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Let $M$ be a finetely connected orientable surface with compact boundary. This means $M$ is homeomorphic to a compact orientable surface $\Sigma$ of genus $g \geq 0$ minus $r \geq 1$ points and minus $k \geq 1$ open discs.

Is there a detailed description of the universal cover of $M$? What if $k = 1$?

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  • $\begingroup$ I think you can see this, to the extent that it's seeable, by going to the universal cover in steps. First start with the compact orientable surface and take its universal cover. Focusing on genus $g\geq 1$, that gives you a plane that you can think of as partitioned into fundamental domains that each map to a dense set of the surface. Now, up to some isotopies, poking holes and cutting out disks can be done in the interior of the fundamental domain, so in the covering space you get a countable number of holes and countable number of removed disks. So now take the universal cover of that. $\endgroup$
    – Greg Friedman
    Commented May 24, 2021 at 10:07
  • $\begingroup$ So up to homotopy this is something like the universal cover an infinite wedge of circles, independent of $r$ and $k$ so long a one of them is positive. $\endgroup$
    – Greg Friedman
    Commented May 24, 2021 at 10:10
  • $\begingroup$ If $g=0$ then you're cutting things out of the sphere, which is basically the same picture with a finite number of holes, so that case is simpler. $\endgroup$
    – Greg Friedman
    Commented May 24, 2021 at 10:13

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