4-manifold obtained from a ribbon disk exterior by attaching a 2-handle is simply-connected if its boundary is a homology sphere

Question

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 $\partial X=Y$ (https://math.stackexchange.com/questions/3432597/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 $J$.

Answer 1

This is false in general, I’ll prove the existence of a counterexample.

Since $D$ is a ribbon disk, $X$ has a handle structure with one $0$-handle, $n$ $1$-handles and $n-1$ $2$-handles (as described in the paper). In particular, $X$ has a 2-complex spine.

Now let $j$ be any embedded loop in $X$. Then by general position $j$ is isotopic to a knot $J$ in $Y =\partial X$ (since we may isotope it off of the 2-complex spine of $X$).

There are ribbon knots whose complements admit loops $j$ generating homology such that $\pi_1(W) \neq 0$ (where as you describe, $\pi_1(W)$ is obtained from $\pi_1(X)$ by killing an element representing $j$). For example, take any non-trivial knot $H$, and let $K=H\# \overline{H}$. Then $K$ is a ribbon knot bounding a disk $D$ (obtained by the spinning construction), and $\pi_1(X)\cong \pi_1(S^3-K)$.

But for any non-trivial knot $H$, $1/2$ surgery on it has non-trivial fundamental group (if you like by the knot complement problem and geometrization theorem). Let $j$ be a loop representing the loop of slope $1/2$ in the peripheral torus of $K$ (and transferred to a knot $J\subset Y$ as above), then attaching a handle along $J$ will give a manifold with non-trivial fundamental group and boundary a homology sphere.

Addendum: I realized in fact in this example that the knot $J$ is easy to describe. $0$-framed surgery on $H\#\overline{H}$ is homeomorphic to the splice of $S^3-\mathcal{N}(H)$ and $S^3-\mathcal{N}(\overline{H})$, obtained by gluing the peripheral tori so that the meridians and longitudes are identified. Take $J$ to be the knot of slope $1/2$ in the torus.