Timeline for Epimorphisms $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ are split
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2017 at 9:33 | answer | added | Jeremy Rickard | timeline score: 4 | |
| Aug 19, 2013 at 10:42 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Aug 16, 2013 at 15:26 | comment | added | Steve D | ... $Hom(A,\mathbb{Z})$ is non-trivial, and so there is a surjection from $A$ to $\mathbb{Z}$, which implies $A=\mathbb{Z}\oplus B$. If one assumes $A$ was finite rank, induction shows $A$ is free. Thus in a Whitehead group, subgroups of finite rank are free. By a (fairly easy) theorem of Pontraygin, this implies countable subgroups are free, which is what Stein really proved. | |
| Aug 16, 2013 at 15:24 | comment | added | Steve D | Have you seen the proof of Stein's theorem in Griffith's book on Infinite Abelian Groups? It uses the exact sequence $0\rightarrow\mathbb{Z}\rightarrow\hat{\mathbb{Z}}\rightarrow D\rightarrow0$, where $\hat{\mathbb{Z}}$ is the completion of $\mathbb{Z}$ and $D$ is the resulting (divisible) quotient. Applying the $Hom$ functor gives $0\rightarrow Hom(A,\mathbb{Z})\rightarrow Hom(A,\hat{\mathbb{Z}})\rightarrow Hom(A,D)\rightarrow0$. Now $Hom(A,D)$ is (non-trivial) divisible - because $D$ is divisible - and $Hom(A,\hat{\mathbb{Z}})$ is reduced - because $\hat{\mathbb{Z}}$ is reduced. Thus... | |
| Aug 16, 2013 at 13:05 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Aug 15, 2013 at 15:31 | comment | added | user9072 | In view of @HJRW comment I cannot resist but mentioning that this is one more reason not to rely that much on formulas and formatting details but to describe things verbally if possible. Not sure if this was the case here, but on a mobile device I use that renders MathJax well in general mathbb does not display as such but in a 'normal' font. | |
| Aug 15, 2013 at 15:01 | comment | added | HJRW | @MartinBrandenburg - apologies - I misread $\mathbb{N}$ for $N$! | |
| Aug 15, 2013 at 14:46 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Aug 15, 2013 at 14:24 | comment | added | Martin Brandenburg | @HJRW: My question deals with $\mathbb{Z}^{\mathbb{N}}$ (which, of course, is not Hopfian) and doesn't have to do with the Hopfian property. | |
| Aug 15, 2013 at 14:12 | comment | added | HJRW | You're asking why $\mathbb{Z}^n$ is Hopfian. One proof is to apply Mal'cev's theorem, which asserts that finitely generated, residually finite groups are Hopfian. | |
| Aug 15, 2013 at 13:24 | comment | added | Jeremy Rickard | @Fernando: If $P\to Q$ is not mono, then there is a non-zero map $\mathbb{Z}\to P$ such that the composition $\mathbb{Z}\to P\to Q$ is zero. Dualizing, there is a non-zero map $P^*\to\mathbb{Z}$ such that the composition $Q^*\to P^*\to\mathbb{Z}$ is zero, and so $Q^*\to P^*$ is not epi. So every epi $Q^*\to P^*$ comes from a free presentation of some Whitehead group. | |
| Aug 15, 2013 at 10:24 | comment | added | Fernando Muro | Martin, why is it so? | |
| Aug 15, 2013 at 10:16 | comment | added | Martin Brandenburg | My question is equivalent to the Whitehead problem for countable abelian groups. | |
| Aug 15, 2013 at 10:01 | comment | added | Asaf Karagila♦ | @Fernando: Yeah, it just occurred to me. Perhaps it's time to go to sleep... :-P | |
| Aug 15, 2013 at 10:00 | comment | added | Fernando Muro | @Asaf, I'm talking about the question, not about Whitehead's problem. | |
| Aug 15, 2013 at 9:56 | comment | added | Asaf Karagila♦ | @Fernando: The Whitehead problem was proved for countable groups. Martin is merely trying an equivalent approach in proving it. | |
| Aug 15, 2013 at 9:50 | comment | added | Fernando Muro | Do you happen to know it's true? | |
| Aug 15, 2013 at 9:01 | history | asked | Martin Brandenburg | CC BY-SA 3.0 |