Timeline for How would you say that a small category is embedded into functors from a large $C'$ to abelian groups?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2015 at 16:56 | vote | accept | Mikhail Bondarko | ||
| Apr 8, 2013 at 6:15 | comment | added | Mikhail Bondarko | 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 $C'$! | |
| Apr 8, 2013 at 5:16 | comment | added | Eric Wofsey | 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. | |
| Apr 8, 2013 at 5:13 | comment | added | Eric Wofsey | 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". | |
| Apr 8, 2013 at 5:07 | comment | added | Mikhail Bondarko | 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 $k$. | |
| Apr 8, 2013 at 4:48 | history | answered | Eric Wofsey | CC BY-SA 3.0 |