# Kerodon

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

Proposition 7.3.6.1. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories, let $\delta : K \rightarrow \operatorname{\mathcal{C}}$ and $F_0: K \rightarrow \operatorname{\mathcal{D}}$ be diagrams, and let $\beta : F_0 \rightarrow F \circ \delta$ be a natural transformation which exhibits $F$ as a left Kan extension of $F_0$ along $\delta$. Then, for every functor $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$, the composite map

$\operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) }( F, G ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(K, \operatorname{\mathcal{D}}) }( F \circ \delta , G \circ \delta ) \xrightarrow { \circ [\beta ] } \operatorname{Hom}_{\operatorname{Fun}(K,\operatorname{\mathcal{D}})}( F_0, G \circ \delta )$

is a homotopy equivalence of Kan complexes.

Proof of Proposition 7.3.6.1. Let $F, G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be functors of $\infty$-categories. Suppose we are given a simplicial set $K$ equipped with diagrams $\delta : K \rightarrow \operatorname{\mathcal{C}}$ and $F_0: K \rightarrow \operatorname{\mathcal{D}}$, together with a natural transformation $\beta : F_0 \rightarrow F \circ \delta$ which exhibits $F$ as a left Kan extension of $F_0$ along $\delta$. Let $\theta$ denote the composite map

$\operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) }( F, G ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(K, \operatorname{\mathcal{D}}) }( F \circ \delta , G \circ \delta ) \xrightarrow { \circ [\beta ] } \operatorname{Hom}_{\operatorname{Fun}(K,\operatorname{\mathcal{D}})}( F_0, G \circ \delta ).$

We wish to show that $\theta$ is a homotopy equivalence.

It follows from Corollary 4.1.3.3 that there exists an inner anodyne morphism $K \hookrightarrow \operatorname{\mathcal{K}}$, where $\operatorname{\mathcal{K}}$ is an $\infty$-category. Since $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are $\infty$-categories, we can extend $\delta$ and $F_0$ to functors $\delta ': \operatorname{\mathcal{K}}\rightarrow \operatorname{\mathcal{C}}$ and $F'_0: \operatorname{\mathcal{K}}\rightarrow \operatorname{\mathcal{D}}$, respectively (Proposition 1.4.6.7). Moreover, the restriction functor $\operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) \rightarrow \operatorname{Fun}(K, \operatorname{\mathcal{D}})$ is a trivial Kan fibration (Proposition 1.4.7.6). We can therefore extend $\beta$ to a natural transformation $\beta ': F'_0 \rightarrow F \circ \delta '$, which induces a map of Kan complexes $\theta ': \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) }( F, G ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F'_0, G \circ \delta ' )$. By construction, the map $\theta$ is obtained (up to homotopy) by composing $\theta '$ with the restriction map $\operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F'_0, G \circ \delta ' ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(K, \operatorname{\mathcal{D}}) }( F_0, G \circ \delta )$, which is a trivial Kan fibration. Consequently, to show that $\theta$ is a homotopy equivalence, it will suffice to show that $\theta '$ is a homotopy equivalence. We may therefore replace $K$ by $\operatorname{\mathcal{K}}$ and thereby reduce to proving Proposition 7.3.6.1 in the special case where $K = \operatorname{\mathcal{K}}$ is an $\infty$-category.

Let $\overline{\operatorname{\mathcal{C}}}$ denote the relative join $\operatorname{\mathcal{K}}\star _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}$. Note that the definition of $\theta$ (as a morphism in the homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}$) depends only on the homotopy class of $\beta$. We may therefore assume without loss of generality that there exists a functor $\overline{F}: \overline{\operatorname{\mathcal{C}}} \rightarrow \operatorname{\mathcal{D}}$ for which $\overline{F}|_{\operatorname{\mathcal{K}}} = F_0$, $\overline{F}|_{\operatorname{\mathcal{C}}} = F$, and the natural transformation $\beta$ is given by the composition

$\Delta ^1 \times \operatorname{\mathcal{K}}\simeq \operatorname{\mathcal{K}}\star _{\operatorname{\mathcal{K}}} \operatorname{\mathcal{K}}\rightarrow \operatorname{\mathcal{K}}\star _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}\xrightarrow {\overline{F}} \operatorname{\mathcal{D}}.$

Let $\overline{G}: \overline{\operatorname{\mathcal{C}}} \rightarrow \operatorname{\mathcal{D}}$ denote the functor given by the composition

$\operatorname{\mathcal{K}}\star _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}\star _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}\simeq \Delta ^1 \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}\xrightarrow {G} \operatorname{\mathcal{D}}.$

Our assumption on $\beta$ guarantees that $\overline{F}$ is left Kan extended from the full subcategory $\operatorname{\mathcal{K}}\subseteq \overline{\operatorname{\mathcal{C}}}$ (Proposition 7.3.2.10). Applying Corollary 7.3.6.9, we deduce that precomposition with the inclusion $\operatorname{\mathcal{K}}\hookrightarrow \overline{\operatorname{\mathcal{C}}}$ determines a trivial Kan fibration

$\varphi _{-}: \operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, G \circ \delta ).$

