Let $(E,\pi , X)$ be a n-dimensional vector bundle over a connected space $X$ and $f:E \to E$ a bundle homomorphism with $f\circ f = Id_{E}$. Show that $ Fix(f):= \{ v\in E | f(v)=v \}$ is a subbundle of $E$.
This is problem 3 from chapter 3 of Bröcker's book Introduction to Differential Topolpgy. The idea I had involved using the rank theorem: For $x\in X$, let $E_x := \pi ^{-1} ({x}) $. By definition of bundle homomorphism, we have that $f_x : E_x \to E_x $ is linear, and since $f$ is an involution, in particular, $f_x$ is invertible, so its rank is $n$, and thus, $f$ has constant rank. Then, by the rank theorem, there exist chart bundles of $E$, $(\phi, U), (\psi , U) $, such that $\psi \circ f \circ \phi ^{-1} : U \times \mathbb{R} ^n \to U \times \mathbb{R} ^n $, $\psi \circ f \circ \phi ^{-1} (u, (v_1,..., v_n)) = (u,(v_1,...,v_n))$, and thus: $\psi \circ f \circ \phi ^{-1} (U\times \mathbb{R} ^n ) = U\times \mathbb{R} ^n$.
Since $f \circ \phi ^{-1} (U\times \mathbb{R} ^n ) =f (\pi ^{-1} (U))$, it would be enough to show that $f (\pi ^{-1} (U))= \pi ^{-1} (U) \cap Fix(f)$.
It is clear to me that $\pi ^{-1} (U) \cap Fix(f) \subseteq f (\pi ^{-1} (U))$, but the other set containment is not clear to me and I don't even know if it is true. Also, I haven't used the fact that the base space $X$ is connected.
Any help would be very appreciated. Thank you so much!
You can write $Fix(f) = \ker(f - Id_{E})$. A kernel of a bundle map $f - Id_{E}$ is known to be a subbundle of $E$, if and only if it has a constant rank.
For each $x \in X$, write $r_{\pm}(x) := rank( f_{x} \mp Id_{E_{x}})$. You have to show that $r_{+}(x)$ is constant in $x$. Since $f_{x}^{2} = Id_{E_{x}}$, it follows that $E_{x} = \ker(f_{x} - Id_{E_{x}}) \oplus \ker(f_{x} + Id_{E_{x}})$. Consequently, one has $r_{+}(x) + r_{-}(x) = rank(E)$.
Finally, $r_{\pm}(x)$ both have the following property: for each $x_{0} \in X$, there is its open neighborhood $U$, such that $r_{\pm}(x) \geq r_{\pm}(x_{0})$ for all $x \in U$. This follows from the fact that $r_{\pm}(x)$ can be expressed as a rank of some square matrix depending smoothly on a $x$, which has this property.
But since $r_{+}(x) + r_{-}(x)$ is constant for all $x \in U$, we must have $r_{\pm}(x) = r_{\pm}(x_{0})$ for all $x \in U$. Since $X$ is connected (thus also path connected, if $X$ is a manifold), then $r_{\pm}(x)$ must be constant on entire $X$.