Characters vanishing on the disk metabolizer extend over the branched cover

If $\chi$ extends to $\tilde{\chi }:H_1(W_{D,q};\mathbb{Z})\to \mathbb{Q}\mathrm{/}\mathbb{Z}$, then for every $x\in P_D$ one has $i(x)=0$, so $\chi (x)=\tilde{\chi }(i(x))=0$. The extension has finite image because its domain is finite.

Conversely, suppose $\chi (P_D)=0$. Then $\chi$ factors uniquely through the subgroup $I=i(H_1({\mathrm{\Sigma }}_q(\mathrm{\partial }D)))$ of the finite group $B=H_1(W_{D,q})$. Let $p$ be the prime underlying the order of $\chi$ as in [F2]. Every element of order coprime to $p$ maps to zero, since its image is killed both by a power of $p$ and by an integer coprime to $p$. Thus the factored map restricts to the $p$-primary subgroup $I_p\le B_p$.

A homomorphism from a subgroup $J$ of a finite abelian $p$-group $B_p$ into $\mathbb{Q}\mathrm{/}\mathbb{Z}[p^{\mathrm{\infty }}]$ extends when one adjoins one element $b\in B_p$: if $p^k$ is the order of $b+J$, choose $c\in \mathbb{Q}\mathrm{/}\mathbb{Z}[p^{\mathrm{\infty }}]$ with $p^{k}c=f(p^{k}b)$, which exists because multiplication by $p^k$ is surjective on this Prüfer group. Set $f^{\mathrm{\prime }}(j+rb)=f(j)+rc$. If $j+rb=j^{\mathrm{\prime }}+r^{\mathrm{\prime }}b$, then $(r-r^{\mathrm{\prime }})b\in J$, so $r-r^{\mathrm{\prime }}$ is divisible by $p^k$; the chosen equation makes the two values equal. The extension is therefore well-defined. Repeating this finite step extends the character from $I_p$ to $B_p$.

Set the extension to zero on the complementary primary summand of $B$ and combine it with the extension on $B_p$. The resulting character on $B$ extends $\chi$ and has finite $p$-power image because $B$ is finite and the values are $p$-primary. This proves the reverse implication.