Duals and direct summands in an abelian monoidal category

Question

This question may be seen as a continuation of Duals and sub-objects in a monoidal category. In an abelian monoidal category, i.e. an abelian category with biadditive monoidal product, if $X \oplus Y$ has a left-dual (or right-dual), do $X$ and $Y$ need to have a left dual (or right-dual)? This is true in the category of modules over a commutative ring, since then a module is dualisable if and only if it is finitely generated projective. But what about more general abelian monoidal categories?

(I have already posted the question on MSE here, but did not get an answer.)

Answer 1

Yes. In an idempotent complete monoidal category $C$ (abelian implies idempotent complete), if $A$ (e.g. your $X\oplus Y$) has a dual and $B$ (e.g. your $X$) is a retract of $A$, then $B$ also does.

Namely, let $i:B\to A, r: A\to B$ be your retraction and $e= ir$ the idempotent corresponding to it, it induces by functoriality an idempotent $e^*$ on $A^*$ which you can split off to get a $B^0$. I claim that then $B^0$ is dual of $B$: you just define the "obvious map" $1 \to A\otimes A^* \to B\otimes B^0$ and $B^0\otimes B\to A^*\otimes A\to 1$ and you check the triangle identities by witnessing them as retracts of the ones for $A$.

There is a less hands on, more abstract proof which consists in embedding $C$ in a closed monoidal category, e.g. $\mathrm{Psh}(C)$, and proving it there using the criterion for dualizability involving internal homs which is clearly closed under retracts, and finally using idempotent completeness to conclude that the dual you found actually lied in $C$ and not only in the bigger category.