Timeline for Objects which can't be defined without making choices but which end up independent of the choice
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 3, 2016 at 13:03 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Jun 5, 2013 at 13:56 | comment | added | user1437 | Tor using torsion products as described in Mac Lane's book "Homology" does not use resolutions at all, and does not make choices. It's just a bit impractical to work with it. | |
| Jun 5, 2013 at 9:06 | comment | added | darij grinberg | Possibly related: mathoverflow.net/questions/50971/… | |
| Jun 5, 2013 at 3:27 | history | edited | José Hdz. Stgo. | CC BY-SA 3.0 |
added 18 characters in body
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| May 21, 2013 at 18:42 | comment | added | Ryan Reich | @zeb Does it count as making choices if you do so by constructing $\operatorname{Ext}^n$ the usual way and showing it's isomorphic to Steven's definition? After all, it's not the definition that contains the choices, but simply a proof that the definition has a certain property. | |
| May 21, 2013 at 2:45 | comment | added | zeb | It seems like now you need to argue that the collection of isomorphism classes of n-step extensions actually forms a set, though. | |
| May 20, 2013 at 22:44 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
edited body
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| May 20, 2013 at 22:44 | comment | added | Steven Landsburg | Toink: I'm pretty sure there's not. | |
| May 20, 2013 at 22:42 | comment | added | Toink | Nice. Is there a construction of this sort for $Tor$? | |
| May 20, 2013 at 22:31 | history | answered | Steven Landsburg | CC BY-SA 3.0 |