Timeline for Baer sums of extensions
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27 at 21:19 | comment | added | Fernando Muro | In the category of abelian groups $\mathbb{Z}$ has no torsion and multiplication by $n$ is not an automorphism. | |
| Nov 27 at 20:13 | comment | added | R. van Dobben de Bruyn | Ah sorry, I guess I didn't mean the Baer sum itself, but rather functoriality in $A$. That tells you that $n[E]$ is represented by the pushout of $A \to E$ along $n \colon A \to A$. This explains why you get something isomorphic to $E$ if $n$ is invertible and a split extension if $n$ acts by $0$ on $A$. | |
| Nov 27 at 18:53 | comment | added | Jeremy Rickard | It's confusing to use $E$ to denote both an element of $\operatorname{Ext}^1$ and an object of $\mathcal {A}$. | |
| Nov 27 at 18:06 | answer | added | kindasorta | timeline score: 1 | |
| Nov 27 at 17:52 | comment | added | kindasorta | As for your question, we should take a direct sum of $n$-fold copies of $E$, which correspond to an extension of the $n$-fold direct sum of $B$ by that of $A$, and then pullback along the $n$-fold diagonal, and quotienting by the kernel of the summation. | |
| Nov 27 at 17:50 | comment | added | kindasorta | I agree about the peculiarity of the $n=0$ case, but assuming $A$ has no $n$ torsion or that the entire category has no torsion, replacing $\eta$ changes the embedding of $A$ in $E$, hence results in a possibly non-isomorphic extension. | |
| Nov 27 at 17:15 | comment | added | R. van Dobben de Bruyn | The question is unclear to me. For $n = 0$, multiplication by $n$ in $\operatorname{Ext}^1(B,A)$ replaces $E$ by the split extension $A \oplus B$, so cannot be obtained in this way. What do you mean by 'isomorphic extensions' — that the new $E'$ you get is abstractly isomorphic to $E$? (Of course you will not get an isomorphic short exact sequence.) Btw, since you know about Baer sum, you certainly know how to compute $n[E]$ in $\operatorname{Ext}^1(B,A)$, right? | |
| Nov 27 at 16:03 | history | edited | kindasorta | CC BY-SA 4.0 |
added 105 characters in body
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| Nov 27 at 15:00 | history | asked | kindasorta | CC BY-SA 4.0 |