# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

Proposition 3.4.1.9 (Transitivity). Suppose we are given a commutative diagram of simplicial sets

$\xymatrix { Z \ar [r] \ar [d]^{h} & Y \ar [r] \ar [d]^{g} & X \ar [d]^{f} \\ U \ar [r] & T \ar [r] & S }$

where the square on the right is homotopy Cartesian. Then the left square is homotopy Cartesian if and only if the outer rectangle is a homotopy Cartesian.

Proof. By virtue of Proposition 3.1.6.1, the morphism $f$ factors as a composition $X \xrightarrow {w_ X} X' \xrightarrow {f'} S$, where $f'$ is a Kan fibration and $w_{X}$ is a weak homotopy equivalence. Using Proposition 3.1.6.1 again, we can factor the induced map $Y \rightarrow T \times _{S} X'$ as a composition $Y \xrightarrow { w_{Y} } Y' \xrightarrow {\overline{g}} T \times _{S} X'$, where $\overline{g}$ is a Kan fibration and $w_{Y}$ is a weak homotopy equivalence. Repeating this argument, we can factor the induced map $Z \rightarrow U \times _{T} Y'$ as a composition $Z \xrightarrow { w_{Z} } Z' \xrightarrow {\overline{h}} U \times _{T} Y'$, where $\overline{h}$ is a Kan fibration and $w_{Z}$ is a weak homotopy equivalence. We then obtain a commutative diagram

$\xymatrix { Z \ar [r] \ar [d]^{w_ Z} & Y \ar [r] \ar [d]^{w_ Y} & X \ar [d]^{w_ X} \\ Z' \ar [r] \ar [d]^{h'} & Y' \ar [r] \ar [d]^{g'} & X' \ar [d]^{f'} \\ U \ar [r] & T \ar [r] & S }$

where the upper vertical maps are weak homotopy equivalences and the lower vertical maps are Kan fibrations. It follows from our assumption (and Remark 3.4.1.8) that the lower right square in this diagram is homotopy Cartesian. To complete the proof, it will suffice (again using Remark 3.4.1.8) to show that the lower left square is homotopy Cartesian if and only if the lower rectangle is homotopy Cartesian. By virtue of the criterion of Example 3.4.1.4, we are reduced to showing that for each vertex $u \in U$ having images $t \in T$ and $s \in S$, respectively, the induced map $Z'_{u} \rightarrow Y'_{t}$ is a homotopy equivalence of Kan complexes if and only if the composite map $Z'_{u} \rightarrow Y'_{t} \rightarrow X'_{s}$ is a homotopy equivalence of Kan complexes. This follows from the two-out-of-three property (Remark 3.1.5.7), since the map of fibers $Y'_{t} \rightarrow X'_{s}$ is a weak homotopy equivalence (by the criterion of Example 3.4.1.4). $\square$