I am reading Introduction to differential topology, T.H Bröcker & K. Janich and I could not solve the following exercise:
Let $(E,\pi, X) $ be a vector bundle over a connected space $X$, let $f: E \rightarrow E$ be a vector bundle homomorphism and $f\circ f=f$. show that $f$ has constant rank.
I tried to find an open neighborhood $U$ s.t on this neighborhood $f$ has constant rank $k$. Then I would like to use connectedness of $X$ to extend this rank everywhere. But obviously this is not working since I have an counterexamples on the $\mathbb{R}\times\mathbb{R}^{2}$ and also I haven't used the fact that $f\circ f=f$.
If you can help me I will appreciate. Thanks in advance.
Let $n$ be the rank of $E$, let $I\subset M_n(\mathbb{R})$ be the set of idempotent matrices, and let $I_k\subset I$ be the set of idempotent matrices of rank $k$ for each $k$ such that $0\leq k\leq n$. By looking at what $f$ looks like in local trivializations of $E$, you can see that it suffices to prove that for each $k$, $I_k$ is clopen as a subset of $I$.
How do you prove that $I_k$ is clopen in $I$? There are several ways you could do this, but the following is probably the easiest. Every rank $k$ idempotent matrix has characteristic polynomial $x^{n-k}(x-1)^k$. The map sending a matrix to the coefficients of its characteristic polynomial is a continuous map $M_n(\mathbb{R})\to\mathbb{R}^n$. It follows that each $I_k$ is closed in $I$, and thus also open, since $I\setminus I_k=\bigcup_{j\neq k} I_j$ is a finite union of closed sets and hence also closed.