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Question. Let $M$ be a triangulated non-orientable 3-manifold with non-orientable boundary. (It is possible to assume that the boundary is the Klein bottle.) Let $\ell$ be a non-orientable loop on the 1-skeleton of the boundary (the regular neighborhood of this loop inside the boundary is the Möbius band and the regular neighborhood of this loop considered in $M$ is the solid Klein bottle). Let me also assume that the resulting space $M/\ell$ after contracting the loop $\ell$ is a contractible space. (Equivalently, after gluing a disk along $\ell$, we get a contractible space.)

Let $M'$ be the orientation double cover of $M$ and $\ell'$ be the loop which covers $\ell$ (this is a single loop by the assumptions on $\ell$). Is it true that after contracting $\ell'$ in $M'$, we get a contractible space?

Remarks and background. This question is kind of easy to formulate special case of something more general that I would like to be true. It regards certain analysis of some contractible singular 3-manifolds. (I am interested in the special cases when the contractibility can be recongized algorithmically.)

In this special case, the only example of $M$ and $\ell$ satisfying the assumptions, I am aware of is the following: $M$ is the solid Klein bottle and $\ell$ is the loop which is homotopic to the core curve of this solid Klein bottle.

Homology of $M'/\ell'$ should be OK. The difficulty, in my opinion, is the fundamental group $\pi(M'/\ell')$. There is a well established theory how to compute the $\pi(M')$ out of the knowledge of $\pi(M)$. However, the trouble is that $\pi(M)$ is not completely known.

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    $\begingroup$ You get more examples as follows: Take a knot $K\subset S^3$ which is invariant under and (orientation-reversing) involution $\tau$ with exactly two fixed points which are both in $K$. Then take $M=Ext(K)/\tau$, where $Ext(K)$ is the complement to an open tubular neighborhood of $K$. However, in this example, $M'/\ell'$ is still contractible. (All examples with $M'/\ell'$ contractible are obtained this way.) $\endgroup$ – Moishe Kohan Feb 13 at 14:12
  • $\begingroup$ Thanks! This is a useful example. $\endgroup$ – Martin Tancer Feb 14 at 11:31

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