ScalingStacks

Proof. Presentability of $\mathcal{V}$ grants the existence of the values $\int_{M}A$. Lemma 4.3.2.13 of [Lu1] states that these expressions define a functor, as indicated. Proposition 4.3.3.7 of [Lu1] states that this functor satisfies the universal property of being a left adjoint to the restriction $\iota^{\ast}$. We thus have the solid diagram among $\infty$-categories

in which the downward functors are restriction to underlying $\infty$-categories, these downward functors are fully faithful. It remains to explain how the $\otimes$-presentability of $\mathcal{V}$ grants the existence of the dashed horizontal functor making the diagram commute.

We must show that, for each symmetric monoidal functor $A\colon\disk_{n}^{B}\to\mathcal{V}$, and for each based map among finite sets $I_{+}\xrightarrow{f}J_{+}$, the diagram of $\infty$-categories

commutes. The map $f\colon I_{+}\to J_{+}$ is canonically a composition of a surjective active map $f^{\sf surj}$ followed by an injective active map $f^{\sf inj}$ followed by an inert map $f^{\sf inrt}$, and so it is enough to verify commutativity of the above diagram for each such class of maps. The case of inert maps is obvious, because then $f_{\ast}$ is projection and $(\iota_{!}A)^{K}$ is defined as the $K$-fold product of functors, for $K=I,J$. The case of injective active maps amounts to verifying that $\iota_{!}A$ carries each monoidal unit to a monoidal unit. This follows because $A$ does so and because the over $\infty$-categories $\disk_{n/\emptyset}^{B}=\{\emptyset\}=\mfld_{n/\emptyset}^{B}$ consist solely of the empty manifold, which is the monoidal unity.

The case of surjective active maps follows from the case that $f\colon I_{+}\to\ast_{+}$ is given by $+\neq i\mapsto\ast$, so that $f_{\ast}=\bigotimes^{I}$ is the $I$-fold tensor product. Well, because $A$ is symmetric monoidal, there is a canonical arrow $\iota_{!}A\circ\bigotimes^{I}\longrightarrow\bigotimes^{I}\circ(\iota_{!}A)^{I}$ between functors $(\mfld_{n}^{B})^{I}\to\mathcal{V}$ that we will argue is an equivalence. This arrow evaluates on $(M_{i})_{i\in I}$ as the horizontal one in the following natural diagram in $\mathcal{V}$:

The arrow labeled by ($\dagger$) is an equivalence precisely because $V\otimes-\colon\mathcal{V}\to\mathcal{V}$ preserves colimits. By inspection, the $I$-fold disjoint union functor $\bigsqcup^{I}\colon\prod_{i\in I}\disk^{B}_{n/M_{i}}\xrightarrow{\simeq}\disk^{B}_{n/\underset{i\in I}{\bigsqcup}M_{i}}$ is an equivalence between $\infty$-categories, for it is essentially surjective and fully faithful. It follows that the arrow labeled by ($\ast$) is an equivalence, after observing the following commutative diagram among $\infty$-categories:

∎

Proof. It suffices to show that the forgetful functor $\disk_{n/M}^{B}\rightarrow\disk_{n/M}$ is an equivalence. By definition, this functor is the projection from the double overcategory:

$$\disk_{n/M}^{B}~:=~(\disk_{n/B})_{/M}\longrightarrow\disk_{n/M}.$$

This functor is a pullback of the likewise functor $\bigl((\spaces_{/\BTop(n)})_{/B}\bigr)_{/M}\to(\spaces_{/\BTop(n)})_{/M}$, which is an equivalence by Lemma 2.5.

∎

Proof. Let $S\subset\disk^{\partial,{\sf or}}_{1/[-1,1]}$ be the full $\infty$-subcategory of embeddings from oriented intervals that surject onto the endpoints. That is, $S$ consists of oriented embeddings among 1-manifolds with boundary of the form $[-1,0)\sqcup\mathbb{R}^{\sqcup\ell}\sqcup(0,1]\hookrightarrow[-1,1]$, for $\ell\geq 0$. To show the inclusion $S\subset\disk^{\partial,{\sf or}}_{1/[-1,1]}$ is final, by Quillen’s Theorem A, we can show the under $\infty$-category $S^{U/}$ has a contractible classifying space for every finite disjoint union of subintervals $U\subset[-1,1]$. This is immediate, because $S^{U/}$ has an initial object, which is a disjoint union of $U$ with connected open neighborhoods of the endpoints not contained in $U$.

