Some preliminaries:
Given a surface $S$ and a covering space $q: \tilde{S} \rightarrow S$, we say that $q$ is abelian if its deck transformation group is abelian. Naturally, the universal abelian cover $p: S_{ab} \rightarrow S$, the cover that covers all abelian ones, has a deck transformation group of $H_1(S)$---as $H_1(S)$ is the largest abelian subgroup of $\pi_1(S, -)$. (Simultaneously, the fundamental group of $S_{ab}$ is the smallest non-abelian subgroup of $\pi_1(S, -)$, the commutator subgroup.)
Question:
I've constructed a $\mathbb{Z}_2$-covering space for a compact, oriented surface with boundary. Call the surface $S$ and its $\mathbb{Z}_2$-covering $q_2: S_2 \rightarrow S$. Because the deck transformation group of $q_2$ is $\mathbb{Z}_2$ and the universal abelian cover $p: S_{ab} \rightarrow S$ covers every abelian cover, we know that there's a unique covering map $r: S_{ab} \rightarrow S_2$ so that $p = q_2 \circ r$. I'm trying to determine whether the induced map on first homology, $r_\star : H_1(S_{ab}) \rightarrow H_1(S_2)$, is surjective. Any suggestions? Thanks!
EDIT:
I didn't specify this before, but I'm working with a surface with nontrivial genus and only one boundary component. (Though, I think that my answer attempt below can apply to any number of boundary components.)
I think I figured it out. I can construct $\mathbb{Z}_6$-covering space for $S$ (which is simultaneously a $\mathbb{Z}_3$-covering space for $S_2$). Call this new covering space $q_6: S_6 \rightarrow S$ and the intermediate covering space $q_3: S_6 \rightarrow S_2$. Because finite-sheeted covers ($q_6$ is a 6-sheeted cover for $S$ and $q_3$, a 3-sheeted one for $S_2$) induce surjective maps on homology groups (see the first answer in Covering map, singular homology for a good explanation), I know that $q_3$ and $q_6$ are surjective on first homology. In particular, I know that there's a division $s: S_{ab} \rightarrow S_2$ and $r: S_{ab} \rightarrow S_6$ so that $s = q_3 \circ r$. Because $(q_3)_\star : H_1(S_6) \rightarrow H_1(S_2)$ is surjective, $s_\star = (q_3 \circ r)_\star = (q_3)_\star \circ s_\star$ is surjective, completing the proof. (Apologies for all the letters---my head is swimming, too!)