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What is the relationship between categories with duals (i.e. rigid categories) and categories equipped with a internal internal Hom functor? Wikipedia seems to suggest they are the same thing, but it's not clear. Does the existence of duals mean that one can define an internal Hom, or does the existence of an internal Hom mean that the category admits dual . . . or is this too naive?

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In a monoidal category in which everything has duals on one side, tensoring by objects on one side will have an adjoint, and in particular in a symmetric monoidal (I think also braided monoidal) category rigidity implies closedness (see the proof below.) Some people find it so much more comfortable to think in terms of duals that they rarely think of internal homs at all.

But such people must like finite-dimensional vector spaces, finite cell complexes, and finitely generated projective modules better than their big analogues, because it is indeed to naive to expect that these notions are equivalent: internal homs do not generally give one duals. Intuitively, you might observe that in a braided monoidal category $A$ is a dual of $A^*$ (just apply the braiding after $\eta$ and before $\varepsilon$) so that objects we know aren't dualizable in classical algebra and topology aren't dualizable in this sense either. More precisely, the internal hom gives you for instance a unit $1\to[A,1\otimes A],$ but for dualizability we need the unit to land in $[A,1]\otimes A$. These are isomorphic just when $A$ is dualizable.

A dualizable object $A$ admits a dual $A^*$ and, picking a side, morphisms $1\to^\eta A^*\otimes A$ and $A\otimes A^*\to^\varepsilon 1$ satisfying the triangle identities $\varepsilon\otimes A\circ A\otimes \eta=id_A,A^*\otimes\varepsilon\circ\eta\otimes A^*=id_{A^*}$.

To see how this relates to internal homs, think of finite-dimensional vector spaces, where the maps from $V$ to $W$ are given by the tensor product $V^*\otimes W$. So if we try to define $[A,B]$ in this way when $A$ is dualizable, let's see whether this gives an adjoint to tensoring by $A$. We need a unit $B\to [A,A\otimes B]=: A^*\otimes A\otimes B$, where for convenience I suppose $\otimes$ is strict. How about $\eta\otimes B$? Similarly, we get a candidate evaluation/counit $\varepsilon\otimes B:A\otimes [A,B]=A\otimes A^*\otimes B\to A$, and we want to check the triangle equalities. $\varepsilon\otimes A\otimes B\circ A\otimes \eta\otimes B:A\otimes B\to A\otimes A^*\otimes A\otimes B\to A\otimes B=id_{A\otimes B}$ and $A^*\otimes\varepsilon\otimes B\circ \eta\otimes A^*\otimes B: A^*\otimes B\to A^*\otimes A\otimes A^*\otimes B\to A^*\otimes B=id_{A^*\otimes B}$ both hold-because they're simply tensoring the triangle identities from the first paragraph with $B$!

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