ScalingStacks

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

$$\operatorname{Mul}_{\mathcal{E}}(X^{\left(m\right)},X)\to\operatorname{Mul}_{\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)}\left(X^{m},X\right)\to\operatorname{Map}_{\operatorname{Alg}_{\mathcal{Q}}\left(\mathcal{C}\right)}\left(X^{\sqcup m},X\right),$$

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. ∎