ScalingStacks

Proof. By 3.2.3, if $\mathcal{P}\not\simeq\mathbb{E}_{0}$, then it is $\left(-1\right)$-connected and in particular $\mathcal{P}\left(2\right)\neq\varnothing$. We shall construct a section to $\underline{f_{A,B}}$ using any binary operation $\mu\in\mathcal{P}\left(2\right)$. Let $i_{A}\colon A\to A\sqcup B$ and $i_{B}\colon B\to A\sqcup B$ be the canonical maps of the coproduct. Define $s$ to be the composition of the following maps:

Now, consider the following diagram in the homotopy category of $\mathcal{C}$:

The upper square commutes since $f_{A,B}$ is a map of algebras. The upper triangle commutes since it is the tensor product of two triangles, which commute by the very definition of $f_{A,B}$. The lower square commutes by the definition of the algebra structure on $A\otimes B$ and the lower triangle also clearly commutes. The composition of the bottom diagonal map and the bottom right map is the identity, since the restriction of $\mu$ to the unit in one of the arguments is homotopic to the identity map of the other argument. The composition of the top diagonal map with the top right map is $s$. It follows that $\underline{f_{A,B}}\circ s\sim\operatorname{Id}_{\underline{A\otimes B}}$. ∎

Proof. For every $\left\langle n\right\rangle\in\mathbf{Fin}_{*}$, the restriction of $F^{\otimes}$ to the fiber over $\left\langle n\right\rangle$ is just $F^{n}\colon\mathcal{C}^{n}\to\mathcal{D}^{n}$, which is clearly a left adjoint. Hence, by A.7.3.2.7, the functor $F^{\otimes}$ is a left adjoint relative to $\mathbf{Fin}_{*}$. Let $G^{\otimes}$ be the right adjoint of $F^{\otimes}.$ Applying A.7.3.2.13, we obtain that $F^{\otimes}$ and $G^{\otimes}$ induce an adjunction:

∎

Proof. For $d=-2$ there is nothing to prove and so we assume that $d\geq-1$. By 4.4.5, it is enough to show that $\underline{f_{A,B}}$ is $d$-connected where $\underline{\left(-\right)}\colon\operatorname{Alg}_{\mathcal{P}}\left(\mathcal{C}\right)\to\mathcal{C}$ is the forgetful functor. By 4.3.5, it is enough to show that $\underline{f_{A,B}}$ has a section and is $\left(d-\frac{1}{2}\right)$-connected. Since $d\geq-1$, we have $\mathcal{P}\neq\mathbb{E}_{0}$ and, therefore, by 5.1.1, $\underline{f_{A,B}}$ has a section. Thus, we are reduced to showing that the image of $\underline{f_{A,B}}$ under the functor $\tau_{\leq d}^{\mathcal{C}}\colon\mathcal{C}\to\tau_{\leq d}\mathcal{C}$ is an equivalence. First, we show that $\tau_{\leq d}^{\mathcal{C}}$ preserves binary products. For $m=\infty$, this follows from T.6.5.1.2. The general case reduces to $m=\infty$ as by T.6.4.1.5 we can embed $\mathcal{C}$ as a full subcategory of an $\infty$-topos spanned by the $\left(m-1\right)$-truncated objects. It follows that we get a symmetric monoidal functor $\tau_{\leq d}^{\times}\colon\mathcal{C}^{\times}\to\left(\tau_{\leq d}\mathcal{C}\right)^{\times}$. By 5.1.2, the functor

induced by $\tau_{\leq n}^{\times}$ is a left adjoint. Consider the following (solid) commutative diagram in the homotopy category of $\mathbf{Cat}_{\infty}$:

where the vertical maps are the forgetful functors and $G$ is induced by restriction along the essentially unique map $\mathcal{P}\to\mathbb{E}_{\infty}$. Since $\tau_{\leq d}\mathcal{C}$ is an essentially $\left(d+1\right)$-category, it follows from 3.1.8 that $G$ is an equivalence. Taking $G^{\prime}$ to be an inverse of $G$ up to homotopy, the outer rectangle is a commutative square in the homotopy category of $\mathbf{Cat}_{\infty}$. Therefore, to show that $\tau_{\leq d}^{\mathcal{C}}\left(\underline{f_{A,B}}\right)$ is an equivalence, it is enough to show that $\underline{G^{\prime}\left(F\left(f_{A,B}\right)\right)}$ is an equivalence. In fact, we shall show that $G^{\prime}\left(F\left(f_{A,B}\right)\right)$ is an equivalence. Note that the composition of the left and then bottom functors preserves binary products and since the right vertical functor preserves products and is conservative, it follows that the top functor $G^{\prime}\circ F$ also preserves binary products. On the other hand, $G^{\prime}\circ F$ also preserves coproducts, since $F$ is left adjoint (by the above discussion) and $G$ is an equivalence. Finally, in $\operatorname{Alg}_{\mathbb{E}_{\infty}}\left(\tau_{\leq d}\mathcal{C}\right)$, the canonical map from the coproduct to the product is an equivalence by A.3.2.4.7. ∎

Proof. The $\infty$-operad $\mathcal{E}=\operatorname{End}_{\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)}^{\operatorname{\scriptsize{red}}}\left(X\right)$ has a unique object, which we call $X$. We need to show that for every $m\in\mathbb{N}$, the multi-mapping space $\operatorname{Mul}_{\mathcal{E}}\left(X^{\left(m\right)},X\right)$ is $\left(d-n-2\right)$-truncated. By 2.2.12 we have a fiber sequence

where the fiber is taken over the fold map $\nabla\colon X^{\sqcup m}\to X$. The fiber is equivalent to the space of lifts for the square

Since $\mathcal{C}$ is an essentially $\left(d+1\right)$-category, so is the Cartesian $\infty$-operad $\mathcal{C}_{\times}$ and, therefore, by 3.1.10, so is $\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)$. In particular, $X$ is $d$-truncated. Hence, by 4.2.8, it is enough to show that the canonical map $X^{\sqcup m}\to X^{m}$ is $n$-connected. Since $\mathcal{Q}$ is $n$-connected, this follows from repeated application of 5.1.3. ∎