The Khovanov-Rozansky factorization of the unknot

The factorization. The circle carries one mark and one arc whose two endpoint labels are both $x$, so the arc factors of [F1] specialize to $(a,x-x)=(a,0)$: the differentials are multiplication by $a$ and by $0$, the middle term carries the shift $\{-1,1\}$, and the square of the differential is $0\cdot a=0$, which is the potential $a(x-x)$ of the closed graph. Hence the complex is exactly $\mathbb{Q}[a,x]\stackrel{a}{\to }\mathbb{Q}[a,x]\{-1,1\}\stackrel{0}{\to }\mathbb{Q}[a,x]$.

Cohomology. In odd inner-factorization parity the cohomology is $\ker (0\text{\hspace{0.17em}⁣}:\mathbb{Q}[a,x]\{-1,1\}\to \mathbb{Q}[a,x])\mathrm{/}\mathrm{im}(a\text{\hspace{0.17em}⁣}:\mathbb{Q}[a,x]\to \mathbb{Q}[a,x]\{-1,1\})=\mathbb{Q}[a,x]\{-1,1\}\mathrm{/}a\mathbb{Q}[a,x]\{-1,1\}\cong \mathbb{Q}[x]\{-1,1\}$, since $\mathbb{Q}[a,x]\mathrm{/}a\mathbb{Q}[a,x]\cong \mathbb{Q}[x]$ and multiplication by $a$ is injective on $\mathbb{Q}[a,x]$. In even inner-factorization parity the cohomology is $\ker (a)\mathrm{/}\mathrm{im}(0)=0$ because multiplication by the nonzerodivisor $a$ is injective. So $H(\mathrm{\Gamma })\cong \mathbb{Q}[x]\{-1,1\}$, all of it in odd inner parity and outer cochain degree $0$, and $a$ acts as zero on it.

Euler characteristic. The graded pieces of $H(\mathrm{\Gamma })$ have $(k,l)=(-1,1+2m)$ for $m\ge 0$, each of dimension $1$ over $\mathbb{Q}$, all in cohomological degree $0$; substituting into the Euler characteristic of [F2] gives $⟨D⟩={\sum }_{m\ge 0}{t}^{-1}{q}^{1+2m}=t^{-1}q\mathrm{/}(1-q^2)$. Since $q^{-1}-q=(1-q^2)\mathrm{/}q$, this is $t^{-1}\mathrm{/}(q^{-1}-q)$; and with $\alpha =-t^{-1}{q}^{-1}$ one has $\alpha \mathrm{/}(1-q^{-2})=-t^{-1}{q}^{-1}\cdot q^2\mathrm{/}(q^2-1)=t^{-1}q\mathrm{/}(1-q^2)$, the same value. This is the unknot normalization used as the base case of the categorification theorem, and it is read off the displayed representative, finite free over $\mathbb{Q}[a,x]$ and infinite free over $\mathbb{Q}[a]$.