Lastly, the result follows because there is an equivalence $S\to\bdelta^{\op}$. On objects this is given by assigning to $(U\hookrightarrow[-1,1])$ the set connected components of the complement $\bigl[[-1,1]\smallsetminus U\bigr]$ together with the linear order inherited from that of $[-1,1]$. That this assignment defines a functor is routine. That this functor is an equivalence of $\infty$-categories follows because each comopnent of the space of morphisms of $\disk^{\partial,\sf or}_{1/[-1,1]}$ is contractible.

∎

Proof. Regard the associative algebra $A$ as a symmetric monoidal functor $A\colon\disk^{\sf or}_{1}\to\mathcal{V}$, as in Section 3.2. Consider the standard collar-gluing $\mathbb{R}\underset{\mathbb{R}\sqcup\mathbb{R}}{\bigcup}\mathbb{R}\cong S^{1}$ by hemispheres. Lemma 3.18, which states that factorization homology staisfies $\otimes$-excision, determines the first of the equivalences in the expression:

$$\int_{S^{1}}A~\simeq~\int_{\mathbb{R}}A\underset{{\displaystyle\int_{{S^{0}\times\mathbb{R}}}\!A}}{\bigotimes}\int_{\mathbb{R}}A~\simeq~A\underset{A\otimes A^{\op}}{\bigotimes}A~\simeq~\hh_{*}(A)~.$$

The second equivalence is by inspecting values, and the final equivalence is definitional.

∎

Proof. After Lemma 2.5, it will suffice to prove the result for the case $B=\BTop(n)$, and so we omit $B$ from the notation and discussion. The functor ${\sf ev}_{0}$ is a Cartesian fibration of $\oo$-categories. Thus, to check finality, by Lemma 4.1.3.2 of [Lu1], it suffices to show that, for each $(V\hookrightarrow M)\in\disk_{n/M}$, the fiber $\infty$-category ${\sf ev}_{0}^{-1}V$ has contractible classifying space. That is, we show that the $\oo$-category $(\disk_{k/N})^{V/}$, of $k$-disks $U$ in $N$ equipped with an embedding $V\hookrightarrow f^{-1}U$, has a contractible classifying space.

There is an identification of spaces

$$\mathsf{B}\bigl(\disk_{k/N}\bigr)^{U/}~{}~\simeq~{}~\underset{(V\hookrightarrow N)\in\disk_{k/N}}{\colim}~\Map_{\mfld_{n/M}}(U,f^{-1}V)~.$$

Formally, the sequence of maps

$$\Map_{\mfld_{n/M}}(U,f^{-1}V)~\longrightarrow~\Map_{\mfld_{n}}(U,f^{-1}V)~\longrightarrow~\Map_{\mfld_{n}}(U,M)$$

is a fiber sequence (here the fiber is taken over any implicit morphism $U\hookrightarrow M$, thereby giving meaning to the lefthand space). So we seek to show the map from the colimit

$$\underset{(V\hookrightarrow N)\in\disk_{k/N}}{\colim}\Map_{\mfld_{n}}(U,f^{-1}V)~{}~\longrightarrow~{}~\Map_{\mfld_{n}}(U,M)$$

is an equivalence of spaces. We recognize this map of spaces as the map of fibers over $U\in\disk_{n}$ of the map of right fibrations over $\disk_{n}$:

$$\underset{(V\hookrightarrow N)\in\disk_{k/N}}{\colim}\disk_{n/f^{-1}V}\longrightarrow\disk_{n/M}~.$$

Being right fibrations, it is enough to show that this functor is an equivalence on maximal $\infty$-subgroupoids. Using Lemma 2.12 which identifies these maximal $\infty$-subgroupoids, this is the problem of showing, for each finite set $J$, that the map of spaces

$$\underset{(V\hookrightarrow N)\in\disk_{k/N}}{\colim}\conf_{J}\bigl(f^{-1}V\bigr)_{\Sigma_{J}}~\longrightarrow~\conf_{J}(M)_{\Sigma_{J}}$$

is an equivalence.

