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Picture below is from 23th page of Do Carmo's Riemannian geometry. I don't know why $\pi_1^{-1}\circ \pi_2$ is coincide with $\varphi_g$ on $x_2(W)$. Since in my view, it is needed that proving $g$ is independent to $p_2$. But there is not proof about this.

enter image description here

Enhao Lan
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1 Answers1

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Assume that there are two different group elements, say $g$ and $g'$, which map $2$ distinct points of $\mathbf{x}_2(W)$ to $2$ points of $\mathbf{x}_1(W)$ respectively, then you get that

$$(g'^{-1}g).(\mathbf{x}_2(W)) \cap \mathbf{x}_2(W) \neq \phi,$$

which gives a contradiction with the definition of a properly discontinuous action (since $g'^{-1}g \neq 1$, by our assumption).

Malkoun
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