Timeline for How do I check if a functor has a (left/right) adjoint?
Current License: CC BY-SA 2.5
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2009 at 11:53 | comment | added | Andrew Stacey | Thinking about this some more, I'd say that this was a good example of what I was trying to get at in the "theorems for nothing" question; particularly in light of Andrew Critch's first comment (dated Nov 17th) above. | |
| Nov 17, 2009 at 9:25 | comment | added | Greg Stevenson | @Andrew: Thanks - I was thinking from the point of view of using the theorem to build an adjoint, but that answers my question. | |
| Nov 17, 2009 at 9:08 | history | edited | Andrew Critch | CC BY-SA 2.5 |
added 41 characters in body
|
| Nov 17, 2009 at 8:57 | history | edited | Andrew Critch | CC BY-SA 2.5 |
compactified
|
| Nov 17, 2009 at 8:48 | history | edited | Andrew Critch | CC BY-SA 2.5 |
added 1 characters in body
|
| Nov 17, 2009 at 8:43 | history | edited | Andrew Critch | CC BY-SA 2.5 |
added 41 characters in body; added 6 characters in body
|
| Nov 17, 2009 at 8:40 | comment | added | Andrew Critch | @Yemen, it is "just" that :) @Greg, not quite: you don't build the unit, you start with it. Then you build a functor to make the unit an actual natural transformation. | |
| Nov 17, 2009 at 8:34 | history | edited | Andrew Critch | CC BY-SA 2.5 |
typesetting
|
| Nov 17, 2009 at 8:16 | history | edited | Andrew Critch | CC BY-SA 2.5 |
regrouped; Post Made Community Wiki
|
| Nov 17, 2009 at 8:12 | comment | added | Greg Stevenson | Well secretly you are building the unit and showing it gives bijections on hom-sets. It is a nice fact, and I am all for using the various definitions of adjunction as the situation warrants, but I feel like this still boils down to a sufficient condition for G to have an adjoint is that you can build one. Am I missing something? (By the way - I don't want this to sound snide - it is a good answer) | |
| Nov 17, 2009 at 8:09 | comment | added | Yemon Choi | Isn't this "just" the characterization in terms of initial objects in comma categories? The work lies in showing uniquness of various morphisms, I'd have thought. In the course I took (waves at TL) it wasn't clear that this was a faster way to construct the free group functor than, say, applying GAFT. It all depends how (over)confident one is that there really is an adjoint pair | |
| Nov 17, 2009 at 8:04 | comment | added | Andrew Critch | No! It says "if you can build just a tiny bit of an adjoint, than the rest of it falls into place." I edited my answer to elaborate on this, because I think this fact doesn't get enough attention in general :) | |
| Nov 17, 2009 at 8:03 | history | edited | Andrew Critch | CC BY-SA 2.5 |
added 415 characters in body
|
| Nov 17, 2009 at 7:50 | history | edited | Andrew Critch | CC BY-SA 2.5 |
added 264 characters in body
|
| Nov 17, 2009 at 7:39 | comment | added | Andrew Stacey | To me, this result says "A functor has an adjoint if you can construct the adjoint." Freyd's adjoint functor theorem is more like "A functor has an adjoint if it looks like it has an adjoint" which seems more useful. | |
| Nov 17, 2009 at 7:37 | history | edited | Andrew Critch | CC BY-SA 2.5 |
typesetting
|
| Nov 17, 2009 at 7:28 | history | edited | Andrew Critch | CC BY-SA 2.5 |
added 57 characters in body; deleted 1 characters in body
|
| Nov 17, 2009 at 7:12 | history | edited | Andrew Critch | CC BY-SA 2.5 |
typo; added 5 characters in body
|
| Nov 17, 2009 at 7:07 | history | edited | Andrew Critch | CC BY-SA 2.5 |
typesetting; deleted 75 characters in body
|
| Nov 17, 2009 at 6:59 | history | edited | Andrew Critch | CC BY-SA 2.5 |
shortened
|
| Nov 17, 2009 at 6:50 | history | answered | Andrew Critch | CC BY-SA 2.5 |