= Solution
There are no <one-handles>, so attaching the three <two-handles> to a four-ball cannot create a <fundamental group> generator: \b[$\pi_1(W)=1$]. The handle <chain complex> has $C_2=\mathbb Z^3$, $C_1=0$ and $C_0=\mathbb Z$, giving
$$
\boxed{H_0(W)=\mathbb Z,\qquad H_2(W)=\mathbb Z^3,\qquad H_j(W)=0\ (j\ne0,2).}
$$
To calculate the boundary, orient the two long components down their central strands, and orient the small component clockwise. The six mutual crossings of the long components have negative sign, giving <linking number> $-3$. Each long component has two positive crossings with the small component, giving <linking number> $1$. The <surgery linking matrix>, in the order of framings $4,3,2$, is consequently
$$
Q=\begin{pmatrix}4&-3&1\\-3&3&1\\1&1&2\end{pmatrix},\qquad\det Q=-7.
$$
Changing a component orientation changes the corresponding row and column signs, and does not change the resulting <homology (mathematics)>. The <long exact sequence of a pair> and <Poincare-Lefschetz duality> identify $H_2(\partial W)=\ker Q$ and $H_1(\partial W)=\operatorname{coker}Q$. The determinant is nonzero, so the kernel vanishes. The gcd of matrix entries is one; the two-by-two minors include $-6$ and $7$, so their gcd is also one. The <Smith normal form> therefore has diagonal $1,1,7$, and
$$
\boxed{H_j(\partial W;\mathbb Z)=\begin{cases}\mathbb Z,&j=0,3,\\\mathbb Z/7,&j=1,\\0,&j=2\text{ or }j\ge4.\end{cases}}
$$
In particular the sign of the braid crossings matters: replacing $-3$ by $+3$ while leaving both meridian linkings positive would represent a different link and would give the wrong determinant.
Back to article page
