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Apr 6, 2020 at 23:57 comment added G. Stefanich @DmitryVaintrob No rigidity needed. I believe it is a general fact for any symmetric monoidal category that if an object $X$ is invertible then it is dualizable with dual $X^{-1}$. The idea is that for any object $Y$ one has that $X^{-1}\otimes Y$ is an internal Hom between $X$ and $Y$. Therefore the map $[X, 1] \otimes X \rightarrow [X, X]$ is an isomorphism (where brackets denote internal hom), which is equivalent to $X$ being dualizable. Full details seem to be supplied in pages.uoregon.edu/ddugger/invcoh.pdf proposition 4.11. As you say, this provides the dual maps that you need.
Apr 6, 2020 at 4:18 comment added Dmitry Vaintrob Thank you German for this insightful and thorough answer! I thought the third localization may follow from the first two but couldn't see how. Does your argument assume rigidity, or does inverting $X$ automatically imply that $X, X^{-1}$ are dual and thus maps $X\to Y$ have dual maps $Y^{-1}\to X^{-1}$? Your implication (2) => (1) is very nice, and thank you for the reference to Ballmer, Hopkins, Neeman and co.
Apr 6, 2020 at 3:26 vote accept Dmitry Vaintrob
Apr 4, 2020 at 23:10 history answered G. Stefanich CC BY-SA 4.0