5
$\begingroup$

Let $\cal{P}$ be a $k$-linear semisimple abelian rigid monoidal category with finite dimensional (over $k$) Hom-spaces (for a field $k$).

By a tensored $\cal{P}$-category we mean a $\cal{P}$-category which admits tensors with objects in $\cal{P}$, i.e. for every two objects $X,Y$ in the category and $P$ in $\cal{P}$, there is an object $P\otimes X$ with an isomorphism: $$[P\otimes X,Y] \cong [P,[X,Y]].$$

Does every enriched functor $F$ between tensored $\cal{P}$-categories preserve tensors? (i.e. the natural induced map $P\otimes F(X)\to F(P\otimes X)$ is isomorphism) What happens if the categories are also abelian and $F$ is exact?

$\endgroup$
6
$\begingroup$

No. The functor "take a vector space to its double dual" is linear but does not preserve tensors with infinite-dimensional vector spaces.

Enriched functors are automatically "lax tensored". An enriched functor $F$ provides a natural map $[X,Y] \overset F\to [FX,FY]$ for any $X,Y$. Consider setting $Y=P\otimes X$, and study: $$ \mathrm{id} \in [P\otimes X,P\otimes X] \cong [P,[X,P\otimes X]] \overset {F\circ} \to [P,[FX,F(P\otimes X)]] \cong [P\otimes FX,F(P\otimes X)].$$ This gives a canonical map $P\otimes FX \to F(P\otimes X)$ which is natural in $P,X$ and compatible with associativity an unit data. It just isn't automatically an isomorphism.


On first reading, I missed the requirement that $\mathcal P$ be rigid. In that case the answer is Yes, enriched functors are automatically tensored. In general, a-priori-lax constructions become strong when there is enough dualizability. Consider the composition $$ F(P\otimes X) \to P \otimes P^* \otimes F(P\otimes X) \to P \otimes F(P^* \otimes P \otimes X) \to P \otimes F(X)$$ where the first map is the unit between $P,P^*$, the second is the lax monoidality constructed above, and the third is $F$ of the counit. This should give an inverse to the lax monoidality.

$\endgroup$
  • $\begingroup$ (I suggest swapping the two halves of the answer, so that the positive answer to the OP comes first, and the negative answer to the more general case comes after) $\endgroup$ – user13113 Feb 26 '17 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.