z$ is both open and closed. Since $f(X)$ is connected, this will imply $Z = f(X)$, so $f$ has constant (maximum) rank on $f(X)$.

Sketch for "closed": The condition $rk_s f = rk_q f$ is defined by the vanishing of determinants (polynomials) of submatrices of $d_s f$, so is closed.

Sketch for "open": Having smaller-than-maximum rank is given by the vanishing of more determinants of submatrices, so having smaller-than-maximum rank is closed, so having maximum rank is an open condition.

Answer 2

OK. Here is one hint:

1. $Rank (S \circ T) \le \min(Rank(S), rank(T))$, where $S$ And $T$ are linear transformations, and the codomain of $T$ is the domain of $S$. Apply this to the case $T = Df_p$ and $S = Df_{f(p)}$.

For the second part ("subsequently that ..."), I don't see how to use the first result in any way. If we let $A = f(X)$, then evidently we have $f | A = id_{A}$, but that doesn't mean that $f$ has constant rank on $A$.