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More on Universal Cover of a Surface with Boundary

  1. Dec 21, 2014 #1

    WWGD

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    I am trying to understand in more detail the answer to:
    http://math.stackexchange.com/questions/673187/universal-cover-of-a-surface-with-boundary
    It is mentioned that the universal cover of a hyperbolic surface ##S## with geodesic boundary is a closed disk ##D^2## with a Cantor set removed from its boundary. I am trying to see what the preimage of an interval of the Cantor set is under this cover, i.e., why are we removing the Cantor set from the boundary . I understand we first construct the surface ##S## by gluing pairs of pants and then embed ##S## in a closed hyperbolic surface ##S^## with boundary, and that the universal cover ##S^ ~## (which we construct by gluing more pants so that we cap all the boundar components.) is the Poincare disk
    ##D^2:= \mathbb H^2 \cup S^1_{\infty} ## (i.e., the last part is the x-axis plus the point at ##\infty## .) But where does the removed Cantor set in the cover of the surface with boundary come from? Are we getting a sort-of infinitely-sheeted cover, and, if so, is there a reasonably-nice expression for the projection map? And if someone had a ref. for explaining why the lift of a hyperbolic surface without boundary is the Poincare disk, that would be great.
     
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  3. Dec 22, 2014 #2

    WWGD

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    Never mind, thanks, it seems I have a lot of reading to do in Hyperbolic geometry.
     
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