Let $(X, d)$ be a metric space and $f : X \to X$ a $C^1$ map. Let $p$ be a periodic point of period $m$ of $f$ and suppose that the differential $D f^m_p$ does not have $1$ as an eigenvalue. Let us introduce local coordinates near $p$ with $p$ as the origin. In these coordinates $D f^m_0$ becomes a matrix.
Can someone explain me why, since $1$ is not among the eigenvalues of $D f^m_0$, the map $F := f^m -Id$ (defined locally in the coordinates) is locally invertible by the Inverse Function Theorem?
P.S The Inverse Function Theorem: Suppose that $f : X \to Y$ is a smooth map whose differential $D f_x$ at the point $x$ is an isomorphism. Then $f$ is a local diffeomorphism at $x$.
Thank you!
We need to check the rank of $DF(p)$. If this rank is maximal, then the inverse function theorem tells you that $F$ is locally invertible on some neighborhood $U \ni p$. In this case, you check rank by looking at the kernel.
$$DF(p) = Df^m(p) - I(p) = 0 \iff Df^m(p) - p = 0 \iff Df^m(p) = p$$
but doesn't this mean you have $1$ as an eigenvalue :). Therefore, the kernel is trivial and so $DF(p)$ has max rank.