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How would you say that a small additive category $C$ embedds (contravariantly) into the category of exact functors from a 'large' abelian $C'$ into abelian groups (this is something like Yoneda's embedding, but $C$ does not map canonically into $C'$)? My problem is that I do not want to consider all functors from $C'$ into abelian groups since this functor category it 'very large' (and I do not want to consider a 'larger universe'). Certainly, I can try to consider a limit of the corresponding functors from small subcategories of $C'$; yet is there a better way to deal with this matters?

Upd. Actually, my $C'$ is just isomorphic to the category of additive functors from $C^{op}$ to abelian groups (though this is not the way how it is defined).

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  • $\begingroup$ Why not just take the Yoneda embedding on $C$ itself, or some appropriate small subcategory of $C'$ that contains $C$? $\endgroup$
    – Eric Wofsey
    Commented Apr 8, 2013 at 3:22
  • $\begingroup$ I don't have any reasonable small subcategory inside $C'$. I also recollected that my $C$ does not canonically map into $C'$; I only have a bifunctor $C\times C'\to Ab$. $\endgroup$
    – Mikhail Bondarko
    Commented Apr 8, 2013 at 4:15

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In your example, there likely exists a cardinal $\kappa$ such that $C'$ is $\kappa$-accessible, as are the functors $C'\to Ab$ associated to each object of $C$ (i.e., they preserve $\kappa$-filtered colimits). In this case, you can consider $C$ embedded into the category of $\kappa$-accessible functors from $C'$ to $Ab$, which is only "large" rather than "very large". This is equivalent to restricting to the subcategory of $\kappa$-compact objects of $C'$.

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  • $\begingroup$ Thank you! Sorry; I only just recollected that my $C'$ is isomorphic to the category of all additive functors from $C$ to abelian groups. So, the objects that come from an embedding of $C$ into $C'$ do yield compact generators. Yet are $k$-directed colimits suffice to obtain $C'$ from $C$? This is probably wrong for any $k$. $\endgroup$
    – Mikhail Bondarko
    Commented Apr 8, 2013 at 5:07
  • $\begingroup$ If $C$ contains less than $\kappa$ morphisms, then it should be easy to write any $C$-shaped diagram in $Ab$ as a $\kappa$-filtered colimit of diagrams in which all the abelian groups have size less than $\kappa$. Given any element of any of the groups of the diagram, just take the subdiagram that it "generates". $\endgroup$
    – Eric Wofsey
    Commented Apr 8, 2013 at 5:13
  • $\begingroup$ This may be easier to understand if you think about additive functors on $C$ as "modules" over $C$, generalizing the case when $C$ has a single object and is a ring. $\endgroup$
    – Eric Wofsey
    Commented Apr 8, 2013 at 5:16
  • $\begingroup$ Yes, this is quite correct! The category of $R$-modules is quite actual for me; yet I don't want to fix this setting. My problem is: I have a theorem that expresses $C'$ in terms of $C$, and I would also like to express $C$ in terms of functors that its objects induce on $C'$. Yet there are 'too many' functors from $C'$! $\endgroup$
    – Mikhail Bondarko
    Commented Apr 8, 2013 at 6:15

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