0NGC
Lemma 4.6 . Let W W B 3 ⊂ ∂ W B^{3}\subset\partial W L 1 L_{1} B 3 B^{3} T 1 T_{1} ∂ T 1 = P p \partial T_{1}=P_{p} L 1 = R ∪ P p T 1 L_{1}=R\cup_{P_{p}}T_{1} T 2 T_{2} L 2 = R ∪ P p T 2 L_{2}=R\cup_{P_{p}}T_{2} gluing
map
𝒮 0 N ( W ; L 1 , 𝕜 ) ⊗ 𝒮 0 N ( B 4 ; T 2 ∪ P p T 1 ¯ , 𝕜 ) { p ( N − 1 ) } → 𝒮 0 N ( W ; L 2 , 𝕜 ) . \mathcal{S}_{0}^{N}(W;L_{1},\mathbbm{k})\otimes\mathcal{S}_{0}^{N}(B^{4};T_{2}\cup_{P_{p}}\overline{T_{1}},\mathbbm{k})\{p(N-1)\}\to\mathcal{S}_{0}^{N}(W;L_{2},\mathbbm{k}).
Moreover, these gluing maps are compatible with composition in 𝒮 0 N ( B 3 , P p , 𝕜 ) \mathcal{S}_{0}^{N}(B^{3};P_{p},\mathbbm{k})
𝒮 0 N ( W ; L 1 , 𝕜 ) ⊗ 𝒮 0 N ( B 4 ; T 2 ∪ P p T 1 ¯ , 𝕜 ) ⊗ 𝒮 0 N ( B 4 ; T 3 ∪ P p T 2 ¯ , 𝕜 ) { 2 p ( N − 1 ) } {\lx@inpgf@ignorespaces\mathcal{S}_{0}^{N}(W;L_{1},\mathbbm{k})\otimes\mathcal{S}_{0}^{N}(B^{4};T_{2}\cup_{P_{p}}\overline{T_{1}},\mathbbm{k})\otimes\mathcal{S}_{0}^{N}(B^{4};T_{3}\cup_{P_{p}}\overline{T_{2}},\mathbbm{k})\{2p(N-1)\}} 𝒮 0 N ( W ; L 2 , 𝕜 ) ⊗ 𝒮 0 N ( B 4 ; T 3 ∪ P p T 2 ¯ , 𝕜 ) { p ( N − 1 ) } {\lx@inpgf@ignorespaces\mathcal{S}_{0}^{N}(W;L_{2},\mathbbm{k})\otimes\mathcal{S}_{0}^{N}(B^{4};T_{3}\cup_{P_{p}}\overline{T_{2}},\mathbbm{k})\{p(N-1)\}} 𝒮 0 N ( W ; L 1 , 𝕜 ) ⊗ 𝒮 0 N ( B 4 ; T 3 ∪ P p T 1 ¯ , 𝕜 ) { p ( N − 1 ) } {\lx@inpgf@ignorespaces\mathcal{S}_{0}^{N}(W;L_{1},\mathbbm{k})\otimes\mathcal{S}_{0}^{N}(B^{4};T_{3}\cup_{P_{p}}\overline{T_{1}},\mathbbm{k})\{p(N-1)\}} 𝒮 0 N ( W , L 3 , 𝕜 ) {\lx@inpgf@ignorespaces\mathcal{S}_{0}^{N}(W;L_{3},\mathbbm{k})}
0NGD
Proof. Straightforward on the level of lasagna fillings. The map descends to the
quotient since skein relations are local.
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