Timeline for functors unique up to self-equivalence of the source category
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2012 at 17:16 | comment | added | Colin McLarty | My answer was wrong. I too quickly identified equivalent categories. I'm voting the question up since I've learned from it already! Really the question should be put one step more generally: is there an intrinsic definition of "a functor up to equivalence of the domain and codomain categories"? | |
| Aug 18, 2012 at 16:24 | answer | added | Tom Leinster | timeline score: 4 | |
| Aug 17, 2012 at 19:06 | answer | added | Buschi Sergio | timeline score: 0 | |
| Aug 17, 2012 at 17:11 | history | edited | o a | CC BY-SA 3.0 |
added 964 characters in body
|
| Aug 17, 2012 at 16:51 | comment | added | o a | Todd Timple: I am interested in the relation between functors : two functors H:S-->T and H':S-->T are "weakly equivalent" iff there is a self-equivalence s:S-->S such that H and $H'\circ s$ are equivalent. I wanted to ask whether there is a higher category theoretic view on this relation between functors? And I wanted to see an example of a class of functors where any two functors are weakly equivalent for non-trivial reasons; in other words, a theorem claiming that functors with certain properties are necessarily weakly equivalent. I shall update the question accordingly..Thank you! | |
| Aug 17, 2012 at 15:45 | comment | added | Todd Trimble | One problem seems to be that every functor is well-defined (in your sense, which you erased, but you mean the special case where $s$ is the identity) by some property. Given a functor $F: C \to D$, define property P on functors $G: C \to D$ by the condition that $G = F$ identically. Any two functors satisfying property P are canonically isomorphic. This silly example indicates that you probably need to sharpen what you really intend by "well-defined". | |
| Aug 17, 2012 at 11:24 | comment | added | o a | Todd Trimble: yes, that is what I mean....But I removed the sentence your comment refers to. | |
| Aug 17, 2012 at 11:06 | history | edited | o a | CC BY-SA 3.0 |
deleted 337 characters in body
|
| Aug 17, 2012 at 11:06 | comment | added | o a | Thanks for your comments. Then I shall just remove the opening sentence: it seems rather to confuse than clarify....Does the second paragraph seems unclear as well ? | |
| Aug 17, 2012 at 10:54 | comment | added | Todd Trimble | I don't think he means "well-defined" in the usual sense. On the other hand, "well-defined" here hasn't been exactly, um, well-defined -- the opening sentence is a bit slippery. But to give an example, I think he means that if a category $C$ has products but not chosen products, then "the product functor" $C \times C \to C$, while not specified, exists and is "defined well enough" since any two choices are canonically isomorphic. A property P of functors would be "well-defining" if any two functors satisfying P are isomorphic, e.g., a universal property P. (But the question needs work.) | |
| Aug 17, 2012 at 8:57 | comment | added | Fernando Muro | A 'functor' which is not well defined is not a functor. | |
| Aug 17, 2012 at 8:19 | history | asked | o a | CC BY-SA 3.0 |