An elementary Seifert enlargement preserves the algebraic concordance class

For each displayed enlargement, its skew part has block form $E_i-E_i^T=\left(\begin{array}{ccc}J& u& 0\\ -u^T& 0& s_i\\ 0^T& -s_i& 0\end{array}\right),$ where $J=V-V^T$, $u=x-y$, $s_0=-1$, and $s_1=1$. If $V$ is knot-derived, [A1] is the choice hypothesis for its Seifert-pairing convention; the following lattice calculation itself is choice-free. Since $J$ is unimodular, choose the integral vector $c=-J^{-1}u$ and replace the penultimate basis vector by itself plus the old-basis combination $c$. This integral basis change removes the $u$ block and leaves $J\oplus \left(\begin{array}{cc}0& s_i\\ -s_i& 0\end{array}\right)$. Both blocks are unimodular, so each enlargement is an integral Seifert lattice.

In $E_i\oplus (-V)$, take the subgroup generated by $(e_j;e_j)$ for the old basis vectors $e_1,\dots ,e_{2g}$ and by $(b;0)$, where $b$ is the last basis vector of the enlarged lattice. Its rank is $2g+1$, half the total rank. The two displayed enlargement matrices have zero old-to-$b$, $b$-to-old, and $b$-to-$b$ entries. The $V$ and $-V$ blocks cancel on the diagonal generators, while all pairings involving $(b;0)$ vanish. Therefore the form vanishes on this subgroup.

The subgroup in step 1.2 is a direct summand: its generators, together with $(a;0)$ and $(0;e_j)$ for every old basis vector, form an integral basis because subtracting $(0;e_j)$ from $(e_j;e_j)$ gives $(e_j;0)$. It is therefore a metabolizer. By [F2], $[E_i]+[-V]=0$, and [F3] gives $[E_i]=[V]$ in $G_{\mathbb{Z}}$.