# Kerodon

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

Proposition 7.1.7.19 (Base Change). Suppose we are given a commutative diagram of $\infty$-categories

7.6
$$\begin{gathered}\label{equation:base-change-relative-limit5} \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}' \ar [rr]^{F'} \ar [dr]^{U'} \ar [dd]^{G} & & \operatorname{\mathcal{D}}' \ar [dl] \ar [dd] \\ & \operatorname{\mathcal{E}}' \ar [dd] & \\ \operatorname{\mathcal{C}}\ar [dr]^{ U } \ar [rr]^{F} & & \operatorname{\mathcal{D}}\ar [dl]^{V} \\ & \operatorname{\mathcal{E}}& } \end{gathered}$$

where each square is a pullback and the diagonal maps are inner fibrations. Let $\overline{f}: K^{\triangleright } \rightarrow \operatorname{\mathcal{C}}'$ be a morphism of simplicial sets. Then:

$(1)$

If $G \circ \overline{f}$ is an $F$-colimit diagram in the $\infty$-category $\operatorname{\mathcal{C}}$, then $\overline{f}$ is an $F'$-colimit diagram in the $\infty$-category $\operatorname{\mathcal{C}}'$.

$(2)$

Assume that $U$ and $V$ are cartesian fibrations, and that the functor $F$ carries $U$-cartesian morphisms of $\operatorname{\mathcal{C}}$ to $V$-cartesian morphisms of $\operatorname{\mathcal{D}}$. If $\overline{f}$ is an $F'$-colimit diagram in the $\infty$-category $\operatorname{\mathcal{C}}'$, then $G \circ \overline{f}$ is an $F$-colimit diagram in the $\infty$-category $\operatorname{\mathcal{C}}$.

Proof. Set $f = \overline{f}|_{K}$. By virtue of Corollary 4.3.6.10 and Proposition 5.1.4.17, we can replace (7.6) by the commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}'_{f/} \ar [rr] \ar [dr] \ar [dd] & & \operatorname{\mathcal{D}}'_{ (F' \circ f)/} \ar [dl] \ar [dd] \\ & \operatorname{\mathcal{E}}'_{ (U' \circ f)/} \ar [dd] & \\ \operatorname{\mathcal{C}}_{ (G \circ f)/ } \ar [dr] & & \operatorname{\mathcal{D}}_{ (F \circ G \circ f)/} \ar [dl] \\ & \operatorname{\mathcal{E}}_{ (U \circ G \circ f)/} & }$

and thereby reduce to the special case $K = \emptyset$. In this case, the desired result follows from Proposition 7.1.6.15. $\square$