ScalingStacks

A resolution of a diagram $D$ is always a collection of simple disjoint curves on the plane and is thus a $1$-manifold embedded in the plane. Now the functor $F$ from $(1+1)$ cobordisms to $R$-modules (see Section 2.3) comes into play. To a union of $k$ circles it assigns the $k$-th tensor power of $A.$ The functor $F$, applied to the diagram $D(\mathcal{L}),$ considered as a one-dimensional manifold, produces a graded $R$-module $A^{\otimes k}$ where $k$ is the number of components of $D(\mathcal{L}).$ We raise the grading of $A^{\otimes k}$ by $|\mathcal{L}|,$ the cardinality of $\mathcal{L},$ and assign the $R$-module $F(D(a))\{-|\mathcal{L}|\}$ to the vertex $V_{D}(\mathcal{L})$ of the cube $V_{D}:$

$$V_{D}(\mathcal{L})=F(D(\mathcal{L}))\{-|\mathcal{L}|\}$$

(40)

(Recall from Section 3.1 that the automorphism $\{1\}$ of the category $R{\mbox{-mod}_{0}}$ lowers the grading by $1$.) Let us now define maps between vertices of $V_{D}.$ Choose $(\mathcal{L},a)\in r(\mathcal{I}).$ We want to have a map

$$\xi_{a}^{V_{D}}(\mathcal{L}):V_{D}(\mathcal{L})\longrightarrow V_{D}(\mathcal{L}a).$$

(41)

The diagrams $D(\mathcal{L})$ and $D(\mathcal{L}a)$ differ only in the neighborhood of the double point $a$ of $D,$ as an example below (for $n=1$, so that $D$ has one double point, $\mathcal{I}=\{a\}$) demonstrates ($D$ is the leftmost diagram, $D(\emptyset)$ is the diagram in the center and $D(a)$ is depicted on the right):