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In a monoidal category with biproducts, let $A$ and $B$ be objects with right duals. Then does $A \oplus B$ have a right dual?

The question is a bit subtle. Suppose I already know that $A \oplus B$ has a dual. Then certainly, given the data of right duals for $A$ and $B$, I can construct a right dual for $A \oplus B$, because knowing that $A \oplus B$ has a dual implies certain nice properties. But this does not seem useful in the more restricted case described above.

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  • $\begingroup$ So I guess you know that categories with biproducts are canonically enriched in commutative monoids. Do you also assume the compatibility between the monoidal structure and the biproducts that would say the monoidal product as a functor is enriched in commutative monoids? I believe the answer would have to be 'yes' in that case, but would be skeptical otherwise. $\endgroup$ Commented Mar 13, 2016 at 1:36
  • $\begingroup$ Hi Todd, no, I don't assume anything more than specified in the question. If all objects have duals then this implies this good interaction between the semiadditive structure and the monoidal product - this is what I was alluding to in the question. $\endgroup$ Commented Mar 13, 2016 at 8:41

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This is not an answer, just an explanation of the mistake I made in my original (now deleted) answer.

It is very tempting to conjecture that an object $X$ has a right dual $X^{\ast}$ iff there is an adjunction of the form

$$\text{Hom}(X \otimes A, B) \cong \text{Hom}(A, X^{\ast} \otimes B)$$

since among other things there are natural choices of a candidate for the unit and the counit. If this were true, then the desired conclusion would follow assuming the monoidal structure distributes over biproducts. But you can't get the zigzag identities this way; as Todd Trimble informed me, there was a long discussion on the n-category cafe about this in 2008.

Abstractly, the problem is the following. The above condition says that the image of $X$ under the canonical monoidal functor

$$C \ni X \mapsto (A \mapsto X \otimes A) \in \text{End}(C)$$

has a right dual, also in the image. We would be able to conclude that $X$ itself has a right dual if this functor were fully faithful. But it's just not, in general! (In other words, there is no "monoidal Yoneda lemma" in general.)

Here is an explicit counterexample. Let $G$ be a finite group and let $C = \text{Rep}(G)$ be the monoidal category of complex $G$-representations. When we talk about endofunctors of $C$ let's restrict our attention to cocontinuous $\mathbb{C}$-linear endofunctors. Then $\text{End}(C)$ is a "matrix algebra": it's equivalent to the category of matrices $M_{ij}$ of vector spaces, where $i, j \in \hat{G}$ runs over the complex irreps of $G$, acting via

$$\bigoplus_i n_i V_i \mapsto \bigoplus_{i, j} n_j M_{ij} \otimes V_i.$$

Natural transformations between two such matrices correspond to tuples of linear maps $M_{ij} \to N_{ij}$, and composition is given by matrix multiplication.

In particular, the tensor product functor $V \mapsto X \otimes V$ always lives in $\text{End}(C)$. As a matrix of vector spaces we have $X_{ij} \cong (X \otimes V_j)_i$ where the subscript $i$ denotes taking the $i$-isotypic component. It follows that there is a nontrivial natural transformation $X_{ij} \to Y_{ij}$ iff there are some $i, j$ such that $V_i$ occurs in both $X \otimes V_j$ and $Y \otimes V_j$. But there is a nontrivial morphism $X \to Y$ iff there is some $i$ such that $V_i$ occurs in both $X$ and $Y$. If $G$ is nonabelian then this second condition is strictly stronger.

Probably you can use this to get a counterexample to the original claim about right duals vs. right adjoints but I haven't worked out what the dualizable objects in "matrix algebras" are yet.

In the positive direction, I believe the "monoidal Yoneda lemma" holds if $C$ is generated under iterated colimits by invertible objects.

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  • $\begingroup$ Say an object X has a weak dual when it satisfies the property at the top of your post. Then another funny aspect to this is that nobody seems to know a finite-dimensional semisimple k-linear tensor category with an object that has a weak dual, but not a dual. $\endgroup$ Commented Mar 15, 2016 at 23:35

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