ScalingStacks

Proof. It is clear that $\theta_{d}$ is essentially surjective since it is surjective on objects. Let $X,Y\in\mathcal{C}$ be two objects. Since the map

$$\operatorname{Map}_{\mathcal{C}}\left(X,Y\right)\to\operatorname{Map}_{h_{d}\mathcal{C}}\left(\theta_{d}\left(X\right),\theta_{d}\left(Y\right)\right)$$

is represented by the map

$$\theta\colon\hom_{\mathcal{C}}^{R}\left(X,Y\right)\to h_{d-1}\hom_{\mathcal{C}}^{R}\left(X,Y\right),$$

it will be enough to show that for every Kan complex $X$, the map $X\to h_{d-1}X$ is a $\left(d-1\right)$-truncation map. We prove this by induction. For $d\leq 0$ it is clear. For $d\geq 1$, recall that $\hom_{X}^{R}\left(p,q\right)$ has the homotopy type of the path space $P_{p,q}X$ between $p$ and $q$ in $X$ when viewed as a space. Thus, by induction, $\theta$ is a map of spaces that is surjective on $\pi_{0}$ and induces the $\left(d-2\right)$-truncation map on path spaces

$$P_{p,q}X\to P_{p,q}\left(h_{d-1}X\right)\simeq h_{d-2}\left(P_{p,q}X\right).$$

It follows that $\theta$ is a $\left(d-1\right)$-truncation map. ∎