Khovanov-Rozansky homology categorifies the HOMFLYPT polynomial

Finiteness. Retain the finitely many mark variables not removed by linear exclusion; a closed circle, for example, retains its polynomial variable. After separating the universal $(a,0)$ row, [F3] gives a finite Koszul complex over this finite-variable polynomial ring. The generators have finitely many first internal degrees, and all polynomial variables have second degree $2$, so every fixed bidegree is finite dimensional. The crossing cube is finite; kernels and quotients preserve this degreewise finiteness. The localized rational-series conclusion follows from [F2], whose Hilbert-Serre argument is explicitly AC-qualified.

The exact crossing regrading. Let $e=\mathrm{\mid }D{\mathrm{\mid }}_{+}-\mathrm{\mid }D{\mathrm{\mid }}_{-}$, $s=s(D)$ and $r=(s-e)\mathrm{/}2$. In a positive published cone the ${\mathrm{\Gamma }}_0$ term has shift $\{-1\mathrm{/}2,3\mathrm{/}2\}$ in degree $-1\mathrm{/}2$ and the ${\mathrm{\Gamma }}_1$ term shift $\{-1\mathrm{/}2,-1\mathrm{/}2\}$ in degree $1\mathrm{/}2$. These are the v2 positive cone shifted by $\{-1\mathrm{/}2,-1\mathrm{/}2\}[-1\mathrm{/}2]$. In the negative cone both published terms have shift $\{1\mathrm{/}2,-3\mathrm{/}2\}$ in degrees $-1\mathrm{/}2,1\mathrm{/}2$, so it is the v2 negative cone shifted by $\{1\mathrm{/}2,1\mathrm{/}2\}[1\mathrm{/}2]$. Tensoring gives $C_{\mathrm{p}\mathrm{u}\mathrm{b}}^{\mathrm{r}\mathrm{a}\mathrm{w}}(D)=C_{\mathrm{v}2}(D)\{-e\mathrm{/}2,-e\mathrm{/}2\}[-e\mathrm{/}2]$, and therefore $C_{\mathrm{p}\mathrm{u}\mathrm{b}}(D)=C_{\mathrm{v}2}(D)\{r,r\}[r]\mathrm{.}$ The equality is read through the canonical totalization identifications; any tensor-shift differential sign is transported by those identifications.

Both stabilizations and oriented-link descent. A positive stabilization increases $s,e$ by $1$, hence leaves $r$ unchanged, and its v2 type IB complex has no shift by [F4]. A negative stabilization increases $s$ by $1$ and decreases $e$ by $1$, hence replaces $r$ by $r+1$. The type IA relation of [F4] is $C_{\mathrm{v}2}(D_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}})\simeq \mathrm{\Pi }{C}_{\mathrm{v}2}(D_{\mathrm{n}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{l}})\{1,1\}[1]$, so the stabilized v2 complex is $\mathrm{\Pi }{C}_{\mathrm{v}2}(D)\{-1,-1\}[-1]$. Its published correction $\{r+1,r+1\}[r+1]$ cancels the trigrading shift, leaving $\mathrm{\Pi }$. Since $r$ increases by one, $\lfloor r\rfloor$ also increases by one, so ${\mathrm{\Pi }}^{\lfloor r\rfloor }{C}_{\mathrm{p}\mathrm{u}\mathrm{b}}(D)$ has no residual parity reversal. Taking termwise cohomology and forgetting its parity also removes $\mathrm{\Pi }$. Inverse moves reverse these equivalences. Conjugation, markings, inverse cancellations and adjacent braid relations preserve the complex by [F5]-[F8]; distant crossings commute by the disjoint-factor signed flip of [F11]. Braid relations, inverse cancellations and conjugations keep $s,e$ and hence $\lfloor r\rfloor$ fixed. Thus both the parity-corrected factorization complex and $C{H}_{\mathrm{p}\mathrm{u}\mathrm{b}}(D)$ with parity forgotten are invariant under every Markov move. Under AC, [F11] supplies a finite Markov sequence for two ambient-isotopic closures, proving oriented-link descent.

