Timeline for Concrete examples of Freyd-Mitchell embedding
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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Nov 14, 2023 at 9:57 | comment | added | Johannes Hahn | No. I mean the direct sum, i.e. the coproduct in the category of $A$-modules. | |
Nov 12, 2023 at 2:23 | comment | added | locally trivial | Even though you write "map $(X_*, \partial)$ to $\bigoplus_{n\in \mathbb{Z}} X_n$" above, do you consider $(...,x_{-1},x_0,x_1,...)$ to be an element of $X$ if infinitely-many $x_i$ are non-zero? i.e., if $A=\mathbb{Q}$ and $X_n=\mathbb{Q}$ for all $n$, then do you consider $(...,1,1,1,...)$ to be an element of $X$? | |
Nov 24, 2019 at 16:08 | vote | accept | Spencer Dembner | ||
Nov 24, 2019 at 0:51 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
Added another example without the condition $\mathbb{Q}\subseteq A$
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Nov 24, 2019 at 0:46 | comment | added | Johannes Hahn | I added another, more general construction that works without this simplifying assumption. | |
Nov 24, 2019 at 0:45 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
Added another example without the condition $\mathbb{Q}\subseteq A$
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Nov 23, 2019 at 21:10 | comment | added | Johannes Hahn | You need that the eigenvalues $n$ all act differently on $X$. If, say $X$ where $2$-torsion, then $X_1$ and $X_3$ are both contained in the $1$-eigenspace. You could of course weaken that and ask that $X$ is torsionfree as an abelian group for example, but then you would get the full category of all chain complexes. | |
Nov 23, 2019 at 18:36 | comment | added | Spencer Dembner | Thanks, I really like this. Could you explain how the argument uses that A is a $\mathbb{Q}$-algebra? | |
Nov 23, 2019 at 16:59 | history | answered | Johannes Hahn | CC BY-SA 4.0 |