Contractibility and orientation double cover

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.

• 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.) – Moishe Kohan Feb 13 at 14:12
• Thanks! This is a useful example. – Martin Tancer Feb 14 at 11:31