ScalingStacks

Proof. Since $\mathcal{C}$ has all pullbacks, every commutative square $q\colon\Delta^{1}\times\Delta^{1}\to\mathcal{C}$ of the form

can be factored as

By 4.1.3, the space of lifts for the original square $q$ is equivalent to the space of lifts in the left square of the above rectangle. Moreover, $n$-truncated morphisms are closed under base change and so to check that $f$ is $n$-connected, we can equivalently restrict ourselves to checking the left orthogonality condition only for squares $q$ in which the map $B\to Y$ is the identity on $B$. Writing $\overline{A}$, $\overline{X}$ and $\overline{B}$ for $A\to B$, $X\to B$ and $\operatorname{Id}\colon B\to B$ as objects of $\mathcal{C}_{/B}$, respectively, we see that by the dual of T.5.5.5.12 the space of lifts fits into a fiber sequence

$$L\left(q\right)=\operatorname{Map}_{\mathcal{C}_{A//B}}\left(\overline{B},\overline{X}\right)\to\operatorname{Map}_{\mathcal{C}_{/B}}\left(\overline{B},\overline{X}\right)\xrightarrow{f^{*}}\operatorname{Map}_{\mathcal{C}_{/B}}\left(\overline{A},\overline{X}\right).$$

Hence, $f$ is $n$-connected if and only if $f^{*}$ is an equivalence for every $n$-truncated morphism $X\to B$. By T.5.5.6.10, a morphism $X\to B$ is $n$-truncated if and only if $\overline{X}$ is an $n$-truncated object of $\mathcal{C}_{/B}$. Hence, we need the above map to be an equivalence for every $n$-truncated object $\overline{X}\in\mathcal{C}_{/B}$. This precisely means that the map $\overline{A}\to\overline{B}$ exhibits $\overline{B}$, the terminal object of $\mathcal{C}_{/B}$, as the $n$-truncation of $\overline{A}$. ∎