ScalingStacks

Proof. By the Yoneda lemma it is enough to show that for every $n$-truncated object $Z$ in $\mathcal{C}$ the induced map

is an equivalence. For this, it is enough to show that for every $g\colon A\to Z$, the fiber of $f_{*}$ over $g$ is contractible. By T.5.5.5.12, the fiber is equivalent to the space of lifts for the square

which is contractible by definition as $f\colon A\to B$ was assumed to be $n$-connected. ∎

Proof. For $\mathcal{C}=\mathcal{S}$, this follows from inspecting the induced map between the long exact sequences of homotopy groups associated with the vertical maps. For $\mathcal{C}=\mathcal{S}^{K}$, this follows from the claim for $\mathcal{S}$, since both truncation and pullbacks are computed level-wise. A general $\infty$-topos is a left exact localization of $\mathcal{S}^{K}$ for some $K$, and left exact colimit-preserving functors between presentable $\infty$-categories commute with truncation by T.5.5.6.28 and with pullbacks by assumption. Finally, by T.6.4.1.5 every $m$-topos is the full subcategory on $\left(m-1\right)$-truncated objects in an $\infty$-topos and this full subcategory is closed under limits. ∎

Proof. To show that $f\colon A\to B$ is $n$-connected, we need to show that the space of lifts for every square

in which the right vertical arrow is $n$-truncated, is contractible. Applying 4.3.3 and 4.1.3, we see that this space is equivalent to the space of lifts in the square

which, by 4.1.4, is equivalent to the space of lifts in the adjoint square

which is contractible since the left vertical arrow is an equivalence. ∎

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 $\mathcal{C}$. ∎