Timeline for A 2-category of chain complexes, chain maps, and chain homotopies?
Current License: CC BY-SA 3.0
15 events
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| Apr 28, 2023 at 7:46 | comment | added | Elías Guisado Villalgordo | Hey, I tried to define horizontal composition using the representation $I\to\operatorname{Hom}^\bullet(A,B)$ for chain homotopies and I arrived to a formula that still doesn't satisfy the law of middle four interchange (so still this would not answer the OP question). I wrote what I did here. Did I do something wrong? | |
| Jan 30, 2012 at 19:42 | comment | added | Anton Fetisov | @Qiauchu, yes, you can factor out higher homotopies, but this is inherently evil. I just wanted to underline that such approach doesn't generalize neatly to higher categories. Also the question was about homotopies per se, and they behave badly. | |
| Jan 30, 2012 at 2:50 | comment | added | Arthur | Sorry I meant *Qiaochu | |
| Jan 30, 2012 at 2:49 | comment | added | Arthur | @Qiauchu, yeah you do get a strict 2-category in this case. I think Anton is probably considering higher chain homotopies because I made a comment of composing j-cells over k-cells. Ralph's comment below shows that the horizontal composition is associative on the nose (in either case). However, I checked that the interchange law holds only up to higher chain homotopy. I haven't checked higher morphisms, but I guess his point is that things don't end up holding on the nose when you consider higher compositions... | |
| Jan 29, 2012 at 23:20 | comment | added | Qiaochu Yuan | @Anton: isn't the point of taking homotopy classes of homotopies to quotient out by the action of those 3-cells? I am pretty sure we get an honest bicategory, perhaps even a (strict) 2-category, this way. | |
| Jan 29, 2012 at 22:38 | comment | added | Anton Fetisov | Strictly speaking, what you get this way isn't a bicategory, it is a $(\infty,1)$-category, as they call it in nLab. Your multiplication of 2-cells is defined and associative only up to a coherent action of 3-cells. And multiplication of 3-cells is ok only up to 4-cells. This is just the same in topology: if you have a homotopy from $f$ to $g$ and from $g$ to $h$, you don't have a uniquely defined homotopy from $f$ to $h$! There are numerous ways to contract $[0;2]$-interval into a $[0;1]$-interval. | |
| Jan 29, 2012 at 4:12 | vote | accept | Arthur | ||
| Jan 29, 2012 at 4:06 | comment | added | Arthur | Not only that, but this also allows you to figure out what the compositions of j-cells over k-cells are (I think) using explicit formulas. Do you end up with all compositions involving some form of addition? And only the composition of k-cells over 0-cells uses the diagonal map? For example, the composition of 2-cells (chain homotopies) over 1-cells (chain maps) is defined by ordinary addition (this is the vertical composition). I figured out the horizontal composition by brute force but using this viewpoint I expect it to come out of using the diagonal map on chain complexes more naturally. | |
| Jan 29, 2012 at 3:49 | comment | added | Arthur | Yeah, this is pretty beautiful. You immediately get the algebraic condition for when two $(n-1)$-chain homotopies are homotopy equivalent by applying a chain map $I \otimes \cdots \otimes I \otimes C \to D$ ($n$ interval objects) to the ``face'' of the $n$-cube and viewing that as an $n$-chain homotopy between the two. This gives a very elegant algebraic expression that looks just like the usual chain homotopy condition for the first level. The differences between the two chain homotopies I wrote above satisfy this requirement, so this solves the problem! Thanks again! :) | |
| Jan 29, 2012 at 1:08 | comment | added | Arthur | Yeah, I completely forgot to input the homotopy condition. Thanks for the reminder and also the quick response. Perhaps this viewpoint will allow me to find this condition as well as the conditions for higher chain homotopies. I can see that it works for the 1-category description (namely, such an $H$ produces the $f,g,$ and $\sigma$ by considering the restriction to 1,0, and $\mathrm{span}(e)$ respectively). This should take care of those leftover terms. For the higher homotopies the idea is very similar--restrict to the appropriate corners, boundaries, and faces. Awesome! | |
| Jan 29, 2012 at 1:04 | comment | added | Dylan Wilson | Equivalently, a homotopy is a map $H: C \otimes I \rightarrow D$, and again you can be guided by the topology :). And $I$ didn't come from nowhere: it is the simplicial chain complex associated to the unit interval. It is also equivalent to what you get when you take the normalized complex of the simplicial abelian group $\mathbb{Z}\Delta^1$... etc. | |
| Jan 28, 2012 at 23:35 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Jan 28, 2012 at 23:30 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Jan 28, 2012 at 23:24 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Jan 28, 2012 at 23:16 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |