ScalingStacks

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. ∎