I'm reading DoCarmo's Riemannian Geometry and I find the following example.
Let $M$ be a differentiable manifold and let $\varphi: G\times M\to M$ be a properly discontinuous action of a group $G$ on $M$. Then $M/G$ has a differentiable structure.
Sketch.
- For each $p\in M$ take the parametrization $x:V\to M$ so that $x(V)\subset U$, where $U\subset M$ is a neighborhood of $p$ such that $U\cap g(U)=\emptyset$, $g\neq e$.
- Define $y=x\circ \pi: V \to M/G$
- Prove that $\{(V, y)\}$ is a differentiable structure.
To prove that $\{(V, y)\}$ covers to $M/G$ i'm able to do it. Prove that coordinates change is differentiable I have a problem. Docarmo to do the following:
Consider $y_{1}:\pi\circ x_{1}:V_{1}\to M/G$ and $y_{2}:\pi\circ x_{2}:V_{2}\to M/G$ with $y_{1}(V_{1})\cap y_{2}(V_{2})\neq \emptyset$. Now let $\pi_{i}$ be the restriction of $\pi$ to $x_{i}(V_{i})$, $i=1,2$. Let $q\in y_{1}(V_{1})\cap y_{2}(V_{2})$ and let $r=x_{2}^{-1}\circ \pi_{1}^{-1}(q)$. Let $W\subset V_{2}$ be a neighborhood of $r$ such that $y_{2}(W)\subset y_{1}(V_{1})\cap y_{2}(V_{2})$. Then the restriction to $W$ is given by $$y_{1}^{-1}\circ y_{2}\vert W = x_{1}^{-1}\circ \pi_{1}^{-1} \circ \pi_{2}\circ x_{2}$$ Therefore is enough to show that $\pi_{1}^{-1}\circ \pi_{2}$ is differentiable at $p_{2}=\pi_{2}^{-1}(q)$. Let $p_{1}=\pi_{1}^{-1}\circ \pi_{2}(p_{2})$. Then $p_{1}$ and $p_{2}$ are equivalent in $M$, hence there is a $g \in G$ such that $gp_{2}= p_{1}$. It follows that the restriction $\pi_{1}^{-1}\circ \pi_{2}\vert x_{2}(W) = \varphi_{g}\vert x_{2}(W)$, thus $\pi_{1}^{-1}\circ \pi_{2}$ is differentiable at $p_{2}$.
My doubt is: Why $\pi_{1}^{-1}\circ \pi_{2}\vert x_{2}(W) = \varphi_{g}\vert x_{2}(W)$? I agree that the domain is the same but the image I'm not sure. I can prove that $\pi_{1}^{-1}\circ \pi_{2}(p_{2})= \varphi_{g}(p_{2})$ but at other points I can't it.