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Suppose that $\mathbb{A},\mathbb{B}$ are finitely complete categories and $F:\mathbb{A}\to\mathbb{B}$ is a functor which reflects finite limits. Does $F$ reflect finitely generated limits?

(Here "finite limits" means limits over finite index categories, and "finitely generated limits" means limits over finitely generated index categories. This terminology is from "Handbook of categorical algebra vol. 1" by Borceux.)

This is basically a question about Proposition 2.9.8 from Borceux's "Handbook of categorical algebra vol. 1". I agree with Borceux about the part of the proposition dealing with the preservation of limits: just construct the limit as an equalizer of two arrows between finite products and observe that $F$ maps the whole construction into $\mathbb{B}$ producing a limit of a composition. However, unfortunately I cannot see how does this help with the reflection of limits.

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I think you can express a finitely generated limit as a finite limit. Given a finitely generated category $C$ with generating subgraph $G = (E \rightrightarrows V)$, let $G^\S$ be the category with objects $V+E$ and two nonidentity morphisms to each edge, from its source and target respectively. There are no nontrivial composites, so finiteness of $G$ implies finiteness of $G^\S$. Now if $F:C\to D$ is a functor, define $F^\S : G^\S \to D$ by $F^\S(v) = F(v)$ for each vertex of $G$, and $F^\S(e) = F(y)$ for each edge $e:x\to y$ of $G$, with the morphisms $F(x) \to F(y)$ and $F(y) \to F(y)$ being $F(e)$ and the identity respectively. Then limits of $F$ coincide with limits of $F^\S$, and the latter are finite; thus if finite limits are reflected, so are limits over $C$.

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  • $\begingroup$ Dear @Mike, thank you. Your construction solves my problem. $\endgroup$
    – ijon
    Mar 17, 2021 at 13:43

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