Fickle's argument for Mazur manifolds

Question

In the page 482 of his article, 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?

Answer 1

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.