Heegaard invariance and well-definedness of the correction term

The minus chain complex is finite free over ${\mathbb{F}}_2[U]$, with generators $[\mathbf{x},-1]$ for the finitely many intersection points in $\mathfrak{t}$. Its homology is finitely generated and its localization is the infinity homology, since localization is exact and the localized differentials are identical. Thus the graded image $L$ of $H{F}^{-}\to R\rho$ is nonzero. Its degrees are bounded above by the largest degree of a minus-chain generator. The set of integers $j$ with $U^j\rho \in L$ is consequently nonempty, bounded below, and closed under addition of nonnegative integers. If $a$ is its minimum, that set is exactly $\{a,a+1,\dots \}$. Every homogeneous element of the Laurent tower is a monomial multiple of $\rho$, and the image is graded; hence $L=U^a{\mathbb{F}}_2[U]\rho$.

The long exact sequence of the actual short exact sequence $C{F}^{-}\to C{F}^{\mathrm{\infty }}\to C{F}^{+}$ identifies the image in plus homology, with its grading, as $R\rho \mathrm{/}{U}^a{\mathbb{F}}_2[U]\rho$. Its nonzero homogeneous elements have exponents $j<a$, so their least degree is $\mathrm{deg}\rho -2(a-1)$, attained at $U^{a-1}\rho$. Sending the standard tower element $U^k$ to $U^{k+a-1}\rho$ identifies the image with $R\mathrm{/}U{\mathbb{F}}_2[U]$ shifted by this least degree. Multiplying any individual element sufficiently many times by $U$ kills it, while decreasing exponents supplies a preimage under any power of $U$. This proves all the tower assertions.

The pointed-move maps preserve absolute grading and the minus/infinity/plus exact sequence, so they carry $L$ onto the corresponding image lattice and induce grading-preserving tower isomorphisms. Therefore the least degree is independent of the diagram and auxiliary data. Choosing $U^b\rho$ instead changes $a$ to $a-b$ and $\mathrm{deg}\rho$ to $\mathrm{deg}\rho -2b$, leaving that degree unchanged. In the genus-zero diagram of $S^3$ there is one generator $[{\mathbf{x}}_0,0]$ of degree zero, no nonconstant disks, and minus lattice $U{\mathbb{F}}_2[U][{\mathbf{x}}_0,0]$. Here $a=1$, so the least grading is zero.