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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 $Vec \boxtimes Vec$?

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  • $\begingroup$ The universal property, that $G \circ \boxtimes = F$, may be too strong. From a 2-categorical point of view you may want there to be a specified isomorphism, not an equality. $\endgroup$ – David Roberts Nov 3 '17 at 6:54
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The universal property isn't a characterization of $C \boxtimes D$, per se: the universal property is a property of the pair $(C \boxtimes D, \boxtimes)$ as an object of the coslice 2-category whose objects are pairs consisting of a category $B$ and a functor $C \times D \to B$. Now, $\mathsf{Vec}$ and $\mathsf{sVec}$ are equivalent just as categories, but there's no way to make them equivalent in this coslice 2-category (being equivalent here is a stronger condition). Does this help?

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  • $\begingroup$ Yes, thank you for your remark. Is that the case that even in the coslice 2-category, as mentions by David above, we can only require that $G \circ \boxtimes \cong F$ (not equals)? $\endgroup$ – heller Nov 3 '17 at 7:26
  • $\begingroup$ @heller Yes: the right level of uniqueness should be that $C \boxtimes D$ is defined up to a unique equivalence in this coslice category, and morphisms in the coslice 2-category are pairs (functor,natural isomorphism). So there should be a unique $G$ with a unique natural equivalence $G \circ \boxtimes \cong F$. I would expect Deligne to be very careful with this sort of details, so it could be useful to read Deligne rather than the EGNO book. $\endgroup$ – Dan Petersen Nov 3 '17 at 8:05
  • $\begingroup$ I actually checked the sec 5 "Produit tensoriel de categories abeliennes" of Deligne. He did not mention the universal property in the definition of the "Deligne" tensor product. Instead he require that $F \in Fun(C \boxtimes D, A) \rightarrow F \circ \boxtimes \in Fun(C \times D, A)$ induces an equivalence of the two categories. That is why I think that $sVec$ can also be viewed as $Vec \boxtimes Vec$. $\endgroup$ – heller Nov 3 '17 at 8:21
  • $\begingroup$ @heller But that's equivalent to the universal property, suitably weakened. Essential surjectivity means any functor on the RHS factors through $\boxtimes$ up to a natural equivalence, etc. $\endgroup$ – Dan Petersen Nov 3 '17 at 8:52

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