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Start with a disk in the zero-handle and attach two untwisted bands passing through the two one-handles. The four passages around its boundary occur in the order $x,y,x^{-1},y^{-1}$, giving an embedded once-punctured torus whose boundary is the attaching circle of the two-handle. Push its interior into the existing four-dimensional handlebody, keeping its boundary on the attaching locus, and cap it with the core disk of a handle. The matching product framing makes the union smooth after rounding its corner. This gives an embedded closed torus $T\subset W$.

The torus uses the core of the two-handle exactly once. Equivalently, it intersects that handle's relative co-core disk transversely in one point. Its cellular two-chain is therefore the generator of $C_2=\mathbb{Z}$, which is a cycle and is not a boundary because there are no three-cells. ThusThis is the zero section in the commutator handlebody of the torus $T^2\times D^2$.

Let $x,y$ be the generators associated with the two pairs of one-handle feet. Following the attaching circle across the identifications marked $a,b$ and $c,d$ gives the word $xy{x}^{-1}{y}^{-1}$, up to reversal and changing generator orientations. Thus the van Kampen theorem givesThe handle decomposition has cellular groups $C_0=\mathbb{Z}$, $C_1={\mathbb{Z}}^2$, $C_2=\mathbb{Z}$, with no higher groups. The boundary of the two-cell is the abelianized attaching word, which is zero. HenceThe zero framing is the product framing of the commutator handlebody of the torus, so the represented manifold is $T^2\times D^2$ and its intersection form is $(0)$. In particular its boundary is $T^2\times S^1=T^3$.

The boundary homology can also be obtained without using the product identification. By Poincare-Lefschetz duality, $H_3(W,\partial W)=H^1(W)={\mathbb{Z}}^2$ and $H_2(W,\partial W)=H^2(W)=\mathbb{Z}$. The map $H_2(W)\to H_2(W,\partial W)$ is its zero intersection form. The long exact sequence of a pair then gives split sequencesThe boundary is connected and oriented, so its zeroth and third homology are both $\mathbb{Z}$. Therefore