Reduced Khovanov cohomology of a knot

If an edge does not meet the basepoint circle, its map is the identity on the basepoint factor and hence commutes with $X_p$. If the edge merges the basepoint circle with another circle, associativity gives $m(Xa\otimes b)=Xm(a\otimes b)$. If it splits the basepoint circle, the Frobenius identity gives $\mathrm{\Delta }(Xa)=(X\otimes 1)\mathrm{\Delta }(a)=(1\otimes X)\mathrm{\Delta }(a)$; on $a=1,X$ these are respectively $X\otimes X$ and $0$ on all three sides. Thus every edge map commutes with $X_p$, so $\mathrm{im}(X_p)$ is a subcomplex.

By [F3], the unreduced differential already squares to zero, and by step 1.1 it preserves $\mathrm{im}(X_p)$; thus the induced quotient differential also squares to zero. In each state the based factor is $\mathcal{A}=\mathbb{Z}\{1,X\}$, and $X_p$ has image $\mathbb{Z}\{X\}$. Therefore the quotient factor is $\mathbb{Z}\{1\}$, with raw quantum degree $+1$; subtracting one gives reduced degree $0$.

After cutting the based circle, every state has one distinguished interval factor $M$. The quotient identification sends the surviving based label $1$ to $1\in M$. For a merge into that interval, $m(1\otimes 1)=1$ and $m(X\otimes 1)=0$ in the quotient, exactly the displayed $m_M$; for a split, $\mathrm{\Delta }(1)=1\otimes X+X\otimes 1$ and the term with $X$ on the distinguished interval is zero in $M$, leaving $X\otimes 1$, exactly ${\mathrm{\Delta }}_M(1)$. Edges away from the interval retain $m$ and $\mathrm{\Delta }$. Hence this statewise identification commutes with every edge and gives the stated based tangle cube isomorphism.