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Proposition 7.5.5.7. Let $\operatorname{\mathcal{C}}$ be a category and let $\overline{\mathscr {F}}: \operatorname{\mathcal{C}}^{\triangleleft } \rightarrow \operatorname{QCat}$ be a functor. The following conditions are equivalent:

$(1)$

The functor $\overline{\mathscr {F}}$ is a categorical limit diagram, in the sense of Definition 7.5.5.1.

$(2)$

For every simplicial set $K$, the functor

\[ \overline{\mathscr {F}}^{K}: \operatorname{\mathcal{C}}^{\triangleleft } \rightarrow \operatorname{QCat}\quad \quad C \mapsto \operatorname{Fun}( K, \overline{\mathscr {F}}(C) ) \]

is a categorical limit diagram.

$(3)$

For every simplicial set $K$, the functor

\[ (\overline{\mathscr {F}}^{K})^{\simeq }: \operatorname{\mathcal{C}}^{\triangleleft } \rightarrow \operatorname{Kan}\quad \quad C \mapsto \operatorname{Fun}( K, \overline{\mathscr {F}}(C) )^{\simeq } \]

is a homotopy limit diagram.

$(4)$

The functor $( \overline{\mathscr {F}}^{\Delta ^1} )^{\simeq }: \operatorname{\mathcal{C}}^{\triangleleft } \rightarrow \operatorname{Kan}$ is a homotopy limit diagram.

Proof. The implication $(1) \Rightarrow (2)$ follows from Remarks 7.5.2.3 and 4.5.1.16, the implication $(2) \Rightarrow (3)$ from Remark 7.5.5.5, and the implication $(3) \Rightarrow (4)$ is immediate. Set $\mathscr {F} = \overline{\mathscr {F}}|_{\operatorname{\mathcal{C}}}$, and let ${\bf 0}$ denote the initial object of $\operatorname{\mathcal{C}}^{\triangleleft }$. Using Remark 7.5.2.3 and Example 7.5.2.8, we see that condition $(4)$ is equivalent to the requirement that the map $\overline{ \mathscr {F} }( {\bf 0} ) \rightarrow \underset {\longleftarrow }{\mathrm{holim}}( \mathscr {F} )$ induces a homotopy equivalence of Kan complexes

\[ \operatorname{Fun}( \Delta ^1, \overline{ \mathscr {F} }( {\bf 0} ))^{\simeq } \rightarrow \operatorname{Fun}( \Delta ^1, \underset {\longleftarrow }{\mathrm{holim}}( \mathscr {F} ))^{\simeq } \simeq \underset {\longleftarrow }{\mathrm{holim}}( ( \overline{\mathscr {F}}^{\Delta ^1} )^{\simeq } ). \]

The implication $(4) \Rightarrow (1)$ now follows from Theorem 4.5.7.1. $\square$