ScalingStacks

Proof. We first prove the case of $m=\infty$. For $n=-2$, there is nothing to prove, and so we assume that $n\geq-1$. Since $f\circ s=\operatorname{Id}_{B}$ we get $\tau_{\leq n}^{\mathcal{C}}\left(f\right)\circ\tau_{\leq n}^{\mathcal{C}}\left(s\right)=\operatorname{Id}_{B}$ and since $\tau_{\leq n}\left(f\right)$ is an equivalence, then so is $\tau_{\leq n}\left(s\right)$ and hence $s$ is $\left(n-\frac{1}{2}\right)$-connected. By 4.3.4, $s$ is $\left(n-1\right)$-connected and hence, by T.6.5.1.20, the map $f$ is $n$-connected (note that $n$-connective means $\left(n-1\right)$-connected).

For a general $m$, by T.6.4.1.5 there exists an $\infty$-topos $\mathcal{D}$ and an equivalence $\mathcal{C}\simeq\tau_{\leq m-1}\mathcal{D}$, and so we may identify $\mathcal{C}$ with the full subcategory of $\left(m-1\right)$-truncated objects of $\mathcal{D}$. If $f\colon A\to B$ is $\left(n-\frac{1}{2}\right)$-connected in $\mathcal{C}$, then it is also $\left(n-\frac{1}{2}\right)$-connected in $\mathcal{D}$, since the restriction of $\tau_{\leq n}^{\mathcal{D}}$ to $\mathcal{C}$ is equivalent to $\tau_{\leq n}^{\mathcal{C}}$. It follows from the case of $m=\infty$ that $f$ is $n$-connected in $\mathcal{D}$. Since $f=\tau_{\leq m-1}^{\mathcal{D}}f$ and $\tau_{\leq m-1}^{\mathcal{D}}$ is a left adjoint functor, by 4.2.5 the map $f$ is also $n$-connected as a map in $\mathcal{C}$. ∎