Kerodon

Remark 3.4.1.12. In the situation of Proposition 3.4.1.11, suppose that the left half and outer rectangle of (3.43) are homotopy pullback squares. If the map $\pi _0(U) \rightarrow \pi _0(T)$ is surjective, then the right half of the diagram is also a homotopy pullback square. To prove this, we can use Proposition 3.1.8.1 to reduce to the case where each of the vertical maps is a Kan fibration. In this case, it will suffice to show that for each vertex $t \in T$ having image $s \in S$, the map of fibers $Y_{t} \rightarrow X_{s}$ is a homotopy equivalence of Kan complexes (Example 3.4.1.4). This condition depends only on the connected component of the vertex $t$, so we may assume without loss of generality that $t$ is the image of a vertex $u \in U$. The desired result now follows by applying the two-out-of-three property (Remark 3.1.7.7) to the diagram