Is it true that given a fusion category $\mathcal{C}$ and its Drinfel'd center $Z(\mathcal{C})$, there is a fully faithful functor $F:\mathcal{C}\hookrightarrow Z(\mathcal{C})$? I.e. can $\mathcal{C}$ can be identified with a full monoidal subcategory of $Z(\mathcal{C})$?
Judging by the mention of a 'restriction functor' $Z(\mathcal{C})\to\mathcal{C}$ here, the answer would be yes -- but are there any references for this result in the literature? Thanks!
The short answer is no.
Suppose you have a fully faithful monoidal functor $(F,J):\mathcal C\to\mathcal B$, where $\mathcal C$ and $\mathcal B$ are fusion and $J_{X,Y}:F(X)\otimes F(Y)\to F(X\otimes Y)$ is the monoidal structure map (sometimes called a tensorator). If $\{b_{U,V}:U\otimes V\to V\otimes U\}_{U,V\in\mathcal B}$ is a braiding for $\mathcal B$, then we can define a braiding $c$ for $\mathcal C$ by the formula
$$F\big(c_{X,Y}\big)=J_{X,Y}^{-1}b_{F(X),F(Y)}J_{Y,X}\,.$$
For any given $X$ and $Y$ in $\mathcal C$, the morphism $c_{X,Y}$ exists because $F$ is full, and it is unique because $F$ is faithful. The map $c_{X,Y}$ is invertible because fully faithful functors reflect isomorphisms. The claim that this does in fact define a braiding on $\mathcal C$ is an exercise in pasting coherence diagrams. The component diagrams to be pasted are the hexagon axioms for $b_{F(X),F(Y)}$, and the monoidality diagram that compares the associators of $\mathcal C$ and $\mathcal B$ with the structure map $J$.
From this construction, we can see that any monoidal subcategory of a braided category can always inherit a braiding from the ambient category. If every fusion category could be monoidally embedded into its Drinfel'd center, then every fusion category would admit a braiding. However, there do exist unbraidable fusion categories such as $G$-graded vector spaces $\text{Vec}_G$ when $G$ is nonabelien. Thus we're forced to conclude that such categories cannot be embedded monoidally into their Drinfel'd centers.
Some more commentary:
As @ATO suggests the induction functor $I:\mathcal C\to\mathcal Z(\mathcal C)$ factors through an equivalence as in
$$\mathcal C\mathop{\longrightarrow}\limits^{\simeq}I(\mathbf 1)\text{-}\text{Mod}_{\mathcal Z(\mathcal C)}\mathop{\longrightarrow}\limits^{\text{Forg}}\mathcal Z(\mathcal C)\,,$$ where the middle category is the category of $I(\mathbf1)$-modules in $\mathcal Z(\mathcal C)$, and the second functor forgets the $I(\mathbf1)$-module structure. This explains why $I$ is typically faithful, but not full.
Unfortunately, this forgetful functor is not monoidal because the product in $I(\mathbf 1)\text{-}\text{Mod}_{\mathcal Z(\mathcal C)}$ is $\otimes_{I(\mathbf 1)}$.
When $\mathcal C$ is braided, $I(\mathbf1)$ is a (braided) Hopf algebra when regarded as an object in $\mathcal C$, and the category of modules $I(\mathbf 1)\text{-}\text{Mod}_{\mathcal C}$ is equivalent to $\mathcal Z(\mathcal C)$. The weird thing is that $I(\mathbf1)\in\mathcal Z(\mathcal C)$ and $I(\mathbf1)\in\mathcal C$ feel like the same algebra, but somehow taking modules can produce $\mathcal C$ or $\mathcal Z(\mathcal C)$ respectively. The difference is that taking modules for a commutative algebra produces a smaller category, while taking modules for a Hopf algebra produces a bigger category.
I assume we are under the assumption the category is braided.
This is mentioned in
Drinfeld, Vladimir; Gelaki, Shlomo; Nikshych, Dmitri; Ostrik, Victor, On braided fusion categories. I, Sel. Math., New Ser. 16, No. 1, 1-119 (2010). ZBL1201.18005.
right after Proposition 2.36 , and in EGNO Proposition 8.6.1 with very similar wording. Both without proofs and without attribution.
The proof seems straightforward. The assignment is $X\to (X,R_{\bullet,X})$ where $R_{\bullet,X}$ is the braiding $Y\otimes X\to X\otimes Y$.
Morphisms in the Drinfeld center between objects $(X,\gamma^{X})$ and $(Y,\gamma^{Y})$ are given by morphisms $f:X\to Y$ such that $\gamma^{Y}_{M}\circ Id_{M}\otimes f = f\otimes Id_{M} \circ \gamma^{X}_{M}$ for an object $M$. But in our case $R_{\bullet, X}$ is the braiding of the category so any morphism $X\to Y$ works.
Edit:
In the nonbraided case there is a right adjoint to the forgetful functor. This is again already mentioned in ENO (5.8) and EGNO (9.2, for example) without attribution and for what I can see, also no proof that this is a right adjoint ( but follows from the adjunction formula for the internal Hom ).
You can however prove that the induction functor $I:C\to Z(C)$ is given by $I(X)\cong \underline{Hom}_{Z(C)}(1,X)$ in EGNO Proposition 8.12.1.
But I don't think you get that this functor is fully faithful, in fact the relevant result is that it induces an equivalence of $C$ with the category of A-modules in $Z(C)$ where $A=I(1)$ (EGNO Proposition 8.12.2 (ii)). When the category is braided you get that this category of modules is equivalent to the center, but I don't think this is true in general. I admit I cannot think of a counterexample, but hope this is of any help.
