MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F: \mathscr{C}\to \mathscr{C}^{op}$, with an adjoint $G$, and $\eta: 1_\mathscr{C} \Rightarrow G\circ F $ and $\varepsilon: F\circ G\Rightarrow 1_{\mathscr{C}^{op}}$ with components (in $\mathscr{C}$):

$\eta_M: M\to G(F(M))$ and $\varepsilon^{op}_N: N\to F(G(N))$ such that the compositions (in $\mathscr{C}$):

$F(M) \xrightarrow{\varepsilon^{op}_{F(M)}} FGF(M) \xrightarrow{F^{op}(\eta_M)}F(M)$ and

$G(N) \xrightarrow{\eta_{G(N)}}GFG(N) \xrightarrow{G(\varepsilon_N)}G(N)\ $ are units.

Moreover suppose $G=F^{op}$ and $\eta$ a isomorphis transformation. Follow that $F$ is full, faithfull and reflect isomorphisms (consider $F(f)\circ \eta_M=\eta_{M'}\circ f$ for $f: M\to M'$), then also $G=F^{op}$ reflect isomorphisms, from above: $\varepsilon$ is Iso. Observe that the components of $\eta: 1_\mathscr{C} \Rightarrow F^{op}\circ F $ and $\varepsilon^{op}: 1_{\mathscr{C}}\Rightarrow F^{op}\circ F$ are of type: $\eta_M: M\to F\circ F(M)$ and $\varepsilon^{op}_M: M\to F\circ F(M)$.

Question: Is true that $\eta =\varepsilon^{op}$ ?

share|cite|improve this question

It's a good question; the answer is no.

Suppose $F \dashv F^{op}$ with unit $\eta: 1_C \to F^{op} F$ and counit $\varepsilon: F F^{op} \to 1_{C^{op}}$. Since $\varepsilon$ is the unique transformation $\theta: F F^{op} \to 1$ such that

$$1_{F^{op}} = (F^{op} \stackrel{\eta F^{op}}{\to} F^{op} F F^{op} \stackrel{F^{op}\theta}{\to} F^{op})$$

the question is whether $F^{op}\eta^{op} = (\eta F^{op})^{-1}$.

Take $C$ to be an commutative group $G$, considered as a category with one object $\bullet$. Here we may simply identify $G^{op}$ with $G$, i.e., the identity $1_G \colon G \to G$ may be seen as a contravariant functor because we have $1_G(g h) = 1_G(h)1_A(g)$ by commutativity. Now take $F$ and therefore $F^{op}$ to be the identity on $G$. Any element $u \in G$ as morphism $\bullet \to \bullet = F^{op} F \bullet$ can serve as the unit transformation (naturality also follows from commutativity). But then, as soon as $u$ is not equal to $u^{-1}$, we have $F^{op}u^{op} = u \neq u^{-1} = (u F^{op})^{-1}$. Therefore taking $G$ to be the additive group $\mathbb{Z}$ and $u = 1 \in \mathbb{Z}$, we reach a counterexample.

share|cite|improve this answer
Nice example! – Tom Leinster Sep 3 '12 at 13:37
Thank you very much. I thinked this problem about the definition of category with duality of "QUadratic and Hermitian Forms over Rings" by Max-ALbert Knus. – Buschi Sergio Sep 3 '12 at 21:16
Next question: suppose we have a functor $F$ that's self-adjoint on the right. Is it always possible to choose an adjunction between $F$ and $F^{op}$ in which the unit is the same as the counit? (Todd's example doesn't settle it, since we could take both unit and counit to be 1.) I can't quite be bothered to pose this formally as a Question, partly because I haven't spent long enough thinking about it -- maybe it's an easy "yes". But anyone else should feel free. – Tom Leinster Sep 3 '12 at 22:07

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.