There is a result, which I have heard is due to Grothendieck, which says that a left exact functor $F : C \to \text{Set}$ is a filtered colimit of representable functors provided that $C$ is essentially small and has finite limits. When I tried proving this, I found that I didn't need to use the hypothesis that $C$ is essentially small, but the diagram that I'm taking a filtered colimit over is not necessarily essentially small as a result.

Neither Wikipedia nor nLab require that filtered categories are essentially small. Is this in fact a requirement in the definition of filtered colimits? Or am I missing a detail in the proof?


The answer to the title question is, of course, no: the category $\mathbf{On}$ of all ordinals is filtered and not essentially small.

If you do not assume $\mathcal{C}$ is essentially small, then it is certainly possible that $F$ is a colimit of a filtered diagram that is not essentially small. The only problem is that colimits of such diagrams need not exist – so in some sense this is an "accidental" colimit.

Here's one way of deducing your generalisation from the essentially small case: let $\mathbf{U}$ be the original set-theoretic universe and let $\mathbf{U}^+$ be a universe so large that $\mathbf{U} \in \mathbf{U}^+$ and $\mathcal{C}$ is locally $\mathbf{U}$-small and essentially $\mathbf{U}^+$-small. The inclusion $\mathbf{Set} \hookrightarrow \mathbf{Set}^+$ preserves finite limits, so $F$ still preserves finite limits considered as a functor $\mathcal{C} \to \mathbf{Set}^+$. Thus $F$ is a colimit of an essentially $\mathbf{U}^+$-small filtered diagram of representable functors $\mathcal{C} \to \mathbf{Set}^+$; but the inclusion $[\mathcal{C}, \mathbf{Set}] \hookrightarrow [\mathcal{C}, \mathbf{Set}^+]$ reflects all colimits, so $F$ is a colimit of a filtered diagram of representable functors $\mathcal{C} \to \mathbf{Set}$ as well. (Obviously, the property of being a filtered category does not depend on the choice of universe.)

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  • $\begingroup$ The proof I had in mind is that $F$ is a colimit over a diagram of shape $\text{el}(F)^{op}$, and under the hypotheses that $C$ has finite limits and $F$ preserves them, $\text{el}(F)^{op}$ is filtered. Is it necessary to say any more than this? I found the additional hypothesis of essential smallness in a book on shape theory but I don't really understand how it was used in the proof. $\endgroup$ – Qiaochu Yuan Jun 23 '13 at 19:46
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    $\begingroup$ That is the standard proof as far as I know. Bear in mind that many results we have about filtered colimits are, strictly speaking, for colimits over small filtered diagrams – so your generalisation may be less useful than at first sight. $\endgroup$ – Zhen Lin Jun 23 '13 at 19:48
  • $\begingroup$ Hmm. Okay, that makes sense. Thanks! $\endgroup$ – Qiaochu Yuan Jun 23 '13 at 20:00
  • $\begingroup$ Ah, apparently what I was confused about is that the definitions of pro-representability and pro-objects require small diagrams. $\endgroup$ – Qiaochu Yuan Jun 23 '13 at 20:38
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    $\begingroup$ Yes, that is a small subtlety. It is similar to the pitfall of saying that $[\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ is the free colimit-completion of $\mathcal{C}$ – this is true when $\mathcal{C}$ is essentially small but can fail when $\mathcal{C}$ is not essentially small. $\endgroup$ – Zhen Lin Jun 23 '13 at 21:43

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