I've had some problems to prove this proposition:
Let $f:\mathbb{R}^{n}\to \mathbb{R}^{n}$ a function of class $C^1$ such that $(f\circ f)(x_0)=x_0$ for some $x_0\in \mathbb{R}^{n}$ and $Df(f(x))Df(x)=I_n$, for all $x\in \mathbb{R}^{n}$, where $D$ denotes the derivative and $I_n$ the identy matrix of size $n$. Show that $f$ is invertible and $f=f^{-1}$.
Thanks.
Note that by the chain rule, $Df^2 \equiv I_n$. Thus, by Lagrange's mean value theorem,
$$ f^2(b) - f^2(a) = Df^2(b-a) = b-a \quad (\forall a,b \in \mathbb{R}^n) $$
In particular, since $f^2(x_0) = x_0$ and $f^2(b) - f^2(x_0) = b - x_0$, we can conclude that
$$ f^2(b) = b, $$
that is, $f^2 \equiv id$. Thus, $f$ is its own left and right inverse. In particular, $f$ is injective: if $f(x) = f(y)$, then
$$ x = f^2(x) = f^2(y) = y. $$
For a similar reason, we also get surjectivity: for each $p \in \mathbb{R}$,
$$ f(f(p)) = p. $$
Finally, since this proves that $f$ is invertible, from $f^2 \equiv id$ we get that
$$ f \equiv f^{-1} $$
by composing with $f^{-1}$ at both sides of the equation.