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Proposition Let $\operatorname{\mathcal{C}}$ be a filtered $\infty $-category. Then $\operatorname{\mathcal{C}}$ is weakly contractible.

Proof. By virtue of Proposition, there exists a functor $Q: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Set_{\Delta }}$ and a natural transformation $u: \operatorname{id}_{\operatorname{Set_{\Delta }}} \rightarrow Q$ with the following properties:

  • The functor $Q$ commutes with filtered colimits.

  • For every simplicial set $X$, the simplicial set $Q(X)$ is a Kan complex.

  • For every simplicial set $X$, the morphism $u_{X}: X \rightarrow Q(X)$ is a weak homotopy equivalence.

To show that $\operatorname{\mathcal{C}}$ is weakly contractible, it will suffice to show that the Kan complex $Q(\operatorname{\mathcal{C}})$ is contractible. Note that $\operatorname{\mathcal{C}}$ is nonempty, so that $Q(\operatorname{\mathcal{C}})$ is also nonempty. It will therefore suffice to show that for every integer $n \geq 0$, every morphism of simplicial sets $\sigma : \Delta ^ n / \operatorname{\partial \Delta }^ n \rightarrow Q(\operatorname{\mathcal{C}})$ is nullhomotopic (Proposition Since the simplicial set $\Delta ^ n / \operatorname{\partial \Delta }^ n$ is finite and the functor $Q$ commutes with filtered colimits, the morphism $\sigma $ factors as a composition $\Delta ^{n} / \operatorname{\partial \Delta }^{n} \rightarrow Q(K) \xrightarrow {Q(\iota )} Q(\operatorname{\mathcal{C}})$, where $K$ is a finite simplicial subset of $\operatorname{\mathcal{C}}$ and $\iota : K \hookrightarrow \operatorname{\mathcal{C}}$ denotes the inclusion map. We will complete the proof by showing that $Q(\iota )$ is nullhomotopic. Since $u_{K}: K \rightarrow Q(K)$ is a weak homotopy equivalence, this is equivalent to assertion that the composite morphism $Q(\iota ) \circ u_{K} = u_{\operatorname{\mathcal{C}}} \circ \iota $ is nullhomotopic. This is clear: our assumption that $\operatorname{\mathcal{C}}$ is filtered guarantees that there exists a natural transformation from $\iota $ to a constant diagram $K \rightarrow \operatorname{\mathcal{C}}$ (Remark $\square$