Timeline for Commutativity of $\operatorname{Hom}$ and $\varprojlim$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2020 at 13:12 | comment | added | LSpice | Re, in particular, as @JeremyRickard says, the key object is in what category you're working. If you embed $\mathbb Q$ in $G = \langle\mathbb Q, x, y \mathrel: [x, y] = 1\rangle$, then the identity map on $\mathbb Q$ does not extend to $G \to \mathbb Q$—we left the category of Abelian groups. | |
| Aug 20, 2020 at 12:52 | answer | added | Jeremy Rickard | timeline score: 9 | |
| Aug 20, 2020 at 12:35 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
typo in the title
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| Aug 20, 2020 at 12:31 | comment | added | Jeremy Rickard | There are various proofs in, or referred to in, this thread. | |
| Aug 20, 2020 at 12:26 | history | edited | Andrey | CC BY-SA 4.0 |
added 13 characters in body
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| Aug 20, 2020 at 12:18 | comment | added | Andrey | Thank you. Yes, I mean an injective abelian group, but I do not understand why is an injective group trivial. For example, groups of rational numbers $\mathbb{Q}$ is an injective or sometimes is called infinity divisible group. Can you tell me any classical book where I can find it in details. | |
| Aug 20, 2020 at 11:07 | comment | added | Jeremy Rickard | Could you clarify what you mean by an injective group? Do you mean an injective abelian group? The only injective objects in the category of (not necessarily abelian) groups are trivial groups, so if that’s what you mean then the answer to your question is trivially “yes”! | |
| Aug 20, 2020 at 9:57 | review | Close votes | |||
| Aug 29, 2020 at 3:04 | |||||
| Aug 20, 2020 at 9:20 | review | First posts | |||
| Aug 20, 2020 at 14:27 | |||||
| Aug 20, 2020 at 9:15 | history | asked | Andrey | CC BY-SA 4.0 |