ScalingStacks

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

$$F\colon\operatorname{Alg}_{\mathcal{P}}\left(\mathcal{C}\right)\to\operatorname{Alg}_{\mathcal{P}}\left(\tau_{\leq d}\mathcal{C}\right)$$

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. ∎