Timeline for How do I check if a functor has a (left/right) adjoint?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2009 at 22:19 | comment | added | Harrison Brown | marc: Actually, yeah, pretty much. | |
| Nov 17, 2009 at 18:16 | comment | added | marc | It's not all that hard to verify the definition directly in a lot of cases. Do you mean "not tedious"-to-check? | |
| Nov 17, 2009 at 7:35 | comment | added | Andrew Stacey | This is usually very easy to check. It's at least as easy as finding the initial morphism in Andrew Critch's answer because it's often easier to prove a general condition than find a specific instance. For example, for the adjoint of the inclusion functor you don't even need to know that abelianisation exists! It's enough that the inclusion AbGrp to Grp preserves the underlying sets and the adjunction follows for free. | |
| Nov 17, 2009 at 6:54 | comment | added | Harrison Brown | I think by "easy" I mean "quickly verifiable?" Analogously to checking that some limit really doesn't commute, to check that it's not. But if you could illustrate an example, yeah, that'd be fantastic. | |
| Nov 17, 2009 at 6:35 | comment | added | Greg Stevenson | It isn't always "easy" although this depends on your definition ;) but they are surprisingly checkable in some situations. I'll try and think of some good examples where this is used and edit my answer to include them if that would help. | |
| Nov 17, 2009 at 6:30 | comment | added | Harrison Brown | I did say "easy-to-check..." :) | |
| Nov 17, 2009 at 6:28 | history | answered | marc | CC BY-SA 2.5 |