is the fixed set of a smooth involution a submanifold?

Question

Let $f:X\rightarrow X$ be a smooth map of a smooth manifold with $f^2=\operatorname{id}$.

Is the subset $\{x\in X\mid f(x)=x\}$ a smooth submanifold?

I tried to find an argument with the implicit function theorem, but I don't have an answer.

Answer 1

If $f$ is involutive, so is its derivative in all points: $(T_xf)^2=Id$. And hence, there is a basis $(b_1,..,b_m,..)$ of $T_xX$ in which $T_xf$ becomes of the form $\pmatrix{I_m & 0\\0&-I_k}$ where $I_k$ is the $k\times k$ identity matrix, and $m+k=\dim X$.

So, if $f(x)=x$, then cutting out the correspondent of $\Bbb R^m\cong\langle b_1,..,b_m\rangle \subseteq T_xX\cong \Bbb R^n$ from the local chart $\phi$ will give you charts. Smoothness guarantees that the change of $b_i$'s along $x$ is smooth.