Fix any \(n \geq -2\). A commutative square of spaces 
in which the maps are truncated and connected as indicated is necessarily a pullback square.
Proof.
Consider the commuting diagram 
where we have used that truncated and connected maps are stable under pullback. By Lemma 5.2.4, the map \(A \rightarrow B \times_D C\) is both \(n\)-connected and \(n\)-truncated, and so is an equivalence. ◻
Original source: arXiv:2401.02956v2
Original source · 2401.02956v2