On the proof of that fixed point set of an involution is a submanifold Let $M$ be a smooth manifold, and let $f:M\to M$ be a smooth involution (i.e. $f^2=\text{id}$).
If we introduce a Riemannian metric on $M$ so that $f$ is isometry, we can prove easily that the fixed point set of $f$ makes a submanifold of $M$. (cf: [Klingenberg, Riemannian Geometry, Theorem 1.10.15]) My question is:
Can we prove this statement without using any Riemannian metric?
That is all of my question. Thank you.
 A: Yes, by linearizing $f$ at a fixed point. 
More explicitly, let $x_0\in M$ be a fixed point of $f$. Let $U$ be an open neigborhood of $x_0\in M$ for which there is a diffeomorphism with an open subset $V$ of $\mathbf R^n$. Replacing $U$ by $f^{-1}(U)\cap U$ if necessary, we may assume that $f(U)\subseteq U$.Then $f(U)=U$ since $f^2=\mathrm{id}$. Therefore, we may assume that $M=V$ is an open subset of $\mathbf R^n$. We may as well assume that $x_0$ is the origin in $\mathbf R^n$.
Now, consider
the map
$$
g\colon V\rightarrow \mathbf R^n
$$
defined by
$$
g(x)=x+D_0f(f^{-1}(x)),
$$
where $D_0f$ is the differential of $f$ at $0$. The map $g$ is of course differentiable. Moreover, one has
$$
g(f(x))=f(x)+D_0f(x)=D_0f(x+D_0f^{-1}(f(x)))=D_0f(x+D_0f(f^{-1}(x)))=D_0f(g(x))
$$
since $f^{-1}=f$. Observe that
$$
D_0g=\mathrm{id}+D_0f\circ D_0f^{-1}=2\mathrm{id}
$$
on $\mathbf R^n$. By the inverse function theorem, there is an open neighbohood $W$ of the origin in $\mathbf R^n$, contained in $V$, such that
$$g\colon W\rightarrow \mathbf R^n$$ is a diffeomorphism onto its image. Replacing $W$ by $f^{-1}(W)$ if necessary, we may assume that $f(W)=W$ like before. Since
$$
g(f(x))=D_0f(g(x)),
$$
for all $x\in W$, we may assume that the action of $f$ on $V$ is the restriction of a linear map $L\colon\mathbf R^n\rightarrow\mathbf R^n$.
Since $f^2=\mathrm{id}$, one has $L^2=\mathrm{id}$. This means that we can diagonalize $L$ over $\mathbf R$, and we may assume that the matrix of $L$ in the standard basis is the diagonal matrix $\mathrm{diag}(1,\ldots,1,-1,\ldots,-1)$, say with $m$ diagonal entries equal to $-1$. Then, the set of fixed point of $f$ on $V$ is equal to $V\cap\mathbf R^m$, where $\mathbf R^m$ is identified with the subset $\mathbf R^m\times\{0\}^{n-m}$ of $\mathbf R^n$. This proves that the set of fixed points of $f$ is a smooth submanifold of $M$.
The argument applies more generally to any finite group action on $M$, or even, any action of a compact group on $M$, adapting the diagonizability part a little bit. The corresponding statement for topological manifolds is false. Note also that it is crucial here not to ask for a smooth manifold to be connected or even nonempty!
