ScalingStacks

Proof. Set $d=d_{1}+d_{2}+2$. By 3.2.6, it is enough to show that for every $\left(d+1\right)$-topos $\mathcal{C}$ with the Cartesian symmetric monoidal structure, the map

induced by pre-composition with $f$, is a homotopy equivalence. Using the tensor-hom adjunction, it is the same as showing that the map

is an equivalence. The underlying category functor gives a commutative diagram:

As $\underline{P}\to\underline{\mathcal{Q}}$ is an equivalence of $\infty$-categories (both are equivalent to $\Delta^{0}$), the bottom map is a homotopy equivalence. Hence, it suffices to show that the induced map on the homotopy fibers is a homotopy equivalence for each choice of a base point. A point in the space $\operatorname{Map}\left(\Delta^{0},\operatorname{Alg}_{\mathcal{R}}\left(\mathcal{C}\right)\right)$ is just an $\mathcal{R}$-algebra $X$ in $\mathcal{C}$. We denote by $\operatorname{Alg}_{\mathcal{R}}\left(\mathcal{C}\right)_{X}$ the $\infty$-operad $\operatorname{Alg}_{\mathcal{R}}\left(\mathcal{C}\right)$ pointed by $X$ viewed as an object of $\mathbf{Op}_{\infty,*}^{\operatorname{\scriptsize{un}}}$. With this notation, we see that the homotopy fiber of the right vertical map is equivalent to

By 2.2.5, the $\infty$-operad $\operatorname{Alg}_{\mathcal{R}}\left(\mathcal{C}\right)$ is unital. Therefore, by the adjunction

the above mapping space is also equivalent to

since $\mathbf{Op}_{\infty}^{\operatorname{\scriptsize{red}}}\subseteq\mathbf{Op}_{\infty}$ is a full subcategory. The induced map on the fibers of the vertical maps in $\left(*\right)$ over $X$, is therefore equivalent to

Finally, since $\mathcal{\mathcal{C}}$ is a $\left(d+1\right)$-topos and $\mathcal{R}$ is $d_{2}$-connected, 5.1.4 implies that the $\infty$-operad $\operatorname{End}_{\operatorname{Alg}_{\mathcal{R}}\left(\mathcal{C}\right)}^{\operatorname{\scriptsize{red}}}\left(X\right)$ is an essentially $\left(d-d_{2}-1=d_{1}+1\right)$-operad. Since $\mathcal{P}\to\mathcal{Q}$ is a $d_{1}$-equivalence, by 3.1.8 the above map is a homotopy equivalence and this completes the proof. ∎

Proof. Since $\mathcal{P}$ is $d_{1}$-connected, the essentially unique map $\mathcal{P}\to\mathbb{E}_{\infty}$ is a $d_{1}$-equivalence. Hence, by 5.3.1, the map $\mathcal{P}\otimes\mathcal{Q}\to\mathcal{P}\otimes\mathbb{E}_{\infty}$ is a $\left(d_{1}+d_{2}+2\right)$-equivalence. Since $\mathbb{E}_{\infty}$ is also $d_{2}$-connected, by the same argument the induced map

is also a $\left(d_{1}+d_{2}+2\right)$-equivalence. The $\left(d_{1}+d_{2}+2\right)$-equivalences are closed under composition, and so the result follows (in fact, we know a posteriori that the map above is actually an equivalence of $\infty$-operads). ∎

Proof. If $\mathcal{P}\not\simeq\mathbb{E}_{0}$, then, by 3.2.3, $\mathcal{P}$ is $d$-connected for some $d\geq-1$. Therefore, by 5.2.2, $\mathcal{P}\otimes\mathcal{P}$ is $\left(2d+2\right)>d$ connected. Since $\mathcal{P}\simeq\mathcal{P}\otimes\mathcal{P}$, we can continue by induction and deduce that $\mathcal{P}$ is $\infty$-connected; hence $\mathcal{P}\simeq\mathbb{E}_{\infty}$. ∎

Proof. By 3.2.3, all $\mathcal{P}_{i}$-s are $\left(-1\right)$-connected. By induction on $k$ and 5.3.3, the $\infty$-operad $\mathcal{P}_{1}\otimes\cdots\otimes\mathcal{P}_{k}$ is $\left(k-2\right)$-connected. For every $n\in\mathbb{N}$ we get

and therefore $\bigotimes\limits_{i=1}^{\infty}\mathcal{P}_{i}\simeq\mathbb{E}_{\infty}$. ∎