ScalingStacks

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:

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

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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

Formally, the sequence of maps

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

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 $\disk_{n}$:

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

is an equivalence.

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

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

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

is an equivalence of spaces, which completes the proof.

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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

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

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

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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 $\mathcal{V}$:

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

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

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.

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

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 $M\xrightarrow{f}[-1,1]$.

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