Kerodon

Proof of Theorem 9.5.1.1 from Theorems 9.5.1.8 and 9.5.1.9. Let $\operatorname{\mathcal{C}}$ be an accessible $\infty $-category. Assume first that $\operatorname{\mathcal{C}}$ is cocomplete; we will show that $\operatorname{\mathcal{C}}$ is complete. Using Theorem 8.3.3.13, we can identify $\operatorname{\mathcal{C}}$ with the full subcategory $\operatorname{Fun}^{ \mathrm{rep}}( \operatorname{\mathcal{C}}^{\operatorname{op}}, \operatorname{\mathcal{S}}) \subseteq \operatorname{Fun}( \operatorname{\mathcal{C}}^{\operatorname{op}}, \operatorname{\mathcal{S}})$ spanned by the representable functors $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \operatorname{\mathcal{S}}$. Since the $\infty $-category of spaces $\operatorname{\mathcal{S}}$ is complete (Corollary 7.4.1.3), the functor $\infty $-category $\operatorname{Fun}( \operatorname{\mathcal{C}}^{\operatorname{op}}, \operatorname{\mathcal{S}})$ is also complete (Proposition 7.1.8.2). It will therefore suffice to show that the full subcategory $\operatorname{Fun}^{\mathrm{rep}}( \operatorname{\mathcal{C}}^{\operatorname{op}}, \operatorname{\mathcal{S}})$ is closed under small limits in $\operatorname{Fun}( \operatorname{\mathcal{C}}^{\operatorname{op}}, \operatorname{\mathcal{S}})$. This follows from the characterization of representable functors given by Theorem 9.5.1.8, together with Remark 7.6.6.23.

We now prove the converse. Assume that $\operatorname{\mathcal{C}}$ is complete; we wish to show that $\operatorname{\mathcal{C}}$ is cocomplete, or equivalently that the opposite $\infty $-category $\operatorname{\mathcal{C}}^{\operatorname{op}}$ is complete. Using Theorem 8.3.3.13, we can identify $\operatorname{\mathcal{C}}^{\operatorname{op}}$ with the full subcategory $\operatorname{Fun}^{ \mathrm{corep}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{S}}) \subseteq \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})$ spanned by the corepresentable functors $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{S}}$. Arguing as above, we see that the $\infty $-category $\operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})$ admits small limits. It will therefore suffice to show that the full subcategory $\operatorname{Fun}^{\mathrm{corep}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})$ is closed under small limits. Suppose we are given a small simplicial set $K$ and a diagram $G: K \rightarrow \operatorname{Fun}^{\mathrm{corep}}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})$; we wish to show that the limit $\varprojlim (G)$ (formed in the $\infty $-category $\operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})$) is also corepresentable. By virtue of Theorem 9.5.1.9, it will suffice to show that $\varprojlim (G)$ is continuous and accessible. As above, the continuity follows from Remark 7.6.6.23. Let us identify $G$ with a functor $g: \operatorname{\mathcal{C}}\rightarrow \operatorname{Fun}(K, \operatorname{\mathcal{S}})$, so that $\varprojlim (G)$ is obtained by composing $g$ with the limit functor $\varprojlim : \operatorname{Fun}(K, \operatorname{\mathcal{S}}) \rightarrow \operatorname{\mathcal{S}}$ (Corollary 7.1.8.4). Note that the functor $\varprojlim $ is accessible: in fact, it preserves small $\kappa $-filtered colimits for any regular cardinal $\kappa $ for which $K$ is $\kappa $-small (Theorem 9.1.5.9). It will therefore suffice to show that the functor $g$ is accessible, which is a special case of Remark 9.4.8.2. $\square$