Timeline for Is the category of cochain complexes with terms in an additive category a 2-category?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2023 at 4:20 | comment | added | მამუკა ჯიბლაძე | Hm I think I do not take into account that one has to compose homotopies between morphisms of complexes rather than between arbitrary elements of $\hom(A,B)$. The difference is that morphisms of complexes are cocycles of $\hom(A,B)$ rather than arbitrary cochains. It is not clear to me whether the 2-structure should also incorporate composition of homotopies between these more general maps. | |
| Apr 27, 2023 at 13:26 | comment | added | Elías Guisado Villalgordo | $\def\hom{\operatorname{hom}}$ @მამუკაჯიბლაძე On that case, how would one exactly compose $I\to\hom(A,B)$ with $I\to\hom(B,C)$? My guess for the tensor product is equation \eqref{eq}, but I don't know what to do with $\vee$. | |
| Apr 27, 2023 at 13:20 | history | edited | Elías Guisado Villalgordo | CC BY-SA 4.0 |
Added computation for possible definition of horizontal composition (and showed that it fails to satisfy the law of the middle four interchange))
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| Apr 27, 2023 at 11:43 | comment | added | მამუკა ჯიბლაძე | Sorry, that was not correct. It should be, I believe, something like this. $I$ is freely generated by two cohomologous cochains. That is, $I$ has generators $x_0,x_1,y$ with the relation $x_0-x_1=dy$. Now when I wrote $I\otimes I$ I should in fact write $I\vee I:=I\oplus I/(0,x_1)\sim(x_0,0)$. The map $\delta:I\to I\vee I$ is given by $\delta(x_0)=(0,x_0)$, $\delta(x_1)=(x_1,0)$, and $\delta(y)=(y,0)+(0,y)$. | |
| Apr 26, 2023 at 18:15 | comment | added | Elías Guisado Villalgordo | @მამუკაჯიბლაძე Do you know what the "appropriate map $I\to I\otimes I$" could be? I computed all such chain maps and it seems to be no canonical choice. | |
| Jun 9, 2022 at 10:16 | comment | added | მამუკა ჯიბლაძე | Yes, I believe - using appropriate $I\to I\otimes I$. | |
| Jun 9, 2022 at 9:36 | comment | added | Elías Guisado Villalgordo | @მამუკაჯიბლაძე And from that is there an obvious way to "compose" a morphism $I\to\operatorname{Hom}(A, B)$ with a morphism $I\to\operatorname{Hom}(B, C)$ to get a morphism $I\to\operatorname{Hom}(A, C)$? | |
| Jun 8, 2022 at 19:24 | comment | added | მამუკა ჯიბლაძე | For Qiaochu Yuan's definition you do not need an interval object in your category. His definition of chain homotopy is a homomorphism of complexes of abelian groups $I\to\operatorname{Hom}(A,B)$, and in complexes of abelian groups you do have an appropriate $I$ | |
| Jun 8, 2022 at 16:59 | history | asked | Elías Guisado Villalgordo | CC BY-SA 4.0 |