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May 24, 2017 at 16:33 comment added Theo Johnson-Freyd There is a sense in which the duality is not canonical: the space of functors $F$ equipped with an isomorphism $F^2 \cong \mathrm{id}$ is not contractible. Indeed, any such functor admits a nontrivial natural automorphism, namely multiplication by $-1$.
May 24, 2017 at 15:48 vote accept Michael Bächtold
May 24, 2017 at 13:47 comment added Arun Debray Ah, I missed that assumption.Thank you.
May 24, 2017 at 13:39 answer added Chris Schommer-Pries timeline score: 10
May 24, 2017 at 13:20 comment added Chris Schommer-Pries @ArunDebray those functors are covariant, not contravariant.
May 24, 2017 at 13:17 comment added Arun Debray $F = \mathrm{id}$ satisfies those requirements, but probably isn't what you're looking for. Also, over $\mathbb{C}$, there's the complex conjugate functor, right?
May 24, 2017 at 12:40 history asked Michael Bächtold CC BY-SA 3.0