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A right closed monoidal category $\mathcal{C}$ is one for which the functor $X \otimes -$ admits a right adjoint, for every $X \in \mathcal{C}$. As explained here

Enrichments vs Internal homs

this gives an enrichement of $\mathcal{C}$ over itself. In this question

Rigid monoidal and closed monoidal categories

it says that rigid monoidal categories are closed monoidal. So what does the self-enrichment of a rigid monoidal category look like?

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    $\begingroup$ $Hom(X,Y)=X^{*}\otimes Y$. BTW you need symmetry in the first line, or that would just be right closed. $\endgroup$
    – Fernando Muro
    Commented Feb 9, 2021 at 10:34
  • $\begingroup$ @Fernando: I have added right thanks! $\endgroup$
    – Jake Wetlock
    Commented Feb 9, 2021 at 11:56
  • $\begingroup$ @Fernando: How does one relate the set $Hom(X,Y)$ with $V^* \otimes Y$? $\endgroup$
    – Jake Wetlock
    Commented Feb 9, 2021 at 12:06
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    $\begingroup$ @TimCromby There's a notational ambiguity. I'll write $Hom(X,Y)$ for the set of morphisms and $[X,Y]$ for the internal hom. As Fernando says, one defines $[X,Y] = X^\ast \otimes Y$. The idea to keep in mind is that if you already had the closed monoidal structure, then you would have defined $X^\ast = [X,I]$. You would have a map $X^\ast \otimes Y = [X,I] \otimes [I,Y] \to [X,Y]$ internalizing the composition operation $Hom(X,I) \times Hom(I,Y)\to Hom(X,Y)$, and using rigidity you'd have shown this map to be an isomorphism. When setting up the closed structure, you run that picture backwards $\endgroup$
    – Tim Campion
    Commented Feb 9, 2021 at 14:24

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