Timeline for Monoidal Forgetful/Free Adjunction for $E_2$-algebras
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 13, 2017 at 23:31 | vote | accept | Jonathan Beardsley | ||
| Jun 13, 2017 at 23:19 | answer | added | Omar Antolín-Camarena | timeline score: 5 | |
| Apr 23, 2015 at 14:50 | comment | added | Lennart Meier | @JonBeardsley I would bet indeed my money on $U$ being lax monoidal. $M\otimes_B N$ should be the coequalizer of the two arrows $M\otimes B \otimes N \to M\otimes N$ and likewise for $M\otimes_A N$. The map $A\to B$ should induce then a map $M\otimes_A N \to M\otimes_B N$. (If you want you can work with bimodules and strictly associative symmetric spectra to check that; then one does not have to say $\infty$ so often). | |
| Apr 23, 2015 at 12:53 | comment | added | Sean Tilson | There is Fausk-Hu-May which addresses this type of situation. It may be worth looking at to see what you could expect. | |
| Apr 22, 2015 at 21:50 | comment | added | Jonathan Beardsley | Fair point. So perhaps we should only ask $U$ to be lax monoidal? | |
| Apr 22, 2015 at 21:14 | history | edited | Jonathan Beardsley | CC BY-SA 3.0 |
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| Apr 22, 2015 at 21:05 | comment | added | Qiaochu Yuan | I'm confused. Suppose we work with just rings instead of ring spectra, so let $A, B$ be two commutative rings and let $f : A \to B$ be a morphism between them. Then $F$ is monoidal, but $U$ isn't. It's easiest to see this when $f$ is a field extension; $F$ preserves dimensions but $U$ multiplies them by the dimension of the extension. | |
| Apr 22, 2015 at 20:46 | history | edited | Jonathan Beardsley | CC BY-SA 3.0 |
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| Apr 22, 2015 at 19:38 | history | asked | Jonathan Beardsley | CC BY-SA 3.0 |