I am reading Lemma 2.1 of this paper (https://arxiv.org/pdf/2012.12587.pdf) and I can't see why $W$ is simply-connected. Here is the situation:

Let $K$ be a ribbon knot in $S^3$; it bounds a ribbon disk $D$ in $B^4$. Let $Y$ be the 3-manifold obtained by 0-surgery on $K$. Let $\nu D$ be a closed tubular neighborhood of $D$ in $D^4$ and let $X=B^4-\text{int} (\nu D)$. Then we have $\partial X=Y$ (Boundary of slice disk exterior is the zero surgery of slice knot). Now let $J$ be an arbitrary knot in $Y$, let $Y'$ be the 3-manifold obtained from $Y$ by an integral Dehn surgery on $J$, and suppose that $Y'$ is an integral homology $S^3$.

The lemma is claiming that $Y'$ bounds a contractible 4-manifold $W$, and in the proof $W$ is constructed by attaching a 2-handle to $X$ along $J$ (with the same framing coefficient with surgery coefficient.) In the second paragraph of the proof, there is the following statement: "$W$ must be simply-connected if the resulting 3-manifold is a homology sphere." Here the resulting 3-manifold is $Y'$ and it is assumed to be a homology sphere, so this means that $W$ is simply-connected, but I can't see why. How can we show this?

What I know is that $W$ is a homology 4-ball, and $X$ has the homology of $S^1\times B^3$. Also from van Kampen's theorem applied to $B^4=X\cup \nu D$, $\pi_1(X)$ is normally generated by the class of $\{\text{pt.}\}\times \partial D^2 \subset D\times D^2 =\nu D$. Finally, by van Kampen's theorem applied to $W=X\cup h$ ($h$ is the 2-handle), $\pi_1(W)$ is the quotient of $\pi_1(X)$ by the subgroup normally generated by the class of the knot $J$.



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