The published weight and the two cone relations. A homogeneous class of $C$ of degree $(j,k,l)$ has degree $(j-n,k+n_1,l+n_2)$ in $C\{n_1,n_2\}[n]$. Its published weight is therefore multiplied by $(-1{)}^{n_1-n}{t}^{2n_1}{q}^{n_1+n_2}$, whenever $n_1-n$ is integral; it is this difference, rather than the sum, that applies to half-integer cones. Applying the explicit shifts of step 1.2 gives $⟨D{\sigma }_i{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}=t^{-1}{q}^{-1}⟨D{e}_i{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}-t^{-1}q⟨D{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}},$ $⟨D{\sigma }_i^{-1}{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}=t{q}^{-1}⟨D{e}_i{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}-t{q}^{-1}⟨D{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}\mathrm{.}$ The strand correction is common to all four diagrams. Also $j+k$ is integral: the common shift $\{r,r\}[r]$ changes this sum by $r-r=0$ from its integer-graded v2 value.

Skein elimination. Multiplying the positive relation by $tq$ gives $⟨D{e}_i{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}=tq⟨D{\sigma }_i{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}+q^2⟨D{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}$. Substitution in the negative relation yields $t⟨D{\sigma }_i{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}-t^{-1}⟨D{\sigma }_i^{-1}{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}=-(q-q^{-1})⟨D{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}\mathrm{.}$ To apply this relation to a general oriented skein triple, smooth all its crossings: the three diagrams have the same oriented Seifert circles, with only one crossing strip distinguished. Apply the finite reducing algorithm of [F12] to that common picture, choosing its finitely many reducing arcs away from the distinguished strip. Endpoints can be moved along the circle arcs and each reducing arc perturbed away from the distinguished point; shrink the local strip disk as needed. Each such reduction is the identical ReidemeisterII move outside that disk in all three diagrams. The resulting common height-zero picture is read as three braid closures differing at the distinguished crossing/smoothing. Link invariance from step 2.1 transports the braid relation back to the original triple. Consequently $G=(1-t^2)⟨-{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}$ has the asserted oriented-link skein relation.

All unlink values. For the crossingless $m$-strand closure, $m\ge 1$, the factorization is the tensor product of $m$ circle rows $(a,0)$. Subtract the first row from each remaining row: it becomes one $(a,0)$ row and $m-1$ zero rows. The cohomology of the first row is $\mathbb{Q}[x_1,\dots ,x_m]\{-1,1\}$; each zero row contributes $\mathbb{Q}\oplus \mathbb{Q}\{-1,1\}$. The outer v2 degree is zero, and the published correction is $\{m\mathrm{/}2,m\mathrm{/}2\}[m\mathrm{/}2]$. Thus $⟨{\mathrm{unlink}}_m{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}=\frac{-t^{m-2}{q}^m(1-t^{-2}{)}^{m-1}}{(1-q^2{)}^m}=\frac{t^{-m}(1-t^2{)}^{m-1}}{(q-q^{-1}{)}^m}\mathrm{.}$ Therefore $G({\mathrm{unlink}}_m)={\delta }^m$, where $\delta =(t^{-1}-t)\mathrm{/}(q-q^{-1})$. For $m=1$ this gives the published unknot series $t^{-1}\mathrm{/}(q-q^{-1})$ and $G(◯)=\delta$.

Finite skein uniqueness and multiplicativity. Fix an ordering and basepoint on each component of a finite regular oriented diagram, away from the crossings. Traverse components in that order, starting at their basepoints; a crossing is bad if its first encounter is on the underpassing branch. Switching the first bad crossing reduces the number of bad crossings by one without changing the others, whereas oriented smoothing reduces the total crossing number. The skein relation expresses the value at this diagram as a unit multiple of the switched-diagram value plus a multiple of the smoothed-diagram value. Induction on the lexicographic pair (crossing number, number of bad crossings) therefore terminates. A diagram with no bad crossing is descending: pull its traversed arcs successively above the remaining arcs, starting at the first component, to isotope it to disjoint unknotted circles. Their prescribed values ${\delta }^m$ determine every value. This proves uniqueness among oriented-link invariants with the skein relation and these unlink values. To check multiplicativity, apply the same induction to a diagram of $L_1$ in a ball disjoint from a fixed $L_2$; then to $L_2$. It reduces both $G(L_1\bigsqcup L_2)$ and $G(L_1)G(L_2)$ to the identical unlink products ${\delta }^{m_1+m_2}$. Thus $G$ is the multiplicative HOMFLYPT function $F$, proving (1).

