ScalingStacks

0N4J

Proof. Proposition 3.7 recognizes factorization homology as a symmetric monoidal left Kan extensions, thereby implementing the left adjoint in an adjunction

i!:𝖠𝗅𝗀𝒟​𝗂𝗌𝗄𝗇𝖡(𝒱)⇄𝖥𝗎𝗇⊗(ℳ​𝖿𝗅𝖽nB,𝒱):i∗.i_{!}\colon\Alg_{\disk_{n}^{B}}(\mathcal{V})\rightleftarrows\Fun^{\otimes}\bigl(\mfld_{n}^{B},\mathcal{V}\bigr)\colon i^{\ast}~.

The unit of this adjunction is an equivalence because 𝒟​𝗂𝗌𝗄𝗇𝖡→ℳ​𝖿𝗅𝖽nB\disk_{n}^{B}\rightarrow\mfld_{n}^{B} is fully faithful, and the Kan extension along a fully faithful functor restricts as the original functor. The counit of this adjunction evaluates on a symmetric monoidal functor ℱ\mathcal{F} as a morphism ∫A→ℱ\int\!A\rightarrow\mathcal{F}, where A=ℱ|ℝnA=\mathcal{F}_{|\mathbb{R}^{n}} is the 𝒟​𝗂𝗌𝗄𝗇𝖡\disk^{B}_{n}-algebra defined by the values of ℱ\mathcal{F} on disjoint unions of BB-framed Euclidean nn-spaces. It remains to verify that this counit is an equivalence.

Since both ℱ\mathcal{F} and ∫A\int\!A are symmetric monoidal and agree on ℝn\mathbb{R}^{n}, the map ∫MA→ℱ⁡(M)\int_{M}A\rightarrow\mathcal{F}(M) is an equivalence for MM isomorphic to a disjoint union of Euclidean spaces, M≅⨆IℝnM\cong\bigsqcup_{I}\mathbb{R}^{n}. Using induction, we will now see that the values of ℱ\mathcal{F} and ∫A\int\!A agree on thickened spheres Sk×ℝn−kS^{k}\times\mathbb{R}^{n-k}, the base case of k=0k=0 just having been shown. In the inductive step, assume the result for Si−1×ℝn−i+1S^{i-1}\times\mathbb{R}^{n-i+1}. Choose a standard collar-gluing Si→𝑓[−1,1]S^{i}\xrightarrow{f}[-1,1] with Si−1=f−1​(0)⊂SiS^{i-1}=f^{-1}(0)\subset S^{i} an equator. There results a collar-gluing of Si×ℝn−iS^{i}\times\mathbb{R}^{n-i}. For ℱ\mathcal{F} a homology theory, we obtain the equivalence ∫Si×ℝn−iA≃ℱ⁡(Si×ℝn−i)\int_{S^{i}\times\mathbb{R}^{n-i}}A\simeq\mathcal{F}(S^{i}\times\mathbb{R}^{n-i}) via the intermediate equvialences

∫Si×ℝn−i​A≃∫ℝ−1i×ℝn−i​A​⨂∫Si−1×ℝn−i+1​A​∫ℝ+1i×ℝn−i​A≃ℱ⁡(ℝ−1i×ℝn−i)​⨂ℱ⁡(Si−1×ℝn−i+1)​ℱ​(ℝ+1i×ℝn−i)≃ℱ⁡(Si×ℝj)\underset{S^{i}\times\mathbb{R}^{n-i}}{\int}\!A\simeq\underset{\mathbb{R}_{-1}^{i}\times\mathbb{R}^{n-i}}{\int}\!A\underset{\underset{S^{i-1}\times\mathbb{R}^{n-i+1}}{\int}\negthinspace A}{\bigotimes}\ \underset{\mathbb{R}_{+1}^{i}\times\mathbb{R}^{n-i}}{\int}\!A\ \simeq\ \mathcal{F}(\mathbb{R}_{-1}^{i}\times\mathbb{R}^{n-i})\underset{\mathcal{F}(S^{i-1}\times\mathbb{R}^{n-i+1})}{\bigotimes}\mathcal{F}(\mathbb{R}_{+1}^{i}\times\mathbb{R}^{n-i})\simeq\mathcal{F}(S^{i}\times\mathbb{R}^{j})

where the first equivalence is by the ⊗\otimes-excision property of factorization homology, the last equivalence is by the assumption that ℱ\mathcal{F} is a homology theory, and the middle equivalence is by induction.

We now restrict to the case of BB-framed nn-manifolds where nn is not equal to 4. By the handlebody theory for topological manifolds ([KS] for n>5n>5, [Qu] for n=5n=5, and [Mo] for n=3n=3) all such manifolds admits a handle decomposition. We now prove the result outside dimension 4 by induction on the handle decomposition. The base case is assured. To verify the inductive step, let MM be obtained from M0M_{0} by adding a handle of index q+1q+1. Therefore MM can be expressed as a collar-gluing M≅M0​⋃Sq×ℝn−q​ℝnM\cong M_{0}\underset{S^{q}\times\mathbb{R}^{n-q}}{\bigcup}\mathbb{R}^{n}, where ℝn\mathbb{R}^{n} is an open neighborhood of the (q+1)(q+1)-handle in MM. The values ℱ\mathcal{F} and ∫A\int\!A agree on the three constituent submanifolds of MM, and they both satisfy ⊗\otimes-excision, so the values ℱ⁡(M)≃∫MA\mathcal{F}(M)\simeq\int_{M}A are equivalent.

This leaves the case of topological 4-manifolds, which do not admit handle decompositions in general. Because both ℱ\mathcal{F} and ∫A\int_{\!}A are symmetric monoidal, we can reduce to the case that MM is connected. Now, any connected topological 4-manifold MM admits a smooth structure on the complement M∖{x}M\smallsetminus\{x\} of a point x∈Mx\in M, [Qu]. Consequently, M∖{x}M\smallsetminus\{x\} admits a handle decomposition, which can be constructed from any Morse function on M∖{x}M\smallsetminus\{x\}, and the preceding argument thereby implies the equivalence ℱ⁡(M∖{x})≃∫M∖{x}A\mathcal{F}(M\smallsetminus\{x\})\simeq\int_{M\smallsetminus\{x\}}A. Applying the ⊗\otimes-excision property to the collar-gluing M∖{x}​⋃Sn−1×ℝ​ℝn≅MM\smallsetminus\{x\}\underset{S^{n-1}\times\mathbb{R}}{\bigcup}\mathbb{R}^{n}\cong M, since ℱ\mathcal{F} and ∫A\int\!A agree on the constituent submanifolds, we obtain the equivalence ℱ⁡(M)≃∫MA\mathcal{F}(M)\simeq\int_{M}A. Therefore every homology theory ℱ\mathcal{F} for nn-manifolds is equivalent to factorization homology with coefficients in ℱ⁡(ℝn)\mathcal{F}(\mathbb{R}^{n}).

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Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

David Ayala, John Francis

Original source: arXiv:1206.5522v6