Timeline for A particular functor on the category of abelian groups?
Current License: CC BY-SA 3.0
20 events
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| Apr 21, 2018 at 22:23 | review | Close votes | |||
| Apr 22, 2018 at 7:31 | |||||
| Apr 17, 2018 at 8:14 | comment | added | YCor | Possibly. Actually I don't know for uncountable groups: it a first step it might be better to first separately ask whether for some infinite cardinal $\kappa$ there exists a group of cardinal $\kappa$ in which every group of cardinal $\le\kappa$ embeds as a subgroup. (I just did an unsuccessful Google search). | |
| Apr 17, 2018 at 7:53 | comment | added | Ali Taghavi | @YCor so it seems that if I remove the word "Abelian"from my question then the question is nontrivial. Yes? | |
| Apr 17, 2018 at 7:51 | comment | added | Ali Taghavi | @YCor I did not pay attention that "Abelian" is not included in the linked question. So I wondered "is there a contradictory situation"?. So i sent you that link because i was surprised by that contradiction!!Any way sorry for my mistake. | |
| Apr 17, 2018 at 7:45 | comment | added | YCor | I know this standard fact: for countable groups instead of countable abelian groups, the result fails. Have you got any particular reason to point me to this link? | |
| Apr 17, 2018 at 7:30 | comment | added | Ali Taghavi | @YCor but please read the linked question and its answer. | |
| Apr 17, 2018 at 7:23 | comment | added | Ali Taghavi | @YCor yes. I was thinking to direct product. | |
| Apr 17, 2018 at 7:21 | comment | added | YCor | Sure it's countable (countable direct sum of countable groups). $A^{(B)}$ denotes $\bigoplus_{b\in B}A$. | |
| Apr 17, 2018 at 7:17 | comment | added | Ali Taghavi | @YCor but this group is not countable when k is countable. | |
| Apr 17, 2018 at 7:08 | comment | added | YCor | Yes forgot to say "of the same cardinal". Such a group is $(\mathbf{Q}\oplus\mathbf{Q}/\mathbf{Z})^{(\kappa)}$. | |
| Apr 17, 2018 at 6:57 | comment | added | Ali Taghavi | @YCor According to your argument, do not we need that the universal group you pointed out has the same cardinality $k$? And is it always possible? Please see mathoverflow.net/questions/28999/… | |
| Apr 16, 2018 at 4:36 | comment | added | Ali Taghavi | @YCor what is that group for the countable case? | |
| Apr 15, 2018 at 18:09 | comment | added | YCor | It's again not research level, because for every infinite cardinal $\kappa$ there's an abelian group in which every abelian group of cardinal $\le\kappa$ embeds. | |
| Apr 15, 2018 at 18:01 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 15, 2018 at 17:45 | comment | added | Ali Taghavi | @TomGoodwillie Thank you very much for your comment and your hint. I am sorry if I posted a non research question. According to your comment I try to improve it as follows. "Is there a functor $F$ on the category of infinite abelian groups which does not increase the cardinality but $F(G)$ can not be embedded in $G$? | |
| Apr 15, 2018 at 17:34 | history | undeleted | Ali Taghavi | ||
| Apr 15, 2018 at 3:02 | history | deleted | Ali Taghavi | via Vote | |
| Apr 15, 2018 at 2:43 | comment | added | Tom Goodwillie | Hint: First solve the problem for sets instead of abelian groups. | |
| Apr 15, 2018 at 2:37 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 15, 2018 at 2:25 | history | asked | Ali Taghavi | CC BY-SA 3.0 |