TITLE: Does the functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ have adjoints?
QUESTION [4 upvotes]: 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).

REPLY [7 votes]: 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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