Timeline for Every monoidal category is equivalent to a strict monoidal category
Current License: CC BY-SA 4.0
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| Jun 19, 2020 at 19:59 | vote | accept | CommunityBot | ||
| Jun 19, 2020 at 19:03 | comment | added | Zahlendreher | Also, if $\mathcal{C}$ is a strict monoidal category and $F\colon \mathcal{C}\to \mathcal{D}$ a strict monoidal functor that is part of an equivalence of categories, then $\mathcal{D}$ is also strict monoidal. This follows using the defining diagram of a monoidal functor (4.1) in Kassel. | |
| Jun 19, 2020 at 18:56 | comment | added | Zahlendreher | I agree that the monoidal functor $G$ is not strict. For the proof, it doesn't need to be strict. We only need a tensor equivalence, i.e. an equivalence of categories involving monoidal functors (in Kassel's terminology). The way Kassel defines a monoidal functor, it is what others (as Mike Shulman remarked), call a strong monoidal functor, i.e. the structural natural transformation $(G_2)_{U,V}\colon G(U)\otimes G(V)\to G(V\star U)$ consists of isomorphisms. | |
| Jun 19, 2020 at 18:29 | answer | added | Noah Snyder | timeline score: 3 | |
| Jun 19, 2020 at 18:24 | comment | added | user159891 | @NoahSnyder And do you see how to prove that $G$ is a tensor functor then? What transformations should I use? | |
| Jun 19, 2020 at 18:14 | comment | added | Noah Snyder | Strong means the natural transformations are natural isomorphisms, that is strong is named because it's stronger than "lax" or "oplax." Strong does not mean strict! Strong is weaker than strict. I think Kassel just calls strong tensor functors "tensor functors." | |
| Jun 19, 2020 at 16:56 | history | edited | user159891 | CC BY-SA 4.0 |
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| Jun 19, 2020 at 15:13 | history | edited | user159891 | CC BY-SA 4.0 |
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| Jun 19, 2020 at 14:05 | history | edited | user159891 | CC BY-SA 4.0 |
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| Jun 19, 2020 at 13:17 | comment | added | user159891 | @MikeShulman In the book, strict tensor functor $F$ is defined as a tensor functor $(F, \varphi_0, \varphi_2)$ where $\varphi_0$ and $\varphi_2$ are identities, which is not the case here? What transformations should I use then instead of identities? | |
| Jun 19, 2020 at 13:11 | history | edited | user159891 | CC BY-SA 4.0 |
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| Jun 19, 2020 at 13:09 | comment | added | Mike Shulman | I assume from what's said that the objects of $C^{\mathrm{str}}$ are finite sequences of objects of $C$ and that the functor $F$ multiplies a sequence of objects together using the tensor product of $C$. In this case I think the answer is just that $G$ is a strong monoidal functor. | |
| Jun 19, 2020 at 13:00 | review | First posts | |||
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| Jun 19, 2020 at 12:58 | history | asked | user159891 | CC BY-SA 4.0 |