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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 this 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.

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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 this 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.

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 $s_1s_2s_1^{-1}s_2^{-1}=t_1...t_p$, s_1s_2s_1^{-1}s_2^{-1}=t_1...t_p$(Ore result) and take$s_3=s_4$- cycles of length 2,$d$, while all other permutations$s_i$should be trivial. This solves Clearly the problemaction on${1,...d}$is transitive. Now the existence of a cover follows by standard arguments. If you manage to make the proof very short it is worth to put it in the article, or at least give a hint. Otherwise, indeed, as Pete said it would be nice to find a reference. 4 added 856 characters in body I don't know the reference for the answerthis question, so but I am pretty sure that it should follow from some known statement. Anyway let me propose give the asnwer answer in the case when S has negative Euler characterisitccharacteristic, orientable and CONNECTED. At least you can compare with your own , answer, hope that it is correct. Let p(sDenote by p(S) denote the number of punctures. Moreover, in the case Genus(S)=0 you have an addtional additional condition$p(S)d-p(S')>d-2$. This condition comes if we want S' to be connected. If assures that S' is alowd to be disconnected, this condition desapearsconnected. I hope think that the answer is correct. All the these conditions seem to be are necessary (and apearently sufficient)sufficient. It is obvious that 1), 2) are necessary. Condition 3) comes from the fact, that the permutation corresponding to going around all punctures on S is a commutator, so it should be even.I think In order to show that this conditions are sufficient, you could use the old result of Ore that tells that every even permutation is a commutator of two permutations. Oystein Ore. Some remarks on commutators. Proc. Amer. Math. Soc., 2:307–314, 1951. Let me prove that conditions are sufficient in the case when genus of S is two. Sketch of a proof. We want to show that there exists a collection of permutation in$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$, and take$s_3=s_4$- cycles of length 2, while all other permutations$s_i\$ should not be too involvedtrivial. Just using some standard facts about permutation groupsThis solves the problem. I am not 100% sure if

If you need manage to include it..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.

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