Extension of covering map

The question is the following: suppose I have manifolds $N$ and $M$ both with boundary, and I have a covering map $\phi$ from $\partial N$ to $\partial M.$ The question is: when is there a covering map $\tilde{\phi}: N \rightarrow M$ which restricts to $\phi?$ When $N, M$ are surfaces, the covering map of the boundaries is given by its monodromy, and there is a $\mathbb{Z}/2 \mathbb{Z}$ obstruction (the sign of the monodromy permutation should be $+1.$ (put a different way: each component of the boundary gets mapped to a collection of cycles. The covering can be extended if and only if the sum of the numbers of even cycles over all boundary components is even). This was proved by Husemoller and Gleason in Husemoller's thesis in the late fifties. I know of no general result in higher dimensions (even in dimension three), from which I conjecture that the question is known to be hard.

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I don't understand something about the result on surfaces you cite. Let M=N=unit disc. The covering of the boundary circle $z\mapsto z^n$ evidently extends to a covering of the discs only when $n=\pm 1$. What does this have to do with the sign of the permutation (which is a cycle in our case)? – Alexandre Eremenko Oct 8 '12 at 0:30
Your point is well-taken. The surfaces should have positive genus, otherwise the problem is either very easy but with a different answer (like your example) or extremely hard (like a sphere with three holes). – Igor Rivin Oct 8 '12 at 0:44

There's an algorithm to tell if such a map exists if $N$ and $M$ are compact 3-manifolds. Given a covering map $\partial N\to \partial M$ of degree $n$, any extension to a covering $N\to M$ must be a covering of the same degree $n$. Construct all covers of $M$ of degree $n$ (this amounts to finding all reps. of $\pi_1 M$ to the symmetric group $S_n$), and determine for which of these covers there is a lift of the maps $\partial N\to \partial M$. This reduces the question to whether given a homeomorphism $\partial N \to \partial M$ extends to a homeomorphism $N\to M$ by replacing $M$ with the $n$-fold covers together with the lifts of the boundary maps (in fact this part of the argument works in any dimensions). So we will now restrict to the case that the map $\partial N\to \partial M$ is a homeomorphism.
I'll now assume that $N$ and $M$ are 3-dimensional with non-trivial boundary (otherwise you're just asking for the homeomorphism problem), and irreducible (one may reduce to the irreducible case in the usual fashion using the Kneser-Milnor decomposition). Then $N$ and $M$ are Haken. Put a boundary pattern in $\partial N$ which has no non-trivial automorphisms, and use the homeomorphism to $\partial M$ to transfer the boundary pattern to $\partial M$.
By Theorem 6.1.6 of Matveev's book, there is an algorithm to tell if there is a homeomorphism of $N$ with $M$ which preserves the boundary pattern. This algorithm will then tell if the homeomorphism $\partial N\to \partial M$ extends to a homeomorphism $N\to M$ since the boundary pattern has no automorphisms.
That's certainly enlightening, and it is good to know that the solution is algorithmic, but not very satisfying, on a number of levels (e.g. the first step of checking all possible homomorphisms to $S_n$ does not thrill...) – Igor Rivin Oct 8 '12 at 1:06
@Igor: agreed. There may be some shortcuts given the extra information you have about the covering space induced of the boundary. For example, if you're considering $M, N$ handlebodies, then the cover $\partial M\to \partial N$ will determine a unique cover $M\cong \tilde{N}\to N$ which induces the cover of the boundary (if it exists - one just has to check that curves in $N$ bounding disks lift). In this case, I think a reasonable algorithm might exist, basically reducing to Haken's algorithm, or Whitehead's algorithm, depending on how the handlebodies are given. – Ian Agol Oct 8 '12 at 4:06