Timeline for Is the center of an abelian rigid monoidal category, abelian?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 10, 2020 at 21:48 | answer | added | AMaths | timeline score: 1 | |
| May 7, 2020 at 22:48 | comment | added | AMaths | Thank you! I can’t believe how easily one forgets these facts | |
| May 7, 2020 at 22:31 | comment | added | R. van Dobben de Bruyn | Any left exact functor between finitely complete categories preserves monomorphisms, because $X \to Y$ is a monomorphism if and only if $\Delta_{X/Y} \colon X \to X \times_Y X$ is an isomorphism. See for example Exercise III.4.4 in Mac Lane. | |
| May 7, 2020 at 19:56 | comment | added | AMaths | Am I wrong in saying that an exact functor between abelian categories preserves Monos because it preserves kernels? But here we don’t know yet that the center is abelian to use that argument. | |
| May 7, 2020 at 19:43 | comment | added | R. van Dobben de Bruyn | Once you check that the centre is finitely complete and cocomplete and the forgetful functor is exact, this shows that $f \colon (A,\sigma) \to (A',\sigma')$ is a monomorphism if and only if it is the kernel of $(A',\sigma') \to \operatorname{coker} f$, because this is true in $\mathscr C$. | |
| May 7, 2020 at 19:35 | comment | added | AMaths | I agree that Ker/Cokers have induced braidings, but does this imply that every monomorphism in the center is the kernel of another morphism? | |
| May 7, 2020 at 18:14 | comment | added | R. van Dobben de Bruyn | I think the point is that in a rigid tensor category, the tensor product is exact. This allows you to equip the cokernel of $A \to A'$ with a natural half braiding induced by the ones on $A$ and $A'$. | |
| May 7, 2020 at 12:12 | history | edited | YCor | CC BY-SA 4.0 |
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| May 7, 2020 at 12:06 | history | asked | AMaths | CC BY-SA 4.0 |