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R2 (i.e. the plane) is a covering map of a donut. R2 is simply connected so different elements in fundamental group of donut will have different lifting correspondence in R2.

Now punch a hole on the donut surface and punch holes on the corresponding places on R2, now the punched R2 is a covering map of punched donut, but the above property won't hold since punched R2 is not simply connected.

Now when you loop around this punched hole on the donut, the corresponding path on punched R2 will loop too. But when you loop around the center 3D hole or around the body center circle of the donut, the corresponding path on punched R2 won't be loop.

So the punched hole seems to be sort of different than the center 3D hole. This difference seems also depends on the covering map, i.e. if another covering map of punched donut is simply connected, this different will disappear.

The question is:(1) what is the essential of the difference and (2) Is there a simply connected space that covering maps punched donut?

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  • $\begingroup$ Intuitively a covering map may only cover some of the holes. Loop around holes that is coverd will become beam in the covering space and loop around holes that is not covered will still be loop in the covering space. The body hole and the center hole is covered and the punched hole is not covered in this case. $\endgroup$
    – jw_
    Jan 6, 2020 at 3:01

1 Answer 1

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The point is that $\mathbb R^2$ was not just a covering, it was the universal covering of $\mathbb T^2$. When you punch holes in it, it becomes not simply-connected and so isn't the universal covering of the punched donut.

Now for any covering space $p :\tilde X\to X$, $p_*\pi_1(\tilde X)$ is a subgroup of $\pi_1(X)$, and in fact if $X$ is nice enough (which is the case of the (punched) donut), any subgroup of $\pi_1(X)$ has a corresponding covering space; it just turns out that the punched $\mathbb R^2$ corresponds to the subgroup generated by the loop around the new hole in the donut, not the "old" loops, but you could do the opposite as well.

Forget about punching holes and just take the regular old donut $\mathbb T^2= S^1\times S^1$. $\mathbb R^2$ is a covering space, but $\mathbb R\times S^1$ is as well, and for this one, one of the two generating loops of the donut will loop in the covering, and the other won't : it's the same phenomenon.

When you take the universal covering space (if it exists, which it does in nice situations) the "asymmetry" disappears because all nontrivial loops below lift to paths that aren't loops above. For the punched donut, this universal covering space would just look a bit different (if you think hard about it, you'll see that the punched donut is homotopy equivalent to a wedge of two circles, so its universal covering space is homotopy equivalent to a tree - if you want to see a picture, type "cayley graph of free group" on some search engine)

So to answer specifically :

(1) The essential difference is that your new covering space is not universal and so is now "biased" towards some loops, and that just happens for a lot of covering spaces

(2) Yes there is, it's easy to describe up to homotopy equivalence, but not so easy up to homeomorphism

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