Timeline for Why pullback only defined up-to-isomorphism but nevertheless presented as functor?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 28, 2020 at 14:08 | history | edited | Almeo Maus | CC BY-SA 4.0 |
added 4 characters in body
|
| Jun 27, 2020 at 20:09 | history | edited | Almeo Maus | CC BY-SA 4.0 |
added 1 character in body
|
| Dec 13, 2012 at 6:16 | vote | accept | Almeo Maus | ||
| Dec 12, 2012 at 3:18 | history | edited | Almeo Maus | CC BY-SA 3.0 |
edited body
|
| Dec 11, 2012 at 23:44 | answer | added | David Roberts♦ | timeline score: 13 | |
| Dec 11, 2012 at 20:15 | comment | added | Andrej Bauer | Oh I see, there is functoriality of $h^{*}$ and then there is functoriality of ${}^{*}$. I was talking about the latter, and you were talking about the former (which is functorial). | |
| Dec 11, 2012 at 17:49 | history | edited | Charles Staats | CC BY-SA 3.0 |
improved commutative diagram formatting
|
| Dec 11, 2012 at 16:25 | comment | added | Todd Trimble | I agree, Zhen. I have just posted an answer regarding this. | |
| Dec 11, 2012 at 16:24 | answer | added | Todd Trimble | timeline score: 18 | |
| Dec 11, 2012 at 16:07 | comment | added | Zhen Lin | @Andrej: Yes, the map $h \mapsto h^*$ is only pseudofunctorial – but that's not what Awodey is talking about here, nor what Almeo is asking about. | |
| Dec 11, 2012 at 15:53 | vote | accept | Almeo Maus | ||
| Dec 11, 2012 at 16:44 | |||||
| Dec 11, 2012 at 15:12 | vote | accept | Almeo Maus | ||
| Dec 11, 2012 at 15:38 | |||||
| Dec 11, 2012 at 14:43 | comment | added | Steven Landsburg | Andrej: "No you won't get a functor"......you're right of course. | |
| Dec 11, 2012 at 14:27 | history | edited | Almeo Maus | CC BY-SA 3.0 |
added 1 characters in body; added 1 characters in body
|
| Dec 11, 2012 at 14:22 | comment | added | Andrej Bauer | @Zhen: no, he write "a functor" because there are many functors, and pullback is one of them. Had he written "let $F$ be a pullback functor" then your comment would be relevant, as in this case we would be talking about one of many different pullback functors (all of which are naturlly isomorphic). I think it is not Awodey who is sloppy here. | |
| Dec 11, 2012 at 14:19 | comment | added | Andrej Bauer | No you won't get a functor, because functoriality will in general hold only up to isomorphism. | |
| Dec 11, 2012 at 14:17 | answer | added | Andrej Bauer | timeline score: 18 | |
| Dec 11, 2012 at 14:17 | history | edited | Almeo Maus | CC BY-SA 3.0 |
added 10 characters in body; deleted 4 characters in body
|
| Dec 11, 2012 at 14:16 | comment | added | Steven Landsburg | Zhen Lin: Yes, he writes a functor, but he also writes the pullback, so Almeo's point is well taken. The writing is indeed sloppy. Almeo: Zhen's solution does work, though: For each diagram, arbitrarily choose a pullback --- then you'll get a (non-uniquely-defined) functor. | |
| Dec 11, 2012 at 14:08 | comment | added | Zhen Lin | Just choose one pullback for each cospan, and show that any family of choices defines a functor. (That is why Awodey writes "a functor"!) | |
| Dec 11, 2012 at 14:03 | history | asked | Almeo Maus | CC BY-SA 3.0 |