Kerodon

Proof. If condition $(2)$ is satisfied, then the composite functor $(F \circ G): \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}$ also admits a right adjoint, given by the composition $(H \circ G): \operatorname{\mathcal{C}}_{/X} \rightarrow \operatorname{\mathcal{C}}_{/X}$ (Remark 6.2.1.8). Consequently, the implication $(2) \Rightarrow (1)$ follows from Remark 7.7.3.12 (and does not require the assumption that $\operatorname{\mathcal{C}}$ admits finite limits). We now prove the converse. Assume that condition $(1)$ is satisfied; we wish to show that the functor $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}_{/X}$ admits a right adjoint. Let $\operatorname{\mathcal{D}}\subseteq \operatorname{\mathcal{C}}_{/X}$ denote the full subcategory spanned by those objects $\widetilde{Y}$ for which the functor $C \mapsto \operatorname{Hom}_{ \operatorname{\mathcal{C}}_{/X} }( G(C), \widetilde{Y} )$ is representable by an object of $\operatorname{\mathcal{C}}$. By virtue of Corollary 6.2.4.2, it will suffice to show that $\operatorname{\mathcal{D}}$ contains every object $\widetilde{Y}$ of $\operatorname{\mathcal{D}}$.

Let us identify $\widetilde{Y}$ with a morphism $u: Y \rightarrow X$ in the $\infty $-category $\operatorname{\mathcal{C}}$. Then $u$ lifts to a morphism $\widetilde{u}: \widetilde{Y} \rightarrow \widetilde{X}$ of $\operatorname{\mathcal{C}}_{/X}$, corresponding to the degenerate $2$-simplex

in the $\infty $-category $\operatorname{\mathcal{C}}$. Let $\eta : \operatorname{id}_{ \operatorname{\mathcal{C}}_{/X} } \rightarrow G \circ F$ and $\epsilon : (F \circ G) \rightarrow \operatorname{id}_{\operatorname{\mathcal{C}}}$ be compatible unit and counit for an adjunction between $G$ and $F$. The we have a commutative diagram

in the $\infty $-category $\operatorname{\mathcal{C}}_{/X}$. Note that the image of (7.84) under the functor $F$ can be identified with the left side of a the commutative diagram

where the left square and outer rectangle are pullback squares. It follows from Proposition 7.6.2.28 that the left square is also a pullback, so that (7.84) is a pullback square in the $\infty $-category $\operatorname{\mathcal{C}}_{/X}$ (Corollary 7.1.6.19). Our assumption that $\operatorname{\mathcal{C}}$ admits finite limits guarantees that the subcategory $\operatorname{\mathcal{D}}\subseteq \operatorname{\mathcal{C}}_{/X}$ is closed under finite limits (Proposition 7.4.1.18). Consequently, to prove that $\widetilde{Y}$ belongs to $\operatorname{\mathcal{D}}$, it will suffice to show that the objects $\widetilde{X}$, $(G \circ F)( \widetilde{Y} )$, and $(G \circ F)( \widetilde{X})$ are contained in $\operatorname{\mathcal{D}}$. Note that each of these objects belongs to the essential image of the functor $G$ (the object $\widetilde{X}$ can obtained by applying $G$ to a final object of $\operatorname{\mathcal{C}}$). We are therefore reduced to proving that, for each object $Z \in \operatorname{\mathcal{C}}$, the functor

is representable by an object of $\operatorname{\mathcal{C}}$, which is a reformulation of assumption $(1)$. $\square$

Proof. Let $S$ be a Kan complex; we wish to show that the slice $\infty $-category $\operatorname{\mathcal{S}}_{/S}$ is cartesian closed. Let $(\operatorname{Set_{\Delta }})_{/S}$ denote the category of simplicial sets $X$ equipped with a map $f: X \rightarrow S$ and let $\operatorname{KFib}(S)$ denote the full subcategory of $(\operatorname{Set_{\Delta }})_{/S}$ spanned by those objects where $f$ is a Kan fibration. We regard $\operatorname{KFib}(S)$ as a (locally Kan) simplicial category, whose homotopy coherent nerve can be identified with $(\operatorname{\mathcal{S}})_{/S}$ (see Example 5.5.2.23). We will show that the simplicial category $\operatorname{KFib}(S)$ satisfies the hypotheses of Remark 7.7.3.9.

Note that the category $(\operatorname{Set_{\Delta }})_{/S}$ is cartesian closed. If $X$ and $Y$ are simplicial sets equipped with morphisms $f: X \rightarrow S$ and $g: Y \rightarrow S$, then there is an exponential $M = Y^{X}$ in the category $(\operatorname{Set_{\Delta }})_{/S}$, whose $n$-simplices are given by pairs $(\sigma , \rho )$, where $\sigma : \Delta ^ n \rightarrow S$ is an $n$-simplex of $S$ and $\rho : \Delta ^ n \times _{S} X \rightarrow \Delta ^ n \times _{S} Y$ is a morphism which is compatible with the projection to $\Delta ^ n$. To complete the proof, it will suffice to show that if $f$ and $g$ are Kan fibrations, then the projection map $M \rightarrow S$ is also a Kan fibration. Let $\iota : A \hookrightarrow B$ be an anodyne morphism of simplicial sets; we wish to show that every lifting problem

admits a solution. Invoking the universal property of $M$, we can rewrite (7.85) as a lifting problem

Since $g$ is a Kan fibration, we are reduced to showing that the morphism $\iota _{X}$ is a weak homotopy equivalence. This follows from Proposition 3.4.0.2, since $\iota $ is a weak homotopy equivalence and $f$ is a Kan fibration. $\square$

Proof. By virtue of Remark 7.7.3.15, we may assume that $\operatorname{\mathcal{C}}$ admits pullbacks. Then, for every object $Y \in \operatorname{\mathcal{C}}$, the slice $\infty $-category $\operatorname{\mathcal{C}}_{/Y}$ admits finite limits (Example 7.6.2.31). Applying Proposition 7.7.3.13, we conclude that $\operatorname{\mathcal{C}}_{/Y}$ is cartesian closed if and only if, for every morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$, the pullback functor $f^{\ast }: \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}_{/X}$ admits a right adjoint. Proposition 7.7.3.19 now follows by allowing the object $Y$ to vary. $\square$

Proof. By virtue of Corollary 7.7.2.33, it will suffice to show that for every morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$, the pullback functor $f^{\ast }: \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}_{/X}$ preserves $K$-indexed colimits. This follows from Corollary 7.1.4.22, since the functor $f^{\ast }$ has a right adjoint (Proposition 7.7.3.19). $\square$

Proof. Combine Corollary 7.7.3.20 with Theorem 7.7.2.8. f $\square$

Proof. Combine Proposition 7.7.3.17 with Corollary 7.7.3.21. $\square$