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Let $\mathcal{C}$ be a braided monoidal category. We have a canonical functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ from $\mathcal{C}$ to the Drinfeld center $\mathcal{Z}(\mathcal{C})$ sending an object $V$ in $\mathcal{C}$ to $(V,c_{V,\,\_})$. Here, $c$ is the braiding in $\mathcal{C}$.

When does this functor admit left/right adjoints, and how do they look like? You are free to assume as much as you want on the category $\mathcal{C}$ (abelian, finite, factorizable, etc).

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    $\begingroup$ 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.) $\endgroup$
    – skd
    Commented Sep 17, 2020 at 20:19
  • $\begingroup$ @skd Sure, but does this mean that the forgetful functor Z(C) -> C is an adjoint of the above? Sorry, maybe this is trivial.. $\endgroup$
    – Bipolar Minds
    Commented Sep 17, 2020 at 20:28
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    $\begingroup$ 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. $\endgroup$
    – Noah Snyder
    Commented Sep 17, 2020 at 22:49
  • $\begingroup$ @Noah Snyder Thx, that's what I thought. I wasn't sure what skd was trying to tell me $\endgroup$
    – Bipolar Minds
    Commented Sep 18, 2020 at 6:23

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Short answer: Yes, it can possibly have an adjoint.

Longer answer: Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists. It is a coalgebra. Your assumptions on $\mathcal{C}$ were that it is braided, and in that case, it is well-known that $L$ is even a bialgebra. Moreover, we know that ${}_L\mathcal{C} = \mathcal{Z}(\mathcal{C})$, i.e. the center of $\mathcal{C}$ is isomorphic to the category of modules over $L$.

Under this isomorphism, your "free central object" $(V, c_{V, -})$ is sent to the trivial $L$-module on $V$, i.e. the action is $\varepsilon \otimes V \colon L \otimes V \to V$, where $\varepsilon \colon L \to 1$ is the counit of $L$. It is an algebra morphism. Thus, walking everything through the isomorphisms, the inclusion functor can actually be interpreted as the pullback functor \begin{align} \varepsilon^* \colon {}_1\mathcal{C} = \mathcal{C} \to {}_L\mathcal{C} \ . \end{align} A sufficient condition for pullbacks along algebra morphisms to have adjoints was identified in my answer to my own question over on M.SE.

Translating to our situtation, $\varepsilon^*$ has a left adjoint if $\mathcal{C}$ has coequalizers and $L$ is coflat (i.e. $L \otimes - $ preserves coequalizers). Then the left adjoint sends an $L$-module $(V, r)$ to the coequalizer of $$ r,\ \varepsilon \otimes id_V \colon L \otimes V \to V \ . $$

So for a particular situation where it works: take $\mathcal{C}$ to be a braided finite tensor category in the sense of EGNO. Then in particular, $\mathcal{C}$ is abelian, so it has coequalizers, and the tensor product is exact, so every object is coflat. Moreover, it's well-known that for these kinds of categories, the coend $L$ indeed does exist.

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