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:

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.

∎