1
$\begingroup$

Let $C$ be a braided monoidal category. The category $\text{Aut}_{\text{br}}(C)$ of braided monoidal autoequivalences of $C$ is monoidal with tensor product functor given by the composition $\circ$.

Is it a braided category as well?

$\endgroup$
7
$\begingroup$

If $G$ is an abelian group, let $C$ be the category whose objects are the elements of $G$ and with only identity morphisms ($C$ is a discrete category). $C$ is symmetric monoidal via the product in $G$. The category $\text{Aut}_\text{br} (C)$ is again discrete, it's objects are the automorphisms of $G$ ($\text{Aut}_\text{br} (C)$ is constructed out of $\text{Aut}(G)$ in the same way as $C$ out of $G$). Being descrete, it can be braided monoidal only if $\text{Aut}(G)$ is commutative, but for a general abelian $G$ it is not true. So the answer to your question in no.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.