Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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 $F\neq G^{op}$.

share|improve this answer
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.

share|improve this answer
Thank you for the thorough explanation! –  Fujita Tomomi Jul 11 '13 at 21:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.