Lemma 2.19 implies the functor $\ddisk_{k/N}\to\disk_{k/N}$ is final, and so the forgetful map

$$\underset{(V\hookrightarrow N)\in\ddisk_{k/N}}{\colim}\conf_{J}\bigl(f^{-1}V\bigr)_{\Sigma_{J}}~\xrightarrow{~\simeq~}~\underset{(V\hookrightarrow N)\in\disk_{k/N}}{\colim}\conf_{J}\bigl(f^{-1}V\bigr)_{\Sigma_{J}}$$

is an equivalence of spaces. Now notice that, for each $(V\hookrightarrow N)\in\ddisk_{k/N}$, the map $\conf_{J}(f^{-1}V)\to\conf_{J}(M)$ an open embedding. Also, for each point $c\colon J\hookrightarrow M$ the image $f\bigl(c(J)\bigr)\subset N$ has cardinality at most $J$. So there is an object $(V\hookrightarrow N)$ of $\ddisk_{k/N}$ whose image contains the subset $f\bigl(c(J)\bigr)$. We see then that the collection of open embeddings

$$\Bigl\{\conf_{J}(f^{-1}V)_{\Sigma_{J}}\hookrightarrow\conf_{J}(M)_{\Sigma_{I}}\mid(V\hookrightarrow N)\in\ddisk_{k/N}\Bigr\}$$

forms an open cover.

Because $N$ is a manifold, the collection of open embeddings from Euclidean spaces into $N$ form a basis for the topology of $N$. It follows that the collection of (at most) $|J|$-tuples of disjoint open disks in $N$ forms an open cover of $N$ in such a way that any finite intersection of such is again covered by such. This is to say that this collection of open embeddings forms a hypercover of $\conf_{J}(M)_{\Sigma_{J}}$. Corollary 1.6 of [DI] gives that the map

$$\underset{(V\hookrightarrow N)\in\ddisk_{k/N}}{\colim}\conf_{J}\bigl(f^{-1}V\bigr)_{\Sigma_{J}}~\xrightarrow{~\simeq~}~\conf_{J}(M)_{\Sigma_{J}}$$

is an equivalence of spaces, which completes the proof.

∎

Proof. The $\oo$-category $\disk_{n/M}^{B}$ is evidently nonempty, as it contains the object $(\emptyset\hookrightarrow M)$. We must then prove that the diagonal functor $\disk_{n/M}^{B}\to\disk_{n/M}^{B}\times\disk_{n/M}^{B}$ is final. This diagonal functor fits into a diagram among $\infty$-categories

that we now explain. The upper left $\infty$-category is that of Definition 3.20 applied to the fold map $\nabla\colon M\sqcup M\to M$; it is equipped with the indicated projection functors. The right vertical arrow is induced by the symmetric monoidal structure on $\mfld_{n}^{B}$, which is disjoint union. This right vertical arrow is an equivalence; an inverse is given by declaring its projection to each factor to be given by intersecting with the corresponding cofactor of the disjoint union. Therefore, to prove that the diagonal functor is final it is sufficient to prove that both of the projection functors ${\sf ev}_{1}$ and ${\sf ev}_{0}$ are final. The finality of ${\sf ev}_{0}$ is Lemma 3.21.

We explain that ${\sf ev}_{1}$ is final. Note that the functor $\nabla^{-1}\colon\disk_{n/M}^{B}\to\mfld_{n/M\sqcup M}^{B}$ factors through the full $\infty$-subcategory $\disk_{n/M\sqcup M}^{B}$. As so, there is a canonical identification between $\infty$-categories

$$\disk_{\nabla}~\simeq~\disk_{n/M}^{B}\underset{\disk_{n/M\sqcup M}^{B}}{\times}{\sf Ar}(\disk_{n/M\sqcup M}^{B})$$

over $\disk_{n/M}^{B}$. Through this identification, the composite functor

$$\disk_{n/M}^{B}\xrightarrow{~\nabla~}\disk_{n/M\sqcup M}^{B}\xrightarrow{~\sf const~}{\sf Ar}(\disk_{n/M\sqcup M}^{B})$$

determines a right adjoint to the functor ${\sf ev}_{1}$. The finality of ${\sf ev}_{1}$ thereby follows.

