Timeline for How does one rigorously lift a map $Sp \rightarrow Sp$ of spectra to equivariant spectra?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2020 at 8:35 | comment | added | Maxime Ramzi | Very generally, if $C$ is a category with finite (co)products and $G$ is a finite group, $CoInd^G(X) = X^G$ with permutation action; here the underlying object of $X^G$ is just $\prod_{g\in G}X$ and $G$ really acts by permuting factors. To prove this, you may want to use the fact that co-induction (resp. induction) is right (resp. left) Kan extension along $*\to BG$, and the fibers are just $G$. | |
| Nov 17, 2020 at 0:06 | comment | added | Bryan Shih | Wow, I've been learning a lot from these posts. @MaximeRamzi do you mind elaboraitng on CoInd computation? or perhaps a reference? I updated my question. | |
| Nov 16, 2020 at 12:11 | history | edited | Maxime Ramzi | CC BY-SA 4.0 |
deleted 35 characters in body
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| Nov 16, 2020 at 12:10 | comment | added | Maxime Ramzi | @HarryGindi : rats, you are absolutely right ! I didn't even notice, ha - let me try to rephrase that | |
| Nov 16, 2020 at 12:06 | comment | added | Harry Gindi | Oh man, Ind(C) for this is not the best notation lol. I did upvote though =]. | |
| Nov 16, 2020 at 11:58 | history | answered | Maxime Ramzi | CC BY-SA 4.0 |