Timeline for Does the functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ have adjoints?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2022 at 8:17 | vote | accept | Bipolar Minds | ||
| Jan 21, 2022 at 11:52 | answer | added | Jo Mo | timeline score: 7 | |
| Sep 18, 2020 at 6:23 | comment | added | Bipolar Minds | @Noah Snyder Thx, that's what I thought. I wasn't sure what skd was trying to tell me | |
| Sep 17, 2020 at 22:49 | comment | added | Noah Snyder | No, the forgetful functor Z(C) --> C is defined even when C is not braided, it has a right adjoint (often called induction) which again doesn't depend on the braiding. For example, it sends 1 to $\bigoplus_x x \otimes x*$ in the semisimple case (and a "canonical coend" in general). It's adjoint can't be the inclusion C --> Z(C) which depends on the braiding for its construction. | |
| Sep 17, 2020 at 20:29 | history | edited | Bipolar Minds | CC BY-SA 4.0 |
added 14 characters in body
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| Sep 17, 2020 at 20:28 | comment | added | Bipolar Minds | @skd Sure, but does this mean that the forgetful functor Z(C) -> C is an adjoint of the above? Sorry, maybe this is trivial.. | |
| Sep 17, 2020 at 20:19 | comment | added | skd | Abstractly: C is an E_2-monoidal category, so by the universal property of the Drinfeld center (as the Hochschild cohomology of C, at least if one works in the derived setting), the identity C -> C factors through the canonical functor Z(C) -> C. (The functor Z(C) -> C sends a pair (x, phi) to x.) | |
| Sep 17, 2020 at 20:15 | history | asked | Bipolar Minds | CC BY-SA 4.0 |