Consider $\mathcal{M}$ a monoidal category. Let $V$ be an object that admits a left/right dual. If $U$ is a subject of $V$ then does it also admit a left right dual?
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$\begingroup$ What is the definition of an object with a dual? $\endgroup$– Julius HamiltonCommented Oct 5 at 13:27
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$\begingroup$ @JuliusHamilton Your comment here, like your comment on my answer below, is immediately answerable by googling. For example, putting "dual object category" into google turns up en.wikipedia.org/wiki/Dual_object as the first hit. I recommend that you do spend at least a little time checking such easily-accessible online references. $\endgroup$– Theo Johnson-FreydCommented Oct 5 at 18:28
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2 Answers
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No. Take any meet-semilattice with top element $\top$, considered as a cartesian monoidal category. Of course the left/right dual of $\top$ is $\top$ itself. Also of course, any object is a subobject of $\top$. If the answer were yes, then every element in a meet-semilattice would be dualizable. But there are many examples where this fails (for example, most linear orders).
answered Sep 17 at 11:40
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$\begingroup$ When viewing a meet-semilattice as a Cartesian monoidal category, is the meet operation the monoidal product? $\endgroup$ Commented Oct 5 at 13:30
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No. Pick a nonsemisimple commutative ring $R$, e.g. the dual numbers $\mathbb{C}[\epsilon]/(\epsilon^2)$, and look at the $\mathcal{M} = \mathrm{Mod}(R)$ made into a monoidal category under $\otimes_R$. The dualizable objects are precisely the finitely generated projective modules. But since $R$ is not semisimple, it contains an ideal $I \subset R$ which is not projective; this is a subobject (of the unit, which is definitely dualizable!) which is not dualizable.
answered Sep 17 at 12:00
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$\begingroup$ What does it mean for a commutative ring to be “non-semi-simple”? $\endgroup$ Commented Oct 5 at 13:32
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$\begingroup$ @JuliusHamilton See ncatlab.org/nlab/show/semisimple+ring for many characterizations of "semisimple ring". $\endgroup$ Commented Oct 5 at 18:25