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](https://mathoverflow.net/questions/16031/suppose-c-and-d-are-morita-equivalent-fusion-categories-can-you-say-anything-ab), the answer would be yes -- but are there any references for this result in the literature? Thanks!



Relationship between fusion category and its Drinfel'd center
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.




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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.


