Timeline for Does every retraction of free groups arise from projection to a subset of a freely generating set?
Current License: CC BY-SA 2.5
9 events
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| May 25, 2010 at 23:40 | comment | added | HJRW | Steve, my apologies, I misunderstood you. You're quite right - the point is that H is determined beforehand. | |
| May 25, 2010 at 23:37 | comment | added | Steve D | @Henry: Yes, there is. See Proposition 2.12 in Lyndon-Schupp, pg. 9 , for example. But it's clear this depends on the complement $H$ you choose, as in your example (where the complement <a> would be OK). | |
| May 25, 2010 at 23:32 | comment | added | HJRW | Vipul, as the example I have added shows, that isn't quite the point. In the example, the image of g does freely generate. | |
| May 25, 2010 at 23:29 | comment | added | Vipul Naik | Steve: Here's where I think you're confused. It's true that a subset B of A maps surjectively to H but that doesn't mean it (or its image) freely generates H. | |
| May 25, 2010 at 23:25 | history | edited | HJRW | CC BY-SA 2.5 |
added 303 characters in body
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| May 25, 2010 at 23:18 | comment | added | HJRW | What do you mean by 'surely'? There isn't. | |
| May 25, 2010 at 23:04 | vote | accept | Vipul Naik | ||
| May 25, 2010 at 22:58 | comment | added | Steve D | If $F_1$ maps onto $F_2$ , surely there is a basis $A$ of $F_1$ with subset $B$ such that <$B$> maps isomorphically onto $F_2$ and $A-B$ to the identity. Is the problem then that the subgroup $H$ is being determined beforehand? | |
| May 25, 2010 at 22:11 | history | answered | HJRW | CC BY-SA 2.5 |