I am studying Khovanov homology from five lectures on Khovanov homology
and i want to try to show Khovanov homology is invariant under first reidemester move but i can not understand how we can write
$C^{î,*}(D)=C^{î,*}(D_0) \oplus C^{î-1,*}(D_{1})$.
where $D,D_0,D_1$ as in the picture below
For example again in this paper it takes D as hopf link and i try to write it as a sum but i can not because if we take i=-2 then $C^{-2,*}(D_0)=0 $ and $C^{-3,*}(D_{1})=0$ because they only have 1 negative cross , but if i take $C^{î,*}(D)=C^{î+1,*}(D_0) \oplus C^{î,*}(D_{1})$ it is true. Can you say where is my mistake ?