# Kerodon

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

Proposition 7.1.6.9. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories. Let $B$ be a simplicial set and let $A$ be a simplicial subset, so that $F$ induces a functor

$F': \operatorname{Fun}(B,\operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( A, \operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}(A, \operatorname{\mathcal{D}}) } \operatorname{Fun}( B, \operatorname{\mathcal{D}}).$

Suppose we are given a diagram $\overline{f}: K^{\triangleright } \rightarrow \operatorname{Fun}(B, \operatorname{\mathcal{C}})$ satisfying the following condition:

$(\ast )$

Let $\sigma : \Delta ^ n \rightarrow B$ be an $n$-simplex of $B$ which is not contained in $A$ and set $b = \sigma (0)$. Then the composite map $K^{\triangleright } \xrightarrow { \overline{f} } \operatorname{Fun}(B, \operatorname{\mathcal{C}}) \xrightarrow { \operatorname{ev}_ b } \operatorname{\mathcal{C}}$ is an $F$-colimit diagram in the $\infty$-category $\operatorname{\mathcal{C}}$.

Then $\overline{f}$ is an $F'$-colimit diagram in the $\infty$-category $\operatorname{Fun}(B,\operatorname{\mathcal{C}})$.

Proof. As in the proof of Corollary 7.1.6.6, we can replace $F$ by the restriction functor

$\operatorname{Fun}(K^{\triangleright }, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(K, \operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}(K, \operatorname{\mathcal{D}}) } \operatorname{Fun}(K^{\triangleright }, \operatorname{\mathcal{D}})$

and thereby reduce to the special case $K = \emptyset$ (Proposition 7.1.6.3). In this case, we view $\overline{f}$ as an object of the $\infty$-category $\operatorname{Fun}(B,\operatorname{\mathcal{C}})$, and we wish to show that this object is $F'$-initial.

Using Proposition 4.1.3.2, we can factor $F$ as a composition $\operatorname{\mathcal{C}}\xrightarrow {G} \operatorname{\mathcal{E}}\xrightarrow {U} \operatorname{\mathcal{D}}$, where $U$ is an inner fibration (so that $\operatorname{\mathcal{E}}$ is an $\infty$-category) and $G$ is inner anodyne (and therefore an equivalence of $\infty$-categories). Note that we have a commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(B,\operatorname{\mathcal{C}}) \ar [d]^{G \circ } \ar [r]^-{F'} & \operatorname{Fun}(A,\operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}(A, \operatorname{\mathcal{D}}) } \operatorname{Fun}(B,\operatorname{\mathcal{D}}) \ar [r] \ar [d] & \operatorname{Fun}(A, \operatorname{\mathcal{C}}) \ar [d]^{G \circ } \\ \operatorname{Fun}(B, \operatorname{\mathcal{E}}) \ar [r]^-{U'} & \operatorname{Fun}(A,\operatorname{\mathcal{E}}) \times _{ \operatorname{Fun}(A, \operatorname{\mathcal{D}}) } \operatorname{Fun}(B,\operatorname{\mathcal{D}}) \ar [r] & \operatorname{Fun}(A, \operatorname{\mathcal{E}}), }$

where the vertical maps on the left and right are equivalences of $\infty$-categories (Remark 4.5.1.16). Since the square on the right is a pullback diagram and the right horizontal maps are isofibrations (Corollary 4.4.5.3), it follows that the vertical map in the middle is also an equivalence of $\infty$-categories (Corollary 4.5.2.23). Consequently, to show that $\overline{f}$ is $F'$-initial, it will suffice to show that $G \circ \overline{f}$ is $U'$-initial when viewed as an object of $\operatorname{Fun}(B, \operatorname{\mathcal{E}})$ (Remark 7.1.4.9). Since $U'$ is an inner fibration (Proposition 4.1.4.1), it will suffice to verify that $\overline{f}$ satisfies the criterion of Corollary 7.1.4.17: every lifting problem

7.6
$$\begin{gathered}\label{equation:relative-colimit-pointwise-universal-strong} \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^ n \ar [r]^-{ \sigma _0 } \ar [d] & \operatorname{Fun}(B, \operatorname{\mathcal{E}}) \ar [d]^{ U' } \\ \Delta ^ n \ar@ {-->}[ur] \ar [r] & \operatorname{Fun}(A,\operatorname{\mathcal{E}}) \times _{ \operatorname{Fun}(A, \operatorname{\mathcal{D}}) } \operatorname{Fun}(B,\operatorname{\mathcal{D}}) } \end{gathered}$$

has a solution, provided that $n > 0$ and $\sigma _0(0) = \overline{f}$. Unwinding the definitions, we can rewrite (7.6) as a lifting problem

$\xymatrix@R =50pt@C=50pt{ (\operatorname{\partial \Delta }^ n \times B) \coprod _{(\operatorname{\partial \Delta }^ n \times A) } (\Delta ^ n \times B) \ar [r]^-{ g } \ar [d] & \operatorname{\mathcal{E}}\ar [d]^{U} \\ \Delta ^ n \times B \ar [r] & \operatorname{\mathcal{D}}. }$

Since $n > 0$, every vertex of the simplicial set $\Delta ^ n \times B$ is contained in $\operatorname{\partial \Delta }^ n \times B$. Moreover, if $\tau : \Delta ^ m \rightarrow \Delta ^ n \times B$ is an $m$-simplex which does not belong to $(\operatorname{\partial \Delta }^ n \times B) \coprod _{(\operatorname{\partial \Delta }^ n \times A) } (\Delta ^ n \times B)$, then condition $(\ast )$ (and Remark 7.1.4.9) guarantee that $g$ carries $\tau (0)$ to a $U'$-initial vertex of $\operatorname{\mathcal{E}}$. The existence of the desired solution now follows from Corollary 7.1.6.6 (applied in the special case $K = \emptyset$). $\square$