We claim that $\overline{G}$ is right Kan extended from the full subcategory $\operatorname{\mathcal{C}}\subseteq \overline{\operatorname{\mathcal{C}}}$. To prove this, it will suffice to show that for every object $X \in \operatorname{\mathcal{K}}$, the functor $\overline{G}$ is right Kan extended from $\operatorname{\mathcal{C}}$ at $X$ (see Proposition 7.3.3.5). Let $e_{X}: X \rightarrow \delta (X)$ denote the morphism in $\overline{\operatorname{\mathcal{C}}}$ given by the edge

$\Delta ^1 \simeq \{ X\} \star _{ \{ \delta (X) \} } \{ \delta (X) \} \hookrightarrow \operatorname{\mathcal{K}}\star _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}= \overline{\operatorname{\mathcal{C}}}.$

Note that $e_{X}$ is cocartesian with respect to the projection map $\overline{\operatorname{\mathcal{C}}} \rightarrow \Delta ^1$ (Proposition 5.2.3.15), and therefore exhibits $\delta (X)$ as a $\operatorname{\mathcal{C}}$-reflection of $X$ in the $\infty$-category $\overline{\operatorname{\mathcal{C}}}$ (Lemma 6.2.3.1). It will therefore suffice to show that $\overline{G}$ carries $e_{X}$ to an isomorphism in the $\infty$-category $\operatorname{\mathcal{D}}$, which is clear (by construction, $\overline{G}(e_ X)$ is the identity morphism $\operatorname{id}_{D}$ for $D = G( \delta (X) )$). Applying Corollary 7.3.6.9 again, we deduce that precomposition with the inclusion map $\operatorname{\mathcal{C}}\hookrightarrow \overline{\operatorname{\mathcal{C}}}$ determines a trivial Kan fibration

$\varphi _{+}: \operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) }( F, G).$

Let $\varphi _{\pm }: \operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}})}( F \circ \delta , G \circ \delta )$ be given by precomposition with the functor $\operatorname{\mathcal{K}}\xrightarrow {\delta } \operatorname{\mathcal{C}}\hookrightarrow \overline{\operatorname{\mathcal{C}}}$. Consider the diagram of Kan complexes

7.32
$$\begin{gathered}\label{equation:mapping-property-of-Kan} \xymatrix@R =50pt@C=45pt{ & \operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \ar [dl]_{\varphi _{+}} \ar [d]^{\varphi _{\pm }} \ar [dr]^{\varphi _{-}} & \\ \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) }( F, G ) \ar [r] & \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{C}})}( F \circ \delta , G \circ \delta ) \ar [r]^-{ \circ [\beta ] } & \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, G \circ \delta ). } \end{gathered}$$

Note that the diagonal maps are homotopy equivalences, and the triangle on the left is commutative. Consequently, to show that $\theta$ is a homotopy equivalence, it will suffice to show that the triangle on the right commutes up to homotopy.

Let $\operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, F \circ \delta , G \circ \delta )$ be the Kan complex introduced in Notation 4.6.8.1. To verify the homotopy commutativity of the right triangle in the diagram (7.32), it will suffice to show that there is exists map of Kan complexes $\rho : \operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, F \circ \delta , G \circ \delta )$ satisfying the following conditions:

• The composition

$\operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \xrightarrow {\rho } \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, F \circ \delta , G \circ \delta ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, F \circ \delta )$

is the constant map taking the value $\beta$.

• The composition

$\operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \xrightarrow {\rho } \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, F \circ \delta , G \circ \delta ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, G \circ \delta )$

is equal to $\varphi _{-}$.

• The composition

$\operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \xrightarrow {\rho } \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, F \circ \delta , G \circ \delta ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F \circ \delta , G \circ \delta )$

is equal to $\varphi _{\pm }$.

Let $\sigma$ denote the $2$-simplex of $\Delta ^1 \times \Delta ^1$ given on vertices by the formulae

$\sigma (0) = (0,0) \quad \quad \sigma (1) = (0,1) \quad \quad \sigma (2) = (1,1),$

and let $T: \Delta ^2 \times \operatorname{\mathcal{K}}\rightarrow \Delta ^1 \times \overline{\operatorname{\mathcal{C}}}$ be the functor given by the composition

\begin{eqnarray*} \Delta ^2 \times \operatorname{\mathcal{K}}& \xrightarrow {\sigma \times \operatorname{id}_{\operatorname{\mathcal{K}}} } & \Delta ^1 \times \Delta ^1 \times \operatorname{\mathcal{K}}\\ & \simeq & \Delta ^1 \times ( \operatorname{\mathcal{K}}\star _{\operatorname{\mathcal{K}}} \operatorname{\mathcal{K}}) \\ & \rightarrow & \Delta ^1 \times ( \operatorname{\mathcal{K}}\star _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}) \\ & = & \Delta ^1 \times \overline{\operatorname{\mathcal{C}}}. \end{eqnarray*}

More concretely, the functor $T$ is given on objects by the formulae

$T(0,X) = (0,X) \quad \quad T( 1,X ) = (0, \delta (X) ) \quad \quad T(2,X) = (1, \delta (X) ).$

We conclude by observing that precomposition with $T$ induces a map of Kan complexes

$\rho : \operatorname{Hom}_{ \operatorname{Fun}( \overline{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) }( \overline{F}, \overline{G} ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{K}}, \operatorname{\mathcal{D}}) }( F_0, F \circ \delta , G \circ \delta )$

having the desired properties. $\square$