ScalingStacks

Proof. Since an $\infty$-operad $\mathcal{U}$ is an essentially $d$-operad if and only if it is equivalent to a (strict) $d$-operad, it is enough to prove the strict version. By definition, the $\infty$-category $\operatorname{Alg}_{\mathcal{O}}\left(\mathcal{U}\right)$ is a full subcategory of $\operatorname{Fun}\left(\mathcal{O}^{\otimes},\mathcal{U}^{\otimes}\right)$. For $d\geq 1$, the $\infty$-category $\mathcal{U}^{\otimes}$ is a $d$-category and, therefore, by T.2.3.4.8, the $\infty$-category $\operatorname{Fun}\left(\mathcal{O}^{\otimes},\mathcal{U}^{\otimes}\right)$ is a $d$-category as well. Hence, every full subcategory of it is a $d$-category. For $d=0$, by 3.9 we can assume that $\mathcal{O}^{\otimes}$ is a $0$-operad as well and therefore both $\mathcal{O}^{\otimes}$ and $\mathcal{U}^{\otimes}$ are skeletal 1-categories with faithful projection to $\mathbf{Fin}_{*}$. Observing that $\operatorname{Alg}_{\mathcal{O}}\left(\mathcal{U}\right)$ is a full subcategory of $\operatorname{Fun}_{\mathbf{Fin}_{*}}\left(\mathcal{O}^{\otimes},\mathcal{U}^{\otimes}\right)$ and using the faithfulness of the projections to $\mathbf{Fin}_{*}$, we see that the mapping spaces are either empty or singletons. For $d=-1$, the claim is obvious. ∎