A prime-power branched cover of a smooth proper disk is a rational homology ball

The core circle of $D^2\times S^1$ is the lift of ${\mu }^q$. Lift that loop to the infinite cyclic cover $X_{D,\mathrm{\infty }}$: it is a path from a point $\tilde{x}$ to $t^q\tilde{x}$. The connecting map of the chain sequence for the $q$-fold quotient sends its homology class to $[\tilde{x}]\in H_0(X_{D,\mathrm{\infty }};{\mathbb{F}}_p)$. By the proof of [F2], this connecting map is an isomorphism in degree one, so the overlap map $H_1(D^2\times S^1;{\mathbb{F}}_p)\to H_1(X_{D,q};{\mathbb{F}}_p)$ is an isomorphism.

The filling piece $D^2\times D^2$ is contractible and the overlap has homology only in degrees zero and one. The Mayer–Vietoris sequence [F4], using the isomorphism from step 1.1, gives $H_i(W_{D,q};{\mathbb{F}}_p)=0$ for every $i>0$. The two glued pieces and their overlap are connected, so $W_{D,q}$ is connected; orientation is inherited from the cyclic branched covering map away from the branch set and extends over the local model.

By [F3], each integral homology group $H_i(W_{D,q};\mathbb{Z})$ is finitely generated. The injection $H_i(W_{D,q};\mathbb{Z})\otimes {\mathbb{F}}_p\hookrightarrow H_i(W_{D,q};{\mathbb{F}}_p)$ from [F5] makes that tensor product zero for $i>0$. A finitely generated abelian group with zero tensor product with ${\mathbb{F}}_p$ has rank zero; hence $H_i(W_{D,q};\mathbb{Q})=0$. Its boundary is ${\mathrm{\Sigma }}_q(\mathrm{\partial }D)$ by [F1], proving the statement.