Necessary and sufficient criteria for a surface to cover a surface

Question

Let S and S' be closed [Edit: orientable] surfaces, then it is well known that for S' to cover S it is necessary and sufficient that chi(S)|chi(S'). (Here 'chi' denotes the Euler characteristic).

However, if S and S' are punctured surfaces then the above condition is necessary but no longer sufficient.

Is the question of determining necessary and sufficient criteria for S' to cover S answered in the research or expository literature?

I think that I know such criteria and want to use them a paper that I'm working on but (a) may be deceiving myself and (b) want to know whether I should write up a proof anew or whether there's a suitable reference. Surely the relevant criteria have been rediscovered many times, but I've never seen them discussed in writing.

Edit: Thanks Pete, I should have demanded that my surfaces be orientable

Answer 1

I don't know the reference for this question, but I am pretty sure that it should follow from some known statement. Anyway let me give the answer in the case when S has negative Euler characteristic, orientable and CONNECTED. At least you can compare with your own answer. Denote by p(S) the number of punctures.

1) chi(S')/chi(S)=d with $d$ positive integer

2) $p(S)\le p(S') \le p(S)d$.

3) $p(S)d-p(S')$ should be even.

Moreover, in the case Genus(S)=0 you have an additional condition

$p(S)d-p(S')>d-2$. This condition assures that S' is connected.

I think that these conditions are necessary and sufficient. It is obvious that 1), 2) are necessary. Condition 3) comes from the fact, that the permutation corresponding to going around all punctures on S is a commutator, so it should be even.

In order to show that these conditions are sufficient, you could use the old result of Ore that tells that every even permutation is a commutator of two permutations. Oystein Ore. Some remarks on commutators. Proc. Amer. Math. Soc., 2:307–314, 1951. Let me prove that conditions are sufficient in the case when genus of S is two or more.

Sketch of a proof. We want to show that there exists a collection of permutation in $S_d$, $s_1,...,s_{2g}, t_1,...,t_p$ ($p=p(S)$) that act transitively on ${1,...,d}$ such that $s_1s_2s_1^{-1}s_2^{-1}...=t_1...t_p$, where $t_1,...,t_p$ are given permutations with the product in the alternating group $A_d$. Then we chose $s_1$ and $s_2$ in such a way that $s_1s_2s_1^{-1}s_2^{-1}=t_1...t_p$ (Ore result) and take $s_3=s_4$ - cycles of length $d$, while all other permutations $s_i$ should be trivial. Clearly the action on $\{1,...d\}$ is transitive. Now the existence of a cover follows by standard arguments.

If you manage to make the proof very short it is worth to put it in the article, or at least give a hint. Otherwise, indeed, as Pete said it would be nice to find a reference.

Answer 2

In a recent paper by Calegari, Sun, and Wang, the authors cite

W.S. Massey, Finite covering spaces of 2-manifolds with boundary. Duke Math. J. 41 (1974), no. 4, 875-887.

which proves that Dmitri's conditions (1) and (2) are sufficient if the S and S' are compact, orientable with non-empty boundary.

(Dmitri's condition (3) is redundant as Massey points out in the second page of his article: writing chi(S') = 2 - 2g(S') - p(S') and chi(S) = 2 - 2g(S) - p(S) Dimitri's condition (1) gives

2 - 2g(S') - p(S') = d(2 - 2g(S) - p(S))

and reducing (mod 2) gives that p(S') and dp(S) have the same parity. But the proof therein is longer than the one that Dimitri for the case that he treats (if one takes the Ore result as given).