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Does there exist a category $\mathcal{C}$ and contravariant functors $F:\mathcal{C}\to\mathcal{C}^{op}, G:\mathcal{C}^{op}\to\mathcal{C}$ such that $F$ is left ajoint to $G$ and $F\neq G^{op}$? In the case of $\mathcal{C}=Set$, I proved that there exist no such functors, because if exist $G$ preserve limits, and so is representable by adjoint functor theorem, and that $G=Hom(-,A)$ has a left ajoint $F=Hom(-,A)$ itself. But I can't come up with any answers to other cases. Partial solutions are welcomed. For example, when $\mathcal{C}$ is a topos, or $G$ is monadic, etc.

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up vote 7 down vote accepted

Here's a very simple example. Let $\mathcal{C}$ be any set with more than two elements, considered as a discrete category. Then $\mathcal{C}=\mathcal{C}^{op}$, and any permutation $F:\mathcal{C}\to\mathcal{C}^{op}=\mathcal{C}$ has an adjoint $G$ (on both sides!) given by the inverse. Any permutation that is not its own inverse then gives an example of an adjunction with $F\neq G^{op}$.

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Thank you! It's quite simple! – Fujita Tomomi Jul 11 '13 at 21:41
By the way, if you want an example where $\mathcal{C}$ is a "nicer" category, you should be able to use the same idea (build a category that has as part of its structure a set that you can permute freely). For instance, if you want $\mathcal{C}$ to be a topos, you could take $\mathcal{C}=Set^X$ for some set $X$ and use an $F$ that is an internal Hom-functor composed with an automorphism of $\mathcal{C}$ that permutes $X$ (by a non-involution). – Eric Wofsey Jul 11 '13 at 21:59
Oh, it is exactly the answer for what I'm going to ask for the next. Thank you very much! – Fujita Tomomi Jul 11 '13 at 22:04

Note that your analysis for the case $Set$ uses the fact that the cartesian product is symmetric monoidal. This suggests that you consider an example of a monoidal biclosed category which is non-symmetric. (For example, consider presheaves on a small non-symmetric monoidal category, equipped with the induced Day convolution.)

To give a concrete instance, consider the poset of binary relations on a set $X$, where the monoidal product is relational composition. (Think of such relations as truth-valued functions $R(-, -): X \times X \to \mathbf{2}$.) This product has an adjoint on each side, which we may denote $\Rightarrow$ and $\Leftarrow$:

$$(R \Rightarrow S)(x, y) = \forall_z R(z, x) \to S(z, y)$$

$$(S \Leftarrow R)(x, y) = \forall_z R(y, z) \to S(x, z)$$

Then we have $- \Rightarrow S$ is adjoint to $S \Leftarrow -$, since

$$T \leq R \Rightarrow S \;\; \text{iff} \;\; T \circ R \leq S \;\;\text{iff}\;\; R \leq S \Leftarrow T$$

(modulo the fact that I may have switched the usual order of relational composition). But clearly these adjoint functors are non-isomorphic.

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Thank you for the thorough explanation! – Fujita Tomomi Jul 11 '13 at 21:45

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