ScalingStacks

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

The unit of this adjunction is an equivalence because $\disk_{n}^{B}\rightarrow\mfld_{n}^{B}$ 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 $\mathcal{F}$ as a morphism $\int\!A\rightarrow\mathcal{F}$, where $A=\mathcal{F}_{|\mathbb{R}^{n}}$ is the $\disk^{B}_{n}$-algebra defined by the values of $\mathcal{F}$ on disjoint unions of $B$-framed Euclidean $n$-spaces. It remains to verify that this counit is an equivalence.

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

We now restrict to the case of $B$-framed $n$-manifolds where $n$ is not equal to 4. By the handlebody theory for topological manifolds ([KS] for $n>5$, [Qu] for $n=5$, and [Mo] for $n=3$) 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 $M$ be obtained from $M_{0}$ by adding a handle of index $q+1$. Therefore $M$ can be expressed as a collar-gluing $M\cong M_{0}\underset{S^{q}\times\mathbb{R}^{n-q}}{\bigcup}\mathbb{R}^{n}$, where $\mathbb{R}^{n}$ is an open neighborhood of the $(q+1)$-handle in $M$. The values $\mathcal{F}$ and $\int\!A$ agree on the three constituent submanifolds of $M$, and they both satisfy $\otimes$-excision, so the values $\mathcal{F}(M)\simeq\int_{M}A$ are equivalent.

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

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