Kerodon

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Warning 9.2.5.24. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{S}}$ be a functor of $\infty $-categories. It follows from Corollary 9.2.5.23 that if $F$ is left exact, then it is flat. Moreover, the converse holds if $\operatorname{\mathcal{C}}$ admits finite limits. Beware that the converse does not hold in general. For example, let $R$ be an associative ring, let $\operatorname{\mathcal{E}}$ be (the nerve of) the category of free left $R$-modules of finite rank, and let

\[ F: \operatorname{\mathcal{E}}^{\operatorname{op}} \rightarrow \operatorname{N}_{\bullet }(\operatorname{Set}) \subseteq \operatorname{\mathcal{S}}\quad \quad F(M) = \operatorname{Hom}_{R}(M,R) \]

be the functor represented by $R$. Then $F$ is a flat functor. Using Remark 9.2.5.16, we see that $F$ is left exact if and only if, for every set $S$, the functor

\[ F^{S}: \operatorname{\mathcal{E}}^{\operatorname{op}} \rightarrow \operatorname{N}_{\bullet }(\operatorname{Set}) \subseteq \operatorname{\mathcal{S}}\quad \quad F(M) = \operatorname{Hom}_{R}(M,R^{S}) \]

is also flat. By virtue of Example 9.2.4.3, this is equivalent the requirement that $R^{S}$ is a flat left $R$-module. This condition is satisfied only when the ring $R$ is right coherent: that is, every finitely generated right ideal $I \subseteq R$ is finitely presented as a right $R$-module. See Theorem 2.1 of [chase].