Kerodon

Proof. Using Proposition 4.1.3.2, we can factor $F$ as a composition $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{C}}' \xrightarrow {F''} \operatorname{\mathcal{D}}$, where $F'$ is inner anodyne and $F''$ is an inner fibration. Using Remark 8.4.1.18, we see that the functor $F'$ is dense. Replacing $F$ by $F''$, we can reduce to proving Proposition 8.4.1.23 in the special case where $\operatorname{\mathcal{C}}$ is an $\infty $-category.

Let $\gamma $ denote the identity map $\operatorname{id}_{F}$, which we regard as a natural transformation from $F$ to $\operatorname{id}_{\operatorname{\mathcal{D}}} \circ F$. Our assumption that $F$ is dense guarantees that $\gamma $ exhibits $\operatorname{id}_{\operatorname{\mathcal{D}}}$ as a left Kan extension of $F$ along $F$. To avoid confusion, let us write $F_0$ to denote the functor $F$, regarded as a functor from $\operatorname{\mathcal{C}}$ to $\operatorname{\mathcal{D}}_0$. Let $\iota : \operatorname{\mathcal{D}}_0 \hookrightarrow \operatorname{\mathcal{D}}$ denote the inclusion map, so that $F = \iota \circ F_0$. We can therefore also regard the identity map $\operatorname{id}_{F}$ as a natural transformation $\alpha : F \rightarrow \iota \circ F_0$. Our assumption that $F$ is dense also guarantees that $\alpha $ exhibits $\iota $ as a left Kan extension of $F$ along $F_0$. Note that $\gamma = \operatorname{id}_{F}$ is a composition of $\alpha = \operatorname{id}_{F}$ with $\beta |_{ \operatorname{\mathcal{C}}}$, where $\beta = \operatorname{id}_{ \iota }$ is the identity transformation from $\iota $ to itself. Invoking the transitivity of Kan extensions (Proposition 7.3.8.18), we deduce that $\beta $ exhibits the identity functor $\operatorname{id}_{\operatorname{\mathcal{D}}}$ as a left Kan extension of $\iota $ along itself: that is, the functor $\iota $ is dense. Applying Example 8.4.1.16, we conclude that $\operatorname{\mathcal{D}}_0$ is a dense subcategory of $\operatorname{\mathcal{D}}$. $\square$