Drinfeld center of a Deligne tensor product Let $\mathcal{C}$ and $\mathcal{D}$ be two tensor categories (if necessary, assume they are fusion categories). Is the canonical braided monoidal functor $$\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})\rightarrow\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$$ an equivalence?
NB: The two monoidal categories $\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})$ and $\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$ have the same Frobenius-Perron dimension so it would be enough to show that the above functor is either injective or surjective in the sense of EGNO.
 A: By Cororllary 3.26 of arxiv:1009.2117, any braided tensor functor out of a non-degenerately braided fusion category is automatically fully faithful. Since $Z(\mathcal{C})\boxtimes Z(\mathcal{D})$ is non-degenerate when $\mathcal{C}, \mathcal{D}$ are fusion, the result follows immediately from your FP dimension observation.
A: Here is another way to see this. As noticed by Theo in the comments to the OP, the center of $\mathcal{E}$ is the endomorphism category of $\mathcal{E}$ as and $\mathcal{E}$-$\mathcal{E}$-bimodule category. That it is, its objects consist of the functors $\mathcal{E} \to \mathcal{E}$ equipped with coherence data with respect to the left and right $\mathcal{E}$-actions.
When you take the tensor product of tensor categories $\mathcal{C} \boxtimes \mathcal{D}$, then the left and right actions break apart in the usual way.
For example the $\mathcal{C}$ actions "only act on the $\mathcal{C}$ components". What this means is that we have a formula
$$\mathcal{Z}(\mathcal{C} \boxtimes \mathcal D) = End_{C \boxtimes D, C \boxtimes D} (\mathcal{C} \boxtimes \mathcal{D}) $$
$$ \cong Fun_{C,C}(\mathcal{C}, Fun_{D,D}(\mathcal{D}, \mathcal{C} \boxtimes \mathcal{D}))$$
$$\cong  Fun_{C,C}(\mathcal{C}, \mathcal{C} \boxtimes Fun_{D,D}(\mathcal{D},  \mathcal{D}))$$
$$\cong Fun_{C,C}(\mathcal{C}, \mathcal{C}) \boxtimes Fun_{D,D}(\mathcal{D},  \mathcal{D})) $$
$$ = \mathcal{Z}(\mathcal{C}) \boxtimes \mathcal{Z}(D)$$
I think the only one of these isomorphisms which, perhaps, isn't straigtforward is the second one, where we identify $Fun_{D,D}(\mathcal{D}, \mathcal{C} \boxtimes \mathcal{D}) \cong \mathcal{C} \boxtimes Fun_{D,D}(\mathcal{D},  \mathcal{D}))$. Note that this doesn't use anything about the tensor category structure of $\mathcal{C}$. So for fusion categories this is totally obvious since as underlying categories then are simply finite sums of $Vect$. It also holds for finite linear categories, though this takes more work. I think it is likely to hold for some larger classes of categories as well, but I am not sure - you might have to be careful about what tensor product you are taking.
