In the sec 1.11. "Delignes' tensor product of locally finite abelian categories" of the book "Tensor Categories" of EGNO, the deligne's tensor product $C \boxtimes D$ of two k-linear locally finite abelian categories $C$ ad $D$ is defined by the universal property: $$ \matrix{ C \times D & \xrightarrow{\boxtimes} & C \boxtimes D\\ & F \searrow & \downarrow{ \exists ! G} \\ & & A } $$ i.e., $\boxtimes: C \times D \rightarrow C \boxtimes D$ is a bifuntor which is right exact in both variables and is such that for any right exact bifunctor $F: C \times D \rightarrow A$, where $A$ is k-linear locally finite abelian category, there exists a unique right exact functor $G: C \boxtimes D \rightarrow A$ such that $$ G \circ \boxtimes = F. $$

Let $Vec$ be the category of finite dimensional $k$-vector spaces. By the proof provided in the book, $Vec \boxtimes Vec \cong Vec$ and $\boxtimes$ is give by the tensor product of vector spaces. To my understanding, if $B$ is a category which is equivalent to $C \boxtimes D$, then $B$ can also be viewed as the Delignes tensor product of $C$ and $D$.

Let $sVec$ be the skeleton of $Vec$. Then there should exists a functor $G: sVec \rightarrow Vec$ such that the following diagram commutes: $$ \matrix{ Vec \times Vec & \xrightarrow{\otimes} & sVec \\ & \otimes \searrow & \downarrow{ G} \\ & & Vec }. $$ However, this is clearly impossible. So my question is if we can view $sVec$ as $Vec \boxtimes Vec$?