is the fixed set of a smooth involution a submanifold? 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.
 A: 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.
A: I can't add a comment to Berci's answer, that is why I post this comment as an answer.
The argument in Berci's answer does not work as stated. (Since the question is a local matter, we can assume that $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$, but I'm not sure that this helps.)
The problem is that $f^2=id$ only implies that 
$(df)_{f(p)}\circ (df)_p = id_{T_pX}$. So you cannot conclude that $(df)_p$ is of the given form, since $(df)_p$ and $(df)_{f(p)}$ are different maps.
A different argument using Riemannian geometry is the following:
Choose a Riemannian metric $g$ on $X$, then $\tilde g= g + (f^* g)$ is a metric for which $f$ is an isometry (because $f^2=id$). The fixed point set of an isometry is easily seen to be a totally geodesic submanifold of $X$. 
