Acyclicity makes the simply connected middle-handle matrix unimodular

By the concentration lemma the presentation has handles only in the two adjacent indices $k,k+1$, so by [F1] the relative handle chain complex is $0\to C_{k+1}\stackrel{{\mathrm{\partial }}_{k+1}}{\to }{C}_k\to 0$ with $C_k,C_{k+1}$ free abelian and bases given by the handle cores, and the row-coordinate matrix of ${\mathrm{\partial }}_{k+1}$ in these bases is $M$ (the column-coordinate matrix is $M^T$).

Vanishing of the relative homology [F3] is exactly acyclicity of this complex: $H_{k+1}(C_{\bullet })=\ker {\mathrm{\partial }}_{k+1}=0$ and $H_k(C_{\bullet })=C_k\mathrm{/}\mathrm{im}{\mathrm{\partial }}_{k+1}=0$. Hence ${\mathrm{\partial }}_{k+1}$ is injective and surjective, that is, a bijection of abelian groups.

A bijection between the finitely generated free abelian groups $C_{k+1}$ and $C_k$ forces their ranks to be equal, and the row-coordinate matrix of ${\mathrm{\partial }}_{k+1}$ is a square integer matrix invertible over $\mathbb{Z}$ (its inverse is the matrix of the inverse isomorphism). For $r>0$, view this matrix and its integer inverse over $\mathbb{Q}$. The field determinant identity in [F4] gives $\det M\det M^{-1}=1$; both determinants are integers by the Leibniz formula, so each is $\pm 1$. For $r=0$, set the empty determinant equal to $1$ by the empty-product convention; the empty matrix is its own inverse. Thus $M\in {\mathrm{GL}}_r(\mathbb{Z})$.

By the definition of the middle-handle matrix [F1] this invertible matrix is exactly $M$, so $M$ is unimodular with $\det M=\pm 1$. This is Ranicki's unimodularity proposition specialised to the trivial fundamental group, where the group ring reduces to $\mathbb{Z}$.