In the page 482 of his [article](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.449.3854&rep=rep1&type=pdf), Fickle wrote the following argument:


Let $Y$ be a homology $3$-sphere. Next


* Add a $2$-handle to $Y \times [0,1]$ and produce a $4$-manifold with
boundary $S^1 \times S^2$,
* Cap off the boundary with a $3$-handle and a $4$-manifold,
* Turn the resulting $4$-manifold upside down.


Then we may conclude that $Y$ bounds a Mazur manifold. Why this is true?



Fickle's argument for Mazur manifolds
Call the manifold built in this way $W$. If you turn the handlebody for $W$ relative to $Y$ upside down, it becomes a handlebody with a single handle in indices $0$, $1$, and $2$. This is almost the same as saying it's a Mazur manifold; you need to check that it's contractible. Because $\pi\_1(W)$ is cyclic (it's got just the one generator from the $1$-handle), this is the same as saying that $H\_1(W) = 0$. This in turn follows from the fact that $Y$ is a homology sphere, as follows.


Suppose the $2$-handle (in the upside-down handlebody) goes over the 1-handle $n$ times, and has framing $k$. Then $Y$ has a surgery description where the linking matrix is
$$
\pmatrix{0 & n\\n & k}.
$$
In order for $Y$ to be a homology sphere, this must have determinant $1$, so $n= \pm 1$ and hence $H\_1(W) = 0$.


By the way, this argument predates Fickle's paper. See for instance Casson-Harer, *Some homology lens spaces which bound rational homology balls*, Pacific J. Math. 96 (1981), no. 1, 23–36.


