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As the title suggests, I'm studying the following topological space:

enter image description here

which is a Klein Bottle $K$ with a disk glued along an essential orientation preserving curve.

I need to compute its fundamental group and clearly describe all its connected coverings

Here is my attempt: we are gluing the Disk $D$ along the following line:

enter image description here

which then let us make the following moves in order to obtain an homotopy equivalence:

enter image description here

therefore the fundamental group would be trivially $\Bbb Z$. Then I'm asked to determine ALL connected covering spaces of it. How to do that? I understand the space only up to homotopy, but even if I start directly with the Klein bottle and the glued disk, I don't know how to produce a clear description of its coverings

Any suggestion?

EDIT: I think I came up with the universal cover of this space $K'$. We start with the usual two sheeted cover $T\to K$ (I changed the notation from the preceding drawings)

enter image description here

and we go on in this way infinitely many times:

enter image description here

we then attach a disk along each vertical loop (the $a$'s and the reversed $a$'s-not labelled in the picture-). This clearly covers my space $K'$ and since it's homotopy equivalent to an infinity long necklace of $S^2$'s, should be simply connected (the only possible non-trivial loop would be the thread keeping the spheres united with each other, but since we have infinitely many spheres, there can't be a continuous map from $S^1$ representing such thread, using compactness).

Does it make sense?

Luigi M
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    Here is my idea : you have a projection map $K \to S^1$ projecting to a generator of $\pi_1(K)$ corresponding to $\Bbb Z$ and which extends to a map $K' \to S^1$ where $K'$ is $K$ with the disk glued (all the disk goes to one point). My guess is that any cover $p : X \to K'$ should be the pullback of a cover $p : Y \to S^1$. –  Aug 18 '17 at 20:29
  • @N.H. Are we using (basically) the fact that since $\Bbb Z$ is abelian, every covering of $K'$ will be normal and therefore a principal bundle over $K'$. We know that they are classified by maps into $S^1$ since any non-trivial subgroup of $\Bbb Z$ is $n\Bbb Z$? – Luigi M Aug 18 '17 at 21:45
  • I'm just using that the fundamental group is $\Bbb Z$ so somehow the covering should be in correspondence with the covering of $S^1$ and it was the most natural way for me to build a covering in $K'$ from a covering in $S^1$, but I don't think I would know how to prove it rigorously –  Aug 18 '17 at 22:34
  • I could be wrong, but if $\pi_1(K)=\langle a,b\mid b^{-1}ab=a^{-1}\rangle$, then your disk is adding the relation $a=1$. So to find coverings, you can take the "usual" Klein bottle coverings corresponding to the subgroups $\langle a,b^m\rangle$ (for an $m$-sheeted covering). These are tori/bottles. Now an $m$-sheeted covering of the disk is just $m$ disks. So in this tori/bottle, you glue $m$ evenly spaced disks. – Steve D Aug 18 '17 at 23:30
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    I don't know what happens in the non-compact (universal) covering. It's clearly not contractible based on your homotopy description – Steve D Aug 18 '17 at 23:46
  • @SteveD could you elaborate on that a little bit? why are you considering the subgroups $\langle a, b^m \rangle$? and why after considering the finite coverings of the bottle you can just glue evenly spaced disks to obtain the covering of $K'$? – Luigi M Aug 19 '17 at 01:55
  • Again, I'm not sure this is entirely correct, but I chose those subgroups because they become the proper subgroups of $\mathbb{Z}$ after adding the extra relation. Plus it seems pretty clear that the torus with two disks double covers this space. On gluing the disks, I just used the fact that the disk is simply connected, so it's coverings are just disjoint unions of disks. Restricting the covering to the circle transverse to the disk, it would seem the disk becomes evenly spaced disks in these finite coverings. – Steve D Aug 19 '17 at 02:10
  • @SteveD so we are basically trying to prove that we start with a covering of the Klein Bottle s.t. the loop (representing) $a$ lifts to a family of loops $\tilde{a}$'s and then attach disks along such loops? and that all coverings of our space arise in this way? – Luigi M Aug 19 '17 at 15:27
  • @SteveD Maybe I came up with the universal cover, I will update the question. – Luigi M Aug 19 '17 at 16:07
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    I think you're probably right about the universal cover (based on the universal cover of $S^2\vee S^1$). I think for the coverings I mentioned, the loop $a$ lifts to a single loop (the way the circle self-covers itself). Then we attach the disk at roots of unity (the fiber of 1 in these self-covers). If I'm right about these, then these are all the compact coverings, because $\mathbb{Z}$ only has one subgroup of a given index. – Steve D Aug 19 '17 at 17:33

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