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I am studying Awodey's Category Theory (2006 edition) and in section 5.5 (page 94) there is Proposition 5.27 stating:

Representable functors preserve all limits

and the proof starts with the statement

It suffices to show that Hom($C$, −) preserves products and equalizers.

Am I wrong that this statement presumes that the category $\mathbf{C}$ of which $C$ is an object has products and equalizers? That is, could not a category $\mathbf{C}$ not having arbitrary products and equalizers possess nonetheless a limit of some diagram $D$? In this case can one still demonstrate that, given some object $C$ in $\mathbf{C}$, the representable functor $\text{Hom}(C,-)$ preserves this limit? I can see that arbitrary products and equalizers exist in the codomain ($\mathbf{Sets}$) of the representable functor, but must one also assume the same of the domain $\mathbf{C}$?

creillyucla
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  • These statements always mean "whenever the limit exists." – Randall Feb 13 '25 at 12:21
  • I don't think this answers my question. I am supposing that the limit of some arbitrary diagram $D$ in $\mathbf{C}$ exists, and I want to show that the representable functor preserves this limit. To this end does it help to show that the representable functor preserves products and equalizers if these constructs do not always exist in $\mathbf{C}$? The existence of a limit for some arbitrary diagram $D$ presumably does not imply the existence of general products and equalizers. – creillyucla Feb 13 '25 at 12:27
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    I see: good phrasing of the question. The point is that you will prove that if a product exists then so does the resulting product after applying the Hom, and that if an equalizer exists then so does.... You can then use this to show the all existing limits will be preserved since any existing limit can be rewritten as a product and an equalizer, all of which exist. Since you've already shown those are preserved, you're done. – Randall Feb 13 '25 at 12:37

1 Answers1

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You're right to be cautious. It's possible for a functor to preserve products and equalizers but not all limits - as long as the domain category isn't complete.

The example of the inclusion from finite abelian groups to all abelian groups is given in this answer. The inclusion preserves all coproducts and coequalizers (that exist), but not all colimits. Taking the opposite categories converts this to limits.

Still, the conclusion that representables preserve all limits is true. This property is often just the definition of limits since it allows us to reduce limits in arbitrary categories to limits in $\bf{Set}$, but for some proofs that use the limiting cone definition of limits and don't reduce to the cases of products and equalizers, see this question.

S.C.
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  • Good point that this has a lot to do with the particular category of sets. – Randall Feb 13 '25 at 12:41
  • "This property is often just the definition of limits...". Could you expand just a bit on this please? – creillyucla Feb 13 '25 at 13:07
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    It doesn't have much to do with this question, so I won't put it in the answer. First, limits of functors of the form $\mathcal{J} \to \bf{Set}$ are defined (as a certain set of natural transformations), then for $F: \mathcal{J} \to \mathcal{C}$, $\lim F$ is defined as the representing object of $\lim \hom(c, F(-))$. That is, $\hom(c, \lim F) \cong \lim \hom(c, F(-))$. nlab essentially does this, but in the end writes out the set-limit as that set of natural transformations: $\hom(c, \lim F) \simeq \hom(pt, \hom(c,F(-)))$. – S.C. Feb 13 '25 at 13:23