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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.)

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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.

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  • $\begingroup$ Ah, I also tried to show it in the idempotent complete case, but got stuck at showing the triangle identified. I'll try again and post it as an answer when I got it. What would the dualisability condition be in a closed monoidal category? I know that the dual of $X$ has to be given by the internal Hom from $X \to 1$, but what else needs to be satisfied? $\endgroup$
    – Jannik Pitt
    Commented Nov 21 at 7:14
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    $\begingroup$ In a closed monoidal category, $X$ is dualizable if and only if the canonical map $Y\otimes \hom(X,1) \to \hom(X,Y) $ is an isomorphism for all $Y$ (I am being sloppy about left/right here but there is a way to make this correct), and since this map is natural in $X,Y$, this condition is clearly closed under retracts $\endgroup$
    – Maxime Ramzi
    Commented Nov 21 at 7:48

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