Timeline for Does projective imply flat?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2015 at 9:44 | comment | added | Fernando Muro | @TheoJohnson-Freyd the argument I had in mind wrongly assumed that $\hom(-,I)$ had to be exact, where $\hom$ is the inner $\hom$ and $I$ is an injective object. But asking that is no different to asking projectives to be flat. Sorry for creating wrong expectations ;) | |
| Jul 1, 2015 at 5:34 | comment | added | Alex Degtyarev | @QiaochuYuan OK, I guess you are right: one needs "projective implies flat" to conclude that $\operatorname{Tor}$ does not depend on which argument is resolved. | |
| Jul 1, 2015 at 0:23 | vote | accept | Theo Johnson-Freyd | ||
| Jun 30, 2015 at 23:56 | comment | added | Qiaochu Yuan | @Alex: I think you can only conclude that Tor is zero if one of the arguments is flat; to conclude this from one of the arguments being projective you already need to know that projective implies flat. | |
| Jun 30, 2015 at 23:51 | answer | added | Daniel Schäppi | timeline score: 18 | |
| Jun 30, 2015 at 21:59 | comment | added | Theo Johnson-Freyd | @AlexDegtyarev Do you mind spelling out your abstract nonsense? Eric Wofsey in an answer below seems to provide a counterexample. | |
| Jun 30, 2015 at 19:41 | comment | added | Alex Degtyarev | Sorry. What I had in mind is: enough projectives implies the existence of $\operatorname{Tor}$ and appropriate exact sequence. Furthermore, $\operatorname{Tor}=0$ if one of the arguments is projective. So, in this case, $\otimes$ is exact. (BTW, another piece of general nonsense, with bicomplexes, shows that it doesn't matter which of the variables is resolved: I guess you need to resolve the first to conclude $\operatorname{Tor}=0$, whereas the exact sequence is given by resolving the second.) | |
| Jun 30, 2015 at 19:07 | answer | added | Eric Wofsey | timeline score: 20 | |
| Jun 30, 2015 at 18:35 | comment | added | Theo Johnson-Freyd | @AlexDegtyarev The thing I know how to do is to use the fact that projective implies flat to conclude that Tor groups can be computed by projectively resolving only one of the two variables. Please explain what you have in mind? All the categories I care about have enough projectives, so I don't mind assuming that as an axiom. | |
| Jun 30, 2015 at 17:02 | comment | added | Theo Johnson-Freyd | @FernandoMuro Yes, all the categories I care about have enough injectives, so I'm happy to add that as an axiom. If you have a proof available, I'll be happy to accept it as an answer. | |
| Jun 30, 2015 at 16:56 | comment | added | Fernando Muro | If you have enough injectives, then yes, would you be willing to assume this? | |
| Jun 30, 2015 at 16:15 | history | asked | Theo Johnson-Freyd | CC BY-SA 3.0 |