ScalingStacks

Proof. By A.2.4.2.7, the opposite of a symmetric monoidal $\infty$-category acquires a symmetric monoidal structure, which is Cartesian if and only if the original symmetric monoidal $\infty$-category is coCartesian. Hence, it is enough to prove (2). The unit object $1\in\mathcal{D}$ has a unique map from the initial object $\varnothing\to 1$. Since $f_{\alpha}$ is both symmetric monoidal and preserves finite coproducts, $f_{\alpha}\left(\varnothing\to 1\right)$ is the unique map from the initial object to the unit object of $\mathcal{C}_{\alpha}$, which is an equivalence by assumption. Since the collection of $f_{\alpha}$ is jointly conservative, it follows that the unit of $\mathcal{D}$ is initial in $\underline{\mathcal{D}}$ as well. Namely, $\mathcal{D}$ is unital as an $\infty$-operad. Using 2.2.3 we have a map of $\infty$-operads $G\colon\mathcal{D}\to\mathcal{D}_{\sqcup}$, which is an equivalence on the underlying $\infty$-categories. We need to show that this map is symmetric monoidal. Namely, that it maps coCartesian edges (over $\mathbf{Fin}_{*}$) to coCartesian edges. Since we already know that it is a map of $\infty$-operads and hence preserves inert morphisms, we only need to show that active coCartesian edges map to coCartesian edges. Using the Segal conditions, we are further reduced to considering only coCartesian lifts of the unique active morphism $\mu\colon\left\langle n\right\rangle\to\left\langle 1\right\rangle$. For every collection of objects $X_{1},X_{2},\dots,X_{n}\in\underline{\mathcal{D}}$, let

$$\mu_{\otimes}\colon X_{1}\oplus\cdots\oplus X_{n}\to X_{1}\otimes X_{2}\otimes\dots\otimes X_{n}$$

be a coCartesian lift of $\mu$ to $\mathcal{D}^{\otimes}$. Since $G$ is an equivalence on the underlying $\infty$-categories, $G\left(\mu_{\otimes}\right)$ can be considered as a map

$$X_{1}\oplus\cdots\oplus X_{n}\to X_{1}\otimes X_{2}\otimes\dots\otimes X_{n}$$

in $\mathcal{D}_{\sqcup}$. Now, let

$$\mu_{\sqcup}\colon X_{1}\oplus\cdots\oplus X_{n}\to X_{1}\sqcup X_{2}\sqcup\dots\sqcup X_{n}$$

be a coCartesian lift of $\mu$ to $\mathcal{D}^{\sqcup}$. There exists a unique (up to homotopy) map

$$g_{X_{1},\dots,X_{n}}\colon X_{1}\sqcup X_{2}\sqcup\dots\sqcup X_{n}\to X_{1}\otimes X_{2}\otimes\dots\otimes X_{n},$$

such that $G\left(\mu_{\otimes}\right)=g_{X_{1},\dots,X_{n}}\circ\mu_{\sqcup}$. We need to show that $g_{X_{1},\dots,X_{n}}$ is an equivalence in $\underline{\mathcal{D}}$ for all $X_{1},\dots,X_{n}\in\underline{\mathcal{D}}$. For every $\alpha$, we have a homotopy commutative diagram

in which the vertical and bottom maps are symmetric monoidal. It follows that $f_{\alpha}\left(g_{X_{1},\dots,X_{n}}\right)$ is an equivalence in $\mathcal{C}_{\alpha}$ for all $\alpha$. By joint conservativity, $g_{X_{1},\dots,X_{n}}$ is an equivalence as well. ∎

Proof. By A.2.2.5.4, since $\mathcal{C}_{\times}$ is symmetric monoidal, so is $\operatorname{Alg}_{\mathcal{D}}\left(\mathcal{C}\right)$ and, for every $X\in\mathcal{D}$, the evaluation functor $e_{X}\colon\operatorname{Alg}_{\mathcal{D}}\left(\mathcal{C}\right)\to\mathcal{C}_{\times}$ is a symmetric monoidal functor. On the underlying $\infty$-categories, $e_{X}$ also preserves finite products since it preserves all limits. Finally, we show that the collection of evaluation functors is jointly conservative since they can be presented as the composition of the conservative restriction functor

$$\operatorname{Alg}_{\mathcal{D}}\left(\mathcal{C}\right)\to\operatorname{Fun}\left(\mathcal{D},\mathcal{C}\right)$$

and the collection of evaluation functors

$$e_{X}\colon\operatorname{Fun}\left(\mathcal{D},\mathcal{C}\right)\to\mathcal{C},$$

which are jointly conservative by T.3.1.2.1. Now, by 3.2.4(1), $\operatorname{Alg}_{\mathcal{D}}\left(\mathcal{C}\right)$ is Cartesian. ∎

Proof. $\left(1\right)\implies\left(2\right)$ Consider the commutative diagram

Since $h_{d+1}\left(\mathcal{P}\right)\to h_{d+1}\left(\mathcal{Q}\right)$ is an equivalence of $\infty$-operads, the bottom map is a homotopy equivalence. By 3.1.8, the vertical maps are equivalences as well, and so, by the 2-out-of-3 property, the top map is an equivalence.

