Let $H$ be a 3-dimensional handlebody. What are the finite covers of $H$? I imagine that they must all be themselves handlebodies - in fact, if I think of a handlebody as a thickened wedge of circles are all of the finite coverings just thickened versions of the coverings of the wedge of circles?
2026-02-22 19:10:18.1771787418
Finite covers of handlebodies?
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Yes, this is true. One way to prove this is to use the boundary connected sum (see for instance here for the definition).
You first prove that a 3-dimensional manifold is a handlebody if and only if it connected and equal to a boundary connected sum of a finite disjoint union of closed 3-balls (where we allow a "boundary self-connected sum" for some of the balls). This part is a pleasant proof by induction on the number of 2-disks along which we do the gluing.
Then argue that given a splitting of a connected 3-manifold $M$ as a boundary connected sum of some ball, the lift of such a decomposition to a finite cover $M'\to M$ is again a boundary connected sum of balls. This part is quite easy since an open ball lifts to a finite disjoint union of open balls.