The proof of the quotient space of a manifold with properly discontinuous action is an orbifold. I am reading the proof of the following proposition from Thurston's the geometry and topology of 3-manifolds:

My questions are:


*

*By $U_x=\tilde{U}_x/I_x$, if $\cap_{i=1}^kU_{x_i}\not=\emptyset$, then for the $\gamma_i$'s s.t. $\cap\gamma_i\tilde{U}_{x_i}$, we can select $\gamma_i\in I_{x_i}$, then it seems that we don't need the conjugation $\gamma_iI_{x_i}\gamma_i^{-1}$ but just $I_{x_i}$? And $\tilde{U}_x$ is invariant by $I_x$, then we have $\gamma_i\tilde{U}_{x_i}$ is just $\tilde{U}_{x_i}$. Is that right?

*I can see $\cap \gamma_iI_{x_i}\gamma_i^{-1}$ acts invariantly on $\cap\gamma_i\tilde{U}_{x_i}$, but it's not very clear to me why $\cap_{i=1}^kU_{x_i}=(\cap\gamma_i\tilde{U}_{x_i})/(\cap \gamma_iI_{x_i}\gamma_i^{-1})$.
 A: Regarding your question 1, no, you don't have the freedom to select $\gamma_i$ any way you might: there is a constraint. First you may select a point $x \in \cap_{i=1}^k U_{x_i}$, then you may select $y \in M$ whose image under the projection map $M \mapsto M / \Gamma$ is equal to $x$. Now you may select $\gamma_i \in \Gamma$, but under a constraint, namely that $y \in \gamma_i \tilde U_{x_i}$. It follows that $y \in \cap_{i=1}^k \gamma_i \tilde U_{x_i}$ and so $\cap_{i=1}^k \gamma_i \tilde U_{x_i} \ne \emptyset$.
Regarding your question 2, all that is being asserted is that once we know that the intersection $\gamma_1 \tilde U_{x_1} \cap \cdots \cap \gamma_k \tilde U_{x_k}$ is nonempty, then we can choose the lift (big tilde expression) of $\cap U_{x_i}$ to be equal to that intersection. 
Something to keep in mind here is the meaning of the "tilde" notations. Under an ordinary covering map $p : X \mapsto Y$, given a subset $U \subset Y$, the expression $\tilde U$ is not always defined: $\tilde U$ is only defined when $U$ is evenly covered. And if this holds then $\tilde U$ is still not well defined: $\tilde U$ represents a choice of subset of $p^{-1}(U)$ which maps homeomorphically to $U$. The situation under an orbifold covering map $X \mapsto Y$ is similar: given $U \subset Y$, the expression $\tilde U$ is only defined when $U$ is "evenly covered in the sense of orbifold covering maps"; and if this holds then $\tilde U$ is not well defined: $\tilde U$ is a choice of a subset of $p^{-1}(U)$ having stabilizer $I_{\tilde U}$ such that $p$ induces a homeomorphism $\tilde U / I_{\tilde U} \approx U$.
So to give a fuller answer to your question 2, what the text is saying is that the (big tilde expression) is being chosen to be $\gamma_1 \tilde U_{x_1} \cap \cdots \cap \gamma_k \tilde U_{x_k}$.
Regarding your question in the comments, in the equation $U_{x_1} \cap \cdots \cap U_{x_k} \ne \emptyset$, several objects indexed by $i=1,...,k$ have already been implicitly chosen by the time that expression on the left hand side of that equation has been formed: the point $x_i \in M/\Gamma$; the point $\tilde x_i \in M$; the stabilizer $I_{x_i}$ of $\tilde x_i$; and the $I_{x_i}$-invariant neighborhood $\tilde U_{x_i}$ of $\tilde x_i$, for which the projection $M \mapsto M/\Gamma$ induces a homeomorphism $\tilde U_{x_i} / I_{x_i} \approx U_{x_i}$. So, choosing $x \in \cap U_{x_i}$, you may indeed choose $y \in \tilde U_{x_i}$ which projects to $x$. It follows that $x$ is equal to the projection to $M/\Gamma$ of $y \cdot I_{x_i}$, but it is not accurate to say that $y \cdot I_{x_i} = x$, because the set $y \cdot I_{x_i}$ is a subset of $M$, but $x$ is a point of $M/\Gamma$. The full pre-image of $x$ in $M$ is $y \cdot \Gamma$ which contains $y \cdot I_{x_i}$, but is probably larger than $y \cdot I_{x_i}$. 
