Timeline for Delooping a weak $E_1$ map by bar construction
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2023 at 11:03 | vote | accept | ChesterX | ||
| Jun 14, 2023 at 9:51 | comment | added | Connor Malin | @ChesterX Yes, that's correct provided $Z$ is connected. | |
| Jun 14, 2023 at 4:03 | comment | added | ChesterX | I am still a bit confused about the $A_\infty$-multiplication part! If I understand your argument correctly, then you are saying that if $\mu_Z : \Omega Z \times \Omega Z \to \Omega Z$ is an $A_\infty$ map, then $Z$ is a loop space. In particular, $\Omega Z$ is actually an $E_2$-space. Am I correct? | |
| Jun 13, 2023 at 18:36 | comment | added | Connor Malin | @ChesterX And the easiest reason to see why the bar construction produces an $A_\infty$ map is to just use the rectification theorem for $A_\infty$ monoids which means we can assume every multiplication is strictly associative. Then just apply bar constructions to the associativity of the multiplication $\Omega Z$. | |
| Jun 13, 2023 at 18:15 | history | edited | Connor Malin | CC BY-SA 4.0 |
added 123 characters in body
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| Jun 13, 2023 at 17:11 | comment | added | Connor Malin | @ChesterX You are right, but you have actually added the information needed to correct the argument. Suppose $G$ is a homotopy commutative $A_\infty$ space which is not a double loop space (for example, the loop space of an $H$-space which is not $A_\infty$ should do). Then your observation shows the argument above can be applied to $BG$. I will edit the answer to correct the argument. | |
| Jun 13, 2023 at 11:15 | comment | added | ChesterX | Why is the bar construction producing an $A_\infty$-map? Also, note that I'm assuming that $\theta$ is an $A_2$ map. If I understand it correctly, the multiplication map on $\Omega Z$ is $A_2$ if and only if $\Omega Z$ is homotopy commutative. | |
| Jun 13, 2023 at 10:13 | history | answered | Connor Malin | CC BY-SA 4.0 |