∎

Proof. After Proposition 3.9, we can assume that $B$ is equivalent to $\BTop(n)$, and so we omit it from the notation and discussion. We will explain the string of canonical equivalences in $\mathcal{V}$:

	$\displaystyle\int_{N}f_{\ast}A$	$\displaystyle\underset{\rm Def~\ref{coend}}{\simeq}$	$\displaystyle\colim_{U\in\disk_{k/N}^{\partial,{\sf or}}}f_{\ast}A(U)$
		$\displaystyle\underset{{\rm Def~}f_{\ast}}{\simeq}$	$\displaystyle\colim_{U\in\disk_{k/N}^{\partial,{\sf or}}}\int_{f^{-1}U}A$
		$\displaystyle\underset{\rm Def~\ref{coend}}{\simeq}$	$\displaystyle\colim_{U\in\disk_{k/N}^{\partial,{\sf or}}}\colim_{V\in\disk_{n/f^{-1}U}}A(V)$
		$\displaystyle\underset{(1)}{\simeq}$	$\displaystyle\colim_{(U,V)\in\disk_{f}}A(V)$
		$\displaystyle\underset{(2)}{\xrightarrow{\simeq}}$	$\displaystyle\colim_{V\in\disk_{n/M}}A(V)$
		$\displaystyle\underset{\rm Def~\ref{coend}}{\simeq}$	$\displaystyle\int_{M}A~.$

The only equivalences that are not definitional are (1) and (2). The equivalence (2) is a direct application of Lemma 3.21, which states that the functor ${\sf ev}_{0}\colon\disk_{f}\to\disk_{n/M}$ is final. Consider the left Kan extension (non-commutative) diagram among $\infty$-categories:

which exists because $\mathcal{V}$ is presentable. By construction, the functor ${\sf ev}_{1}\colon\disk_{f}\to\disk^{\partial,\sf or}_{k/N}$ is a coCartesian fibration. In particular, for each object $U\in\disk^{\partial,\sf or}_{k/N}$, the inclusion of the fiber into the over $\infty$-category

$${\sf ev}_{1}^{-1}(U)\longrightarrow(\disk_{f})_{/U}$$

is final. Therefore, the value of ${\sf LKan}$ on $U\in\disk^{\partial,\sf or}_{k/N}$ is the colimit over the fiber:

$$\colim_{V\in\disk_{n/f^{-1}U}}A(V)~\underset{\rm Def~\ref{disk-f}}{\simeq}~\colim_{V\in{\sf ev}_{1}^{-1}U}A(V)\xrightarrow{~\simeq~}{\sf LKan}(U)~.$$

So the colimit of ${\sf LKan}$ is the codomain of (1). The equivalence (1) follows from Proposition 4.3.3.7 of [Lu1], which implies the colimit of a left Kan extension agrees with the colimit because they both satisfy the same universal property.

∎

Proof of Lemma 3.18. Given a collar-gluing $f:M\rightarrow[-1,1]$, there are canonical morphisms in $\mathcal{V}$

$$\int_{M^{\prime}}A\bigotimes_{\displaystyle\int_{M_{0}\times\mathbb{R}}A}\int_{M^{\prime\prime}}A\longrightarrow\int_{[-1,1]}f_{\ast}A\longrightarrow\int_{M}A~~.$$

The lefthand morphism is an equivalence by Lemma 3.12, which shows that factorization homology over the closed interval is equivalent to the bar construction, and inspection of the functor $f_{\ast}A$. The righthand morphism is an equivalence by Proposition 3.23, which grants that factorization homology pushes forward along $M\xrightarrow{f}[-1,1]$.

∎

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

$$i_{!}\colon\Alg_{\disk_{n}^{B}}(\mathcal{V})\rightleftarrows\Fun^{\otimes}\bigl(\mfld_{n}^{B},\mathcal{V}\bigr)\colon i^{\ast}~.$$

The unit of this adjunction is an equivalence because $\disk_{n}^{B}\rightarrow\mfld_{n}^{B}$ 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

$$\underset{S^{i}\times\mathbb{R}^{n-i}}{\int}\!A\simeq\underset{\mathbb{R}_{-1}^{i}\times\mathbb{R}^{n-i}}{\int}\!A\underset{\underset{S^{i-1}\times\mathbb{R}^{n-i+1}}{\int}\negthinspace A}{\bigotimes}\ \underset{\mathbb{R}_{+1}^{i}\times\mathbb{R}^{n-i}}{\int}\!A\ \simeq\ \mathcal{F}(\mathbb{R}_{-1}^{i}\times\mathbb{R}^{n-i})\underset{\mathcal{F}(S^{i-1}\times\mathbb{R}^{n-i+1})}{\bigotimes}\mathcal{F}(\mathbb{R}_{+1}^{i}\times\mathbb{R}^{n-i})\simeq\mathcal{F}(S^{i}\times\mathbb{R}^{j})$$

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 $\mathcal{F}(\mathbb{R}^{n})$.