4
$\begingroup$

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 $\eta =\varepsilon^{op}$ ?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Nice example! $\endgroup$
    – Tom Leinster
    Commented Sep 3, 2012 at 13:37
  • $\begingroup$ 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. $\endgroup$
    – Buschi Sergio
    Commented Sep 3, 2012 at 21:16
  • 1
    $\begingroup$ 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. $\endgroup$
    – Tom Leinster
    Commented Sep 3, 2012 at 22:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .