How to prove that the following quotient is a smooth manifold? Let M be a smooth manifold. Suppose that $f : M \to M$ is a smooth involution
without fixed points, i.e. for all points $x \in M$ the conditions $f^2(x) = x$ and $f(x) \ne x$ are satisfied. Define an equivalence relation on M by $x ∼ y \Leftrightarrow x = f^k(y)$ for some $k \in \mathbb{Z}$. I want to show that the associated quotient space $M/∼$ naturally inherits the structure of a smooth manifold such that the projection $ \pi : M \to M/∼ $ is smooth.
Given an atlas $\{U_\alpha, \phi_\alpha\}_{\alpha \in \mathcal{I}}$ of $M$, choosing symmetrized neighborhoods (that is restricting the size of each $U_\alpha$ such that $U_\alpha \cap f(U_\alpha)= \emptyset$, assuming that can be done), one can construct an atlas for $M/∼$ by considering the covering $ \{ V_\alpha = \pi(U_\alpha), \psi_\alpha = \phi_\alpha \circ (\pi |_{U_\alpha})^{-1} \}_{\alpha \in \mathcal{I}} $ which is obviously a smooth atlas on $M/∼$.
$M/∼$ is compact and connected because M is compact and connected. But I am not sure how to prove Hausdorffness. Moreover, I am not totally convinced that I can always have a choice of symmetrized neighborhoods.
 A: First, pick an arbitrary Riemannian metric $\langle \cdot ,\cdot \rangle _1$ and define a new Riemainnian metric $\langle \cdot ,\cdot \rangle = \langle \cdot, \cdot\rangle_1 + f^\ast \langle \cdot,\cdot \rangle_1$.  Then by construction, $\langle \cdot ,\cdot \rangle$ is $f$ invariant.  Said another way, $f$ is an isometry with respect to this metric.
Now, let's show you can always find a symmetrizing neighborhood.  So, let $x\in M$ and set $d  = d(x,f(x)) > 0$.  Consider the ball $B = B(x, d/2)$ of radius $d/2$ around $x$.  Since $f$ is an isometry, $f(B) = B(f(x), d/2)$.  We claim that $B$ is a symmetrizing neighborhood.  (If it's not a chart, intersect it with a chart, giving a symmetrizing neighborhood).
To see this, assume for a contradiction that there is a $y\in B\cap f(B)$.  Then $$d = d(x,f(x)) \leq d(x,y)  + d(y, f(x) < d/2 + d/2 = d,$$ so $d < d$.
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I'll leave the Hausdorff condition to you, but as a hint, for $x$ distinct from both $y$ and $f(y)$, let $d = \min\{d(x,y), d(x,f(y)\}$, and consider $U = B(x,d/2)\cup B(f(x),d/2)$ and $V = B(y,d/2)\cup B(f(y), d/2)$.
