ScalingStacks

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

$$\mathcal{P}\otimes\mathbb{E}_{\infty}\to\mathbb{E}_{\infty}\otimes\mathbb{E}_{\infty}\simeq\mathbb{E}_{\infty}$$

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). ∎