ScalingStacks

The process of gluing a lasagna filling works as follows. Let $F^{\prime}$ be a lasagna filling of $B^{4}$ with boundary link $L_{T}$ as above; i.e., inside $S^{3}=\partial B^{4}$ we have $m$ pairs of embedded 3-balls $B_{i}\cup\overline{B_{i}}$, such that $L_{T}\cap B_{i}=T_{i}$ and $L_{T}\cap\overline{B_{i}}=\overline{T}_{i}$ for $1\leq i\leq m$. Denote the numbers of boundary points by $2p_{i}:=|\partial T_{i}|$. Now we attach $m$ 1-handles with core-parallel lasagna sheets $I\times T_{i}\subset I\times B^{3}$ along the $B_{i}\cup\overline{B_{i}}\cong S^{0}\times B^{3}$ to obtain a lasagna filling of $W_{1}$ with boundary $L$. Since the relations in $\mathcal{S}_{0}^{N}$ are local, this induces a map:

(17)

$$\mathrm{glue}_{L_{T}}\colon\mathcal{S}_{0}^{N}(B^{4};L_{T},\mathbbm{k})\{(\textstyle\sum_{i}p_{i})(N-1)\}\to\mathcal{S}_{0}^{N}(W_{1};L,\mathbbm{k})$$

The grading shift is there to compensate the change in Euler characteristic of the surfaces in lasagna fillings upon gluing.