Kerodon

Proof. Assume that $\operatorname{\mathcal{E}}$ is weakly contractible; we will show that $\operatorname{\mathcal{E}}$ is filtered (the converse follows from Proposition 9.1.1.13). Suppose that we are given a diagram $e: K \rightarrow \operatorname{\mathcal{E}}$, where $K$ is a finite simplicial set. We wish to show that there exists a natural transformation from $e$ to a constant diagram. Choose a monomorphism $K \hookrightarrow L$, where $L$ is another finite simplicial set which is weakly contractible (for example, we can take $L = K^{\triangleright }$). Let $\operatorname{Ex}^{\infty }( \operatorname{\mathcal{E}})$ be the simplicial set given by Construction 3.3.6.1, so that $\operatorname{Ex}^{\infty }( \operatorname{\mathcal{E}})$ is a Kan complex (Proposition 3.3.6.9). Let $\rho ^{\infty }: \operatorname{\mathcal{E}}\rightarrow \operatorname{Ex}^{\infty }( \operatorname{\mathcal{E}})$ be the weak homotopy equivalence of Proposition 3.3.6.7. Since $\operatorname{\mathcal{E}}$ is weakly contractible, the Kan complex $\operatorname{Ex}^{\infty }(\operatorname{\mathcal{E}})$ is contractible. It follows that the composite map $K \xrightarrow {e} \operatorname{\mathcal{E}}\xrightarrow { \rho ^{\infty } } \operatorname{Ex}^{\infty }( \operatorname{\mathcal{E}})$ can be extended to a map $e^{+}: L \rightarrow \operatorname{Ex}^{\infty }( \operatorname{\mathcal{E}})$. Since the simplicial set $L$ is finite, the morphism $\overline{e}$ factors through $\operatorname{Ex}^{m}( \operatorname{\mathcal{E}})$ for some $m \gg 0$ (see Proposition 3.6.1.9), and therefore corresponds to a diagram $f: \operatorname{Sd}^{m}(L) \rightarrow \operatorname{\mathcal{E}}$ such that $f|_{ \operatorname{Sd}^{m}(K) }$ is given by the composition $\operatorname{Sd}^{m}(K) \twoheadrightarrow K \xrightarrow {e} \operatorname{\mathcal{E}}$. Note that the quotient map $\operatorname{Sd}^{m}(K) \twoheadrightarrow K$ is universally localizing (Proposition 6.3.7.2) and therefore right cofinal (Corollary 7.2.1.11). We may therefore replace $K$ by $\operatorname{Sd}^{m}(K)$ and $L$ by $\operatorname{Sd}^{m}(L)$, and thereby reduce to the case where $m = 0$: that is, the diagram $e$ admits an extension $f: L \rightarrow \operatorname{\mathcal{E}}$. Using Lemma 9.1.3.1, we can choose a natural transformation $\beta : f \rightarrow f'$, where $f': L \rightarrow \operatorname{\mathcal{E}}$ factors through the Kan complex $\operatorname{\mathcal{E}}_{C} = \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$ for some object $C \in \operatorname{\mathcal{C}}$. Since $L$ is weakly contractible, the morphism $f'$ is nullhomotopic (when regarded as a diagram in the Kan complex $\operatorname{\mathcal{E}}_{C}$). We may therefore assume without loss of generality that $f' = \underline{X}$ is the constant diagram associated to some object $X \in \operatorname{\mathcal{E}}_{C}$. Restricting to the simplicial subset $K \subseteq L$, we obtain a natural transformation from $e = f|_{K}$ to the constant diagram $f'|_{K}$. $\square$