Timeline for How to distinguish between natural and unnatural equivalences of categories
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2009 at 21:18 | comment | added | Harry Gindi | The single dual map is not covariant... | |
| Nov 11, 2009 at 5:53 | comment | added | Dinakar Muthiah | On the other hand, you can try and do the same thing with the single dual. By making a lot of choices, you can get maps from each finite dimensional vector space to its single dual. But there is no obvious way to promote this single dual operator to a covariant functor in such a way that the collection of maps we have chosen will give a natural transformation. What I'm asking is whether there is an analogous way to express the difference between strong and weak equivalence? Is there some structure you can add to test whether a given pair of inverse functors is strong equivalence? | |
| Nov 11, 2009 at 5:48 | comment | added | Dinakar Muthiah | What I mean is the following. We can define the double dual of a vector space without mentioning functors. We can define the map from a vector space to its double dual without mentioning functors. And we can do this for all vector spaces at the same time. Now to say it is natural, we're saying that this collection of maps are actually a natural transformation between two functors. So our maps have been promoted to natural transformations because the identity and double dual operator have been promoted to functors. | |
| Nov 5, 2009 at 18:14 | comment | added | Tom Leinster | Dinakar, I wonder if you're getting confused between some different concepts here. It's a matter of fact that the double dual functor from Vect to Vect is isomorphic (=naturally isomorphic) to the identity functor. But that has little to do with the distinction between "natural" and "unnatural", or "strong" and "weak", equivalences of categories. | |
| Nov 5, 2009 at 13:53 | comment | added | Dinakar Muthiah | You don't. I'm saying maybe there is a covariant functor out there that on objects takes vector spaces to their single dual (probably not, but i don't know how to prove that). The map from a vector space to its double dual doesn't make explicit mention of functors, but here there is an obvious covariant functor that is the double dual on objects. | |
| Nov 5, 2009 at 5:52 | comment | added | S. Carnahan♦ | How do you make a natural transformation between a covariant and a contravariant functor? | |
| Nov 5, 2009 at 5:32 | comment | added | Dinakar Muthiah | Suppose I don't want to deny the axiom of choice. Then is there some extra property or structure that the strong equivalence will have that the weak one won't? The example I'm having in mind is the isomorphism from a fin. dim. vector space to it's double dual. These isomorphisms have the nice property of also being natural transforms between two functors. The isomorphism from a vector space to it's single dual probably doesn't have this property, but I'm not interested in proving that. I want a positive test so I can point to a pair of functors and say they give a strong equivalence. | |
| Nov 5, 2009 at 5:15 | comment | added | Mike Shulman | In fact, in the absence of choice, most categories will not even HAVE a skeleton. | |
| Nov 5, 2009 at 3:40 | history | answered | Tom Leinster | CC BY-SA 2.5 |