The double cover branched over a slice disk is a rational homology ball

By [F2] we may take $D$ neat, so the closed tubular neighbourhood $N$ of [F3] is diffeomorphic to $D\times D^2$, with $N\cap S^3=\mathrm{\partial }D\times D^2$ and with $\mathrm{\partial }N$ split into the two solid tori $\mathrm{\partial }D\times D^2\subseteq S^3$ and $D\times S^1$, glued along the torus $\mathrm{\partial }D\times S^1$. Then $Y:=\overline{B^4\setminus \mathrm{int}N}$, with its corners rounded, is a compact $4$-manifold with boundary and $B^4=N\cup Y$ with $N\cap Y=D\times S^1$, which is homotopy equivalent to $S^1$ with generator the meridian circle $\{x\}\times S^1$.

Mayer-Vietoris [F8] for the collar-thickened open cover of $B^4$ by the interiors of enlarged $N$ and $Y$, retracting to $N,Y,N\cap Y$ respectively, with $N$ and $B^4$ contractible and $N\cap Y\simeq S^1$ gives $H_i(Y;\mathbb{Z})=0$ for every $i\ge 2$, because $H_i(N\cap Y)$ vanishes there and $H_i(B^4)$ vanishes for $i\ge 1$; in degree one it makes $H_1(N\cap Y;\mathbb{Z})\to H_1(Y;\mathbb{Z})$ an isomorphism, since the preceding term $H_2(B^4)$ vanishes and $H_1(N)=0$, so $H_1(Y;\mathbb{Z})\cong \mathbb{Z}$ is generated by the meridian; in reduced degree zero all terms of ${\tilde{H}}_0(N\cap Y)\to {\tilde{H}}_0(N)\oplus {\tilde{H}}_0(Y)\to {\tilde{H}}_0(B^4)$ vanish except possibly the middle, so $Y$ is connected.

The connected manifold $Y$ is path-connected, and its ball or half-ball charts give contractible neighbourhoods, so it is locally path-connected and semilocally simply connected. Fix a basepoint in $Y$. By [F5] the composite ${\pi }_1(Y)\to H_1(Y;\mathbb{Z})\cong \mathbb{Z}\to \mathbb{Z}\mathrm{/}2$ (reduction mod $2$) is a surjection, so it determines a connected double cover $p:\tilde{Y}\to Y$. Over ${\mathbb{F}}_2$ the singular chains of the cover form the short exact sequence of [F4], and its long exact homology sequence together with step 2.1 gives: $H_0(Y;{\mathbb{F}}_2)={\mathbb{F}}_2$, $T_{\ast }:H_0(Y;{\mathbb{F}}_2)\to H_0(\tilde{Y};{\mathbb{F}}_2)$ is zero and $P_{\ast }:H_0(\tilde{Y};{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism (the cover is connected), so the connecting map $H_1(Y;{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism; consequently $P_{\ast }=0$ in degree one and $T_{\ast }:H_1(Y;{\mathbb{F}}_2)\to H_1(\tilde{Y};{\mathbb{F}}_2)$ is an isomorphism; and $H_i(\tilde{Y};{\mathbb{F}}_2)=0$ for $i\ge 2$.

The preimage in $\tilde{Y}$ of the solid torus $N\cap Y\cong D\times S^1$ is connected, because the meridian has odd class in $\mathbb{Z}\mathrm{/}2$, and the covering restricts to the model $(x,z)\mapsto (x,z^2)$ of $D\times S^1$ onto itself; naturality of the transfer in step 3.1 shows that its upstairs meridian generates $H_1(\tilde{Y};{\mathbb{F}}_2)$: the transfer of a downstairs circle is the sum of its two lifted half-circle paths, hence the single upstairs circle. Glue a copy of $D\times D^2$ to $\tilde{Y}$ by a diffeomorphism of its boundary solid torus onto that upstairs overlap, identifying the upstairs circle coordinate $z$ with itself; its projection downstairs is $(x,z)\mapsto (x,z^2)$; after rounding corners the result $W$ is a compact connected smooth $4$-manifold, and the gluing map exhibits $W\to B^4$ as a branched double cover whose restriction off $D$ is the covering $p$ and whose local model at $D$ is $(x,z)\mapsto (x,z^2)$ in complex normal coordinates; pulling back the orientation of $B^4$ along this branched cover orients $W$, and the boundary $\mathrm{\partial }W$ is the double cover of $S^3$ branched along $\mathrm{\partial }D$.

Mayer-Vietoris [F8] over ${\mathbb{F}}_2$ for collar-thickened open pieces retracting to the displayed pieces of $W=\tilde{Y}\cup (D\times D^2)$ with overlap $D\times S^1$, whose $H_1$ maps isomorphically onto $H_1(\tilde{Y};{\mathbb{F}}_2)$ by step 4.1, while $H_i(D\times D^2;{\mathbb{F}}_2)=H_i(D\times S^1;{\mathbb{F}}_2)=0$ for $i\ge 2$ and $H_i(\tilde{Y};{\mathbb{F}}_2)=0$ for $i\ge 2$ by step 3.1, yields $H_i(W;{\mathbb{F}}_2)=0$ for every $i>0$ and, by the reduced degree-zero segment, ${\tilde{H}}_0(W)=0$, so $W$ is connected; the exact piece in degrees two and one is $0\to H_2(W)\to H_1(D\times S^1)\stackrel{\cong }{\to }{H}_1(\tilde{Y})\to H_1(W)\to 0$, so $H_2(W)=H_1(W)=0$.

The double $DW=W{\cup }_{\mathrm{\partial }W}W$ of $W$ along its collared boundary is a compact smooth $4$-manifold without boundary, hence by [F7] has the homotopy type of a CW complex whose compact image lies in a finite subcomplex $K^{\mathrm{\prime }}$; the folding retraction $r:DW\to W$ collapsing the second copy onto the first through the collar satisfies $r{\mathrm{\mid }}_W={\mathrm{i}\mathrm{d}}_W$, so composing an equivalence, its inverse and $r$ exhibits $W$ as a homotopy retract of the finite CW complex $K^{\mathrm{\prime }}$. Therefore every $H_i(W;\mathbb{Z})$ is finitely generated. For $i>0$, the universal coefficient sequence of [F6] injects $H_i(W;\mathbb{Z})\otimes {\mathbb{F}}_2$ into $H_i(W;{\mathbb{F}}_2)=0$ by step 5.1, so $H_i(W;\mathbb{Z})$ has no nontrivial free part; tensoring with $\mathbb{Q}$ gives $H_i(W;\mathbb{Q})=H_i(W;\mathbb{Z})\otimes \mathbb{Q}=0$ for $i>0$. This completes the proof; full AC is used for the bundle homotopy-invariance, CW and universal-coefficient suppliers, and supplies the countable choice needed for collaring and tubes.