TITLE: is the fixed set of a smooth involution a submanifold?
QUESTION [9 upvotes]: 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.

REPLY [4 votes]: 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 $X$.