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Is the Drinfeld-Majid center of an abelian rigid monoidal category, abelian? [stated in 1J of On the center of fusion categories" by Bruguières and Virelizier (link at Virelizier's page)]

In particular, I’m not seeing why any monomorphism in the center would have to be a kernel of a morphism? (I’m relatively happy with the other axioms holding, but if anyone has a reference where this is discuss explicitly, it’s appreciated )

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  • $\begingroup$ 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 $A'$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 7, 2020 at 18:14
  • $\begingroup$ I agree that Ker/Cokers have induced braidings, but does this imply that every monomorphism in the center is the kernel of another morphism? $\endgroup$
    – AMaths
    Commented May 7, 2020 at 19:35
  • $\begingroup$ 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 $\mathscr C$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 7, 2020 at 19:43
  • $\begingroup$ 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. $\endgroup$
    – AMaths
    Commented May 7, 2020 at 19:56
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    $\begingroup$ 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. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 7, 2020 at 22:31

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My question got answered in the comments, so I thought best to write a small answer for it here:

If $C$ is a rigid monoidal category, then the forgetful functor $U: Z(C) \rightarrow C $ creates colimits and limits: any (co)limit of the underlying objects of a diagram in $Z(C)$ will have an induced braiding satisfying the nessecary conditions. This follows from $A\otimes - $ and $-\otimes A$ preserving (co)limites. Hence, if $C$ is abelian, then the center is finitely complete and cocomplete and naturally automatically additive. The only thing remaining is to check if any mono is a kernel and any epi is a cokernel which I was stuck at, which just follows from the fact that in a finitely (co)complete category, monos (epis) are just pullbacks and pushforwards, so they are preserved by $U$. If $m:(A,\sigma)\rightarrow (B,\tau) $ is a monomorphism in the center, then $m$ is a mono in $C$ and is the kernel of $f:B\rightarrow coker(f)$ and as mentioned above $coker(f)$ has an induced braiding compatible with $\sigma $ and $\tau $ and $f$.

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  • $\begingroup$ U should go the other way, right? $\endgroup$
    – Noah Snyder
    Commented May 11, 2020 at 3:36
  • $\begingroup$ @NoahSnyder Indeed! Thank you:) $\endgroup$
    – AMaths
    Commented May 11, 2020 at 12:55

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