$\left(2\right)\implies\left(3\right)$ Since $\mathcal{S}_{\leq d}^{K}$ is a $\left(d+1\right)$-topos, this is just a special case.

$\left(3\right)\implies\left(4\right)$ By Yoneda’s lemma applied to $\mathbf{Cat}_{\infty}$, the map

$$\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{S}_{\leq d}\right)\to\underline{\operatorname{Alg}}_{\mathcal{P}}\left(\mathcal{S}_{\leq d}\right)$$

is an equivalence of $\infty$-categories if for every $\infty$-category $\mathcal{E}$, the map

$$\operatorname{Map}_{\mathbf{Cat}_{\infty}}(\mathcal{E},\underline{\operatorname{Alg}}_{\mathcal{Q}}\left(\mathcal{S}_{\leq d}\right))\to\operatorname{Map}_{\mathbf{Cat}_{\infty}}(\mathcal{E},\underline{\operatorname{Alg}}_{\mathcal{P}}\left(\mathcal{S}_{\leq d}\right))$$

is a homotopy equivalence. Using the fully faithful embedding $\mathbf{Cat}_{\infty}\hookrightarrow\mathbf{Op}_{\infty}$, which is left adjoint to the underlying category functor $\mathbf{Op}_{\infty}\to\mathbf{Cat}_{\infty}$ (see A.2.1.4.11), this map is equivalent to

$$\operatorname{Map}_{\mathbf{Op}_{\infty}}\left(\mathcal{E},\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{S}_{\leq d}\right)\right)\to\operatorname{Map}_{\mathbf{Op}_{\infty}}\left(\mathcal{E},\operatorname{Alg}_{\mathcal{P}}\left(\mathcal{S}_{\leq d}\right)\right).$$

By adjointness with the Boardman–Vogt tensor product and the fact that it is symmetric, the map is equivalent to

$$\operatorname{Map}_{\mathbf{Op}_{\infty}}\left(\mathcal{Q},\operatorname{Alg}_{\mathcal{E}}\left(\mathcal{S}_{\leq d}\right)\right)\to\operatorname{Map}_{\mathbf{Op}_{\infty}}\left(\mathcal{P},\operatorname{Alg}_{\mathcal{E}}\left(\mathcal{S}_{\leq d}\right)\right).$$

Since $\mathcal{E}$ is an $\infty$-category, by 3.2.5 the $\infty$-operad $\operatorname{Alg}_{\mathcal{E}}\left(\mathcal{S}_{\leq d}\right)$ is just the $\infty$-category of functors $\left(\mathcal{S}_{\leq d}\right)^{\mathcal{E}}$ endowed with the Cartesian symmetric monoidal structure. Since the functor category is invariant under Joyal equivalences, we can replace $\mathcal{E}$ with any simplicial set $K$.

$\left(4\right)\implies\left(1\right)$ Consider the commutative diagram

By 3.1.8, the vertical maps are equivalences; hence by 2-out-of-3, the top map is an equivalence if and only if the bottom map is. We can therefore assume without loss of generality that $\mathcal{P}$ and $\mathcal{Q}$ are themselves essentially $d$-operads. This implies that $\mathcal{P}\left(n\right)$ and $\mathcal{Q}\left(n\right)$ are $d$-truncated spaces for all $n\geq 0$. Now, consider the commutative diagram

where $U_{\mathcal{P}}$ and $U_{\mathcal{Q}}$ are the corresponding forgetful functors. By 2.4.4, the associated map

$$T_{\mathcal{P}}=\coprod_{n}\left(\mathcal{P}\left(n\right)\times X^{n}\right)_{h\Sigma_{n}}\overset{\sim}{\longrightarrow}\coprod_{n}\left(\mathcal{Q}\left(n\right)\times X^{n}\right)_{h\Sigma_{n}}=T_{\mathcal{Q}}$$

of Construction 2.4.3 is a natural equivalence of functors. On the other hand, by 2.4.6, this map is induced from a map of symmetric sequences $f_{\mathbf{SSeq}}\colon\left\{\mathcal{P}\left(n\right)\right\}\to\left\{\mathcal{Q}\left(n\right)\right\}$. We want to deduce that $f_{\mathbf{SSeq}}$ is an equivalence. For $d=-1$, there is nothing to prove and so we assume that $d\geq 0$. Taking $X=\left[n\right]$, there is a coproduct decomposition

$$\left(\mathcal{P}\left(n\right)\times X^{n}\right)_{h\Sigma_{n}}=\mathcal{P}\left(n\right)\sqcup J,$$

where the summand $\mathcal{P}\left(n\right)$ corresponds to orbits of points whose $X^{n}$ component is a permutation (note that when $d=0$, the homotopy orbits in $\mathcal{S}_{\leq 0}$ are just the orbits as a set). This characterization implies that $f_{\mathbf{SSeq}}\colon\mathcal{P}\left(n\right)\to\mathcal{Q}\left(n\right)$ is an equivalence. Finally, since $\left(-\right)_{\mathbf{SSeq}}$ is conservative, by 2.3.6, we deduce that $f$ is an equivalence. ∎