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I'm currently trying to understand the definition of an orbifold as expressed in Thurston's Geometry and topology of three manifolds (The definition is in chapter 13 p300). I'm confused about the following:

Assume we have a surface $X$, the cover of which includes two open sets $U_a$ and $U_b$ with $U_a \subset U_b$. To $U_a$ we associate the set $\tilde{U}_a \subset \mathbb{R}^2$ and let $\Gamma_a$ be an order three rotation acting on $\tilde{U}_a$. Likewise, to $U_b$ we associate the set $\tilde{U}_b$ and let $\Gamma_b$ be an order six rotation acting on $\tilde{U}_b$. Assume also that the fixed point of $\Gamma_a$ and $\Gamma_b$ are associated with the same point $x$ in $X$. It seems to me that the these charts are consistent with the conditions laid out in the definition, but I'm sure this can't be the case as it would lead to an ambiguity as to whether we have an order three or order six cone point at this point. Can anyone tell me where I'm going wrong?

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The map from $\tilde{U}_a$ to $\tilde U_b$ must be an embedding, not merely equivariant with respect to the group actions. Are you saying that you can construct such a map? –  Kevin Walker Mar 22 '12 at 23:30
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My previous comment misses the mark. I think the problem with your example becomes apparent when looking at Thurston's commutative diagram on $\tilde U_i/\Gamma$. –  Kevin Walker Mar 22 '12 at 23:41
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Kevin, I think you're right. The problem was that I was ignoring the fact that things have to commute with respect to the inclusion map from $U_a$ to $U_b$ as well. If this is the case then my example no longer satisfies the conditions, as the composite map $\phi_j^{-1} \circ f_{ij} \circ \phi_{ij} \circ \phi_i$ now maps $U_i$ two to one into $U_j$. –  uncooltoby Mar 23 '12 at 19:57

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