85 According to the gluing theorem for TQFTs-via-fields, gluing along $B^3 \subset \bd B^4$ |
85 According to the gluing theorem for TQFTs-via-fields, gluing along $B^3 \subset \bd B^4$ |
86 corresponds to taking a coend (self tensor product) over the cylinder category |
86 corresponds to taking a coend (self tensor product) over the cylinder category |
87 associated to $B^3$ (with appropriate boundary conditions). |
87 associated to $B^3$ (with appropriate boundary conditions). |
88 The coend is not an exact functor, so the exactness of the triangle breaks. |
88 The coend is not an exact functor, so the exactness of the triangle breaks. |
89 \item The obvious solution to this problem is to replace the coend with its derived counterpart. |
89 \item The obvious solution to this problem is to replace the coend with its derived counterpart. |
90 This presumably works fine for $S^1\times B^3$ (the answer being to Hochschild homology |
90 This presumably works fine for $S^1\times B^3$ (the answer being the Hochschild homology |
91 of an appropriate bimodule), but for more complicated 4-manifolds this leaves much to be desired. |
91 of an appropriate bimodule), but for more complicated 4-manifolds this leaves much to be desired. |
92 If we build our manifold up via a handle decomposition, the computation |
92 If we build our manifold up via a handle decomposition, the computation |
93 would be a sequence of derived coends. |
93 would be a sequence of derived coends. |
94 A different handle decomposition of the same manifold would yield a different |
94 A different handle decomposition of the same manifold would yield a different |
95 sequence of derived coends. |
95 sequence of derived coends. |