Reduced Khovanov homology over F2 is independent of the chosen link component

On $\mathcal{A}={\mathbb{F}}_2\{1,X\}$ define $N(1)=0$ and $N(X)=1$, and on a state tensor define $N$ as the sum of this operation over its circle factors. For multiplication, direct evaluation on $(1,1),(1,X),(X,1),(X,X)$ gives $Nm=m(N\otimes \mathrm{i}\mathrm{d}+\mathrm{i}\mathrm{d}\otimes N)$. For comultiplication, evaluation at $1$ gives $(N\otimes \mathrm{i}\mathrm{d}+\mathrm{i}\mathrm{d}\otimes N)\mathrm{\Delta }(1)=0=\mathrm{\Delta }N(1)$, and evaluation at $X$ gives $(N\otimes \mathrm{i}\mathrm{d}+\mathrm{i}\mathrm{d}\otimes N)\mathrm{\Delta }(X)=1\otimes X+X\otimes 1=\mathrm{\Delta }N(X)$. Thus $N$ commutes with each saddle map and hence with $d$. Also $N^2=0$: applying $N$ twice to one circle gives zero, and every pair of distinct circles occurs twice, hence cancels over ${\mathbb{F}}_2$.

For a basepoint $p$, multiplication $X_p$ is a chain map and $X_p^2=0$. The local operators on the circle containing $p$ satisfy $N{X}_p+X_{p}N=\mathrm{i}\mathrm{d}$ on that factor; terms of $N$ on all other circles commute with $X_p$ and occur twice in the sum, so the identity holds on every state tensor.

Put $P=X_{p}N$ and $Q=N{X}_p$. The identity in step 2.1 and $X_p^2=N^2=0$ imply $P+Q=\mathrm{i}\mathrm{d}$, $P^2=P$, $Q^2=Q$, and $PQ=QP=0$. Both maps commute with $d$ by step 1.1. Moreover $\mathrm{im}P=\mathrm{im}{X}_p$: one inclusion follows from $P=X_{p}N$, and $P{X}_p=X_{p}N{X}_p=X_p$ gives the reverse inclusion. Hence the chain module splits as $\mathrm{im}{X}_p\oplus \mathrm{im}Q$.

The quotient map identifies $C\mathrm{/}\mathrm{im}{X}_p$ with $\mathrm{im}Q$. The restriction $X_p:\mathrm{im}Q\to \mathrm{im}P$ is an isomorphism with inverse $N$, since $N{X}_p$ is the identity on $\mathrm{im}Q$ and $X_{p}N$ is the identity on $\mathrm{im}P$. The map $X_p$ lowers raw quantum degree by two, while [F3] lowers the quotient grading by one. Thus, for each cohomological degree $i$, if $h_{i,j}$ is the dimension of unreduced cohomology in raw degree $j$ and $r_{i,j}(p)$ the dimension of reduced cohomology in reduced degree $j$, then $h_{i,j}=r_{i,j-1}(p)+r_{i,j+1}(p)$.

The diagram has finitely many cube generators, so the sequences $j\mapsto h_{i,j}$ and $j\mapsto r_{i,j}(p)$ have finite support. Rewrite step 4.1 as $r_{i,k}(p)=h_{i,k+1}-r_{i,k+2}(p)$. For each parity of $k$, start above the largest supported reduced degree, where the latter term is zero, and recurse downward by two; this uniquely determines all reduced dimensions from the unreduced dimensions. Therefore $r_{i,j}(p)=r_{i,j}(q)$ in every bidegree. Finite-dimensional vector spaces with equal dimensions are isomorphic in each bidegree; only finitely many bidegrees occur, so these isomorphisms assemble without an infinite choice principle.