This is Example 6.47 in Saveliev's book *Invariants for homology $3$-spheres*:



> 
> Let us consider a two-component link $\mathcal L = L\_1 \cup L\_2$ in
> $S^3$ such that $\mathrm{lk}(L\_1,L\_2) = \pm 1$ and the component $L\_1$ is an
> unknot. Let $\Sigma\_p (\mathcal L)$ be the integral homology sphere
> obtained by surgery on the link $\mathcal L$ with $L\_1$ framed by $0$
> and $L\_2$ framed by $p$. The manifolds $\Sigma\_p (\mathcal L)$ are
> referred as *Mazur homology spheres* because they generalize the
> original Mazur's example of a homology sphere bounding a smooth
> acyclic $4$-manifold.
> 
> 
> 


1. How we generalize Mazur's construction? Originally, we attach a $2$-handle $B^2 \times B^2$ to $S^1 \times B^3$ along a knot on the boundary $\partial S^1 \times B^3 = S^1 \times S^2$.
2. Do we attach two $2$-handles to obtain an acyclic $4$-manifold?



Mazur homology spheres
Saveliev writes:



> 
> they generalize the original Mazur's *example*
> 
> 
> 


and not "exampleS". The original example he's referring to is a single contractible 4-manifold with a boundary that's not the 3-sphere. Then there is a wealth of Mazur (4-)*manifolds* which are obtained, as you say, by attaching a 2-handle to $S^1\times B^3$. Saveliev talks about Mazur *homology (3-)spheres*, which are the boundaries of such objects. Now that we have cleared this, I'll get to your questions.


1. To me, the key point is that we can skip the 1-handle attachment in the construction of a Mazur manifold, and work directly in $S^1\times S^2$ (which we can then fill in uniquely with $S^1 \times B^3$, thanks to Laudenbach and Poenaru). In this sense, to get a Mazur manifold, we just need a knot in $S^1\times S^2$ that generates the homology but such that doing surgery along it doesn't produce $S^3$ (essentially, thanks to Gabai, we just need a knot that's not isotopic to a fibre). In Saveliev's example, once we do the 0-surgery along $L\_1$ we are in business, since $L\_1$ is unknotted and doing 0-surgery along it this yields $S^1\times S^2$. (Then we need to make sure that $L\_2$ does not become a fibre, but this might be harder to do.)
2. No, attaching the two 2-handles does not yield a homology ball. For instance, because the Euler characteristic is 3 instead of 1. What you need to do is surger out the 2-sphere $S$ of self-intersection 0 that you get by capping off a Seifert surface for $L\_1$ and replace it with an $S^1\times D^3$ (this corresponds to attaching a 5-dimensional 3-handle to $S$).