The v2 normalization and its precise relation to the published one. In integer grading, $⟨C\{u,v\}[n]{⟩}_{\mathrm{v}2}=(-1{)}^n{t}^u{q}^v⟨C{⟩}_{\mathrm{v}2}$. Positive stabilization has factor $1$; negative stabilization has factor $-t^{-1}{q}^{-1}=\alpha$ by step 2.1. Its exponent $e-s+1$ decreases by $2$, whereas the positive stabilization leaves that exponent unchanged, so $\tilde{F}$ is invariant under both moves. The v2 unknot value is the circle tower $t^{-1}q\mathrm{/}(1-q^2)=\alpha \mathrm{/}(1-q^{-2})$ from [F2]. Under the variable substitution $t_{\mathrm{v}2}=-t^{2}q$ and $q_{\mathrm{v}2}=q$, choose $\sqrt{\alpha }=(tq{)}^{-1}$; then step 1.2 and the published weight give $⟨D{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}=(tq{)}^{s-e}⟨D{⟩}_{\mathrm{v}2}(-t^{2}q,q)=tq\tilde{F}(D)(-t^{2}q,q)\mathrm{.}$ Hence $F(D)=(1-t^2)tq\tilde{F}(D)(-t^{2}q,q)$: this is the explicit sense in which the v2 normalization is the same HOMFLYPT invariant after regrading. The local v2 cone relations are $⟨D{\sigma }_i⟩=⟨D{e}_i⟩-q^2⟨D⟩$ and $⟨D{\sigma }_i^{-1}⟩=q^{-2}(⟨D{e}_i⟩-⟨D⟩)$, giving $q^{-1}⟨D{\sigma }_i⟩-q⟨D{\sigma }_i^{-1}⟩=(q^{-1}-q)⟨D⟩$ by direct elimination. This records the consistent sign independently of the discrepant v2 source display. Oriented-link descent again uses AC via [F11], proving (2).

The Hecke coefficient homomorphism. The assignments in (3) obey $(q^{-1}{)}^2=q^{-2}$ and $q^{-2}\frac{(q^2-1)t^2}{q^2(1-t^2)}{t}^{-2}{q}^2=\frac{q^2-1}{q^2(1-t^2)}=\frac{(q^2-1)t^2}{q^2(1-t^2)}+1-q^{-2}\mathrm{.}$ All assigned units are units in $T$, so they define $\phi :R\to T$. It has $\phi (l)=t^{-1}$, $\phi (m)=q^{-1}-q$ and $\phi (\alpha )=\delta$ by the displayed definitions of [F10]. Thus $\psi =\phi (P)$ is an oriented-link invariant with the same skein relation and unknot value $1$. Apply its skein relation at a small kink on an unlink: both crossing choices are isotopic to that unlink, and the smoothing adds one unknotted component. Since $q-q^{-1}$ is a unit in $T$, this gives $\psi ({\mathrm{unlink}}_{m+1})=\delta \psi ({\mathrm{unlink}}_m)$, hence $\psi ({\mathrm{unlink}}_m)={\delta }^{m-1}$. The invariant $\delta \psi$ therefore has the unlink values ${\delta }^m$ and the skein relation of $F$. Finite uniqueness from step 4.1 gives $F=\delta \phi (P)$ and $⟨D{⟩}_{\mathrm{p}\mathrm{u}\mathrm{b}}=\delta \phi (P(\widehat{D}))\mathrm{/}(1-t^2)$. This comparison also uses the explicit AC assumption of the Hecke supplier [F10], proving (3).