Duality 2-functor on adjunctions

Question

This question is about the definition of the duality 2-functor in Hovey's book on Model categories, Section 1.4.

There he defines the 2-category of categories with adjunctions as follows:

• objects are categories
• 1-morphisms are adjunctions $(F,U,\varphi)$, where $F$ is the left adjoint to $U$ and $\varphi:\operatorname{Hom}(F-,*)\to \operatorname{Hom}(-,U*)$.
• 2-morphisms are natural transformations $F\to F'$.

He then goes on defining the duality 2-functor as $D(\mathcal{C})=\mathcal{C}^{op}$ on objects, $D(F,U,\varphi)=(U,F,\varphi^{-1})$ on 1-morphisms and for $\sigma:F\to F'$, he defines $D(\sigma)=U\varepsilon'\circ U\sigma\circ \eta:(U,F,\varphi)\Rightarrow (U',F',\varphi')$.

The claim now is that $D^2=\operatorname{Id}$.

On objects and 1-morphisms this is obvious. On 2-morphisms I don't see why this follows. From my calculations, I think this should be the equality:

$$\varepsilon\circ F(U\varepsilon'\circ U\sigma\circ \eta)\circ F\eta'=\sigma$$

which doesn't seem to hold. I probably have mixed up some direction of arrows, but I don't know which one.

Answer 1

The definition of $D(\sigma)$ is not correct, but almost, it might be a typo in the book:

Say, we have $\mathcal A\overset{F,F'}\to\mathcal B$, then the natural transformation $U\sigma\circ\eta$ goes $1_{\mathcal A}\to UF\to UF'$ while the domain of $U\varepsilon'$ is $UF'U'$. So we are missing an $U'$ here, by syntax, and the correct version would be $$D(\sigma)\ :=\ U\varepsilon'\circ U\sigma U'\circ \eta U'\,,$$

Now, this $D(\sigma)$ goes $U'\Rightarrow U\ $ in the functor category $Fun(\mathcal B, \mathcal A)$, but if we converse the arrows in $\mathcal A$, we get the opposite direction (for the exactly same mappings $U,U'$):
$\quad D(\sigma):U\Rightarrow U'$ in $Fun(\mathcal B^{op},\mathcal A^{op})$, $\ $ exactly in the wished direction.

Finally, we will have $$D^2(\sigma)=\varepsilon F'\circ\ F\,D(\sigma)\,F' \circ F\eta'\,.$$

---

I personally prefer to work with double categorical notation, i.e, arranging the 2-cells into squares. E.g., the unit of the adjunction $\eta$ can be drawn as

$\ \ \ \, \overset{1_{\mathcal A}}\longrightarrow$
$F\! \downarrow \,\eta\,\downarrow \!\! 1_{\mathcal A} \quad\quad$ read in direction $\ \swarrow$
$\ \ \ \, \underset{U}\longrightarrow$

Then, $D(\sigma)$ will be just three squares ($\varepsilon',\ \sigma$ and $\eta$) pasted together.

All in all, $D^2(\sigma)$ will be the composite of the following squares:

$\def\-{-\!\!-\!\!-} \ \ \ \ \-$
$ \ \ \ \ \vert\ \ \eta' \ \vert$
$ \ \ \ \ \- \, \- \, \- $
$ \ \ \ \ \vert\ \ \varepsilon'\ \ \vert\ \ \ \sigma\ \ \vert\ \ \ \eta\ \ \vert$
$ \ \ \ \ \- \, \- \, \- $
$\hspace{4.2pc}\vert\ \ \ \varepsilon\ \ \vert$
$\hspace{4.4pc}\-$

which indeed gives back $\sigma$ as expected, by the adjunction properties.

(All you need to verify is that horizontal and vertical compositions of 2x2 squares commute with each other in this context, and then extend the picture by identity squares.)

Answer 2

The following calculation shows the claim using just vertical composition of morphisms (horizontal one only implicitly): $$\begin{align*} (D^2\sigma)_X&=\varepsilon_{F'X}\circ F(D(\sigma)_{F'X})\circ F\eta'_X &\text{by definition}\\ &=\varepsilon_{F'X}\circ F(U\varepsilon'_{F'X}\circ U\sigma_{U'F'X}\circ \eta_{U'F'X})\circ F\eta'_X&\text{by definition}\\ &=(\varepsilon_{F'X}\circ FU\varepsilon'_{F'X})\circ FU\sigma_{U'F'X}\circ F\eta_{U'F'X}\circ F\eta'_X&\text{because $F$ is a functor}\\ &=\varepsilon'_{F'X}\circ (\varepsilon_{F'U'F'X}\circ FU\sigma_{U'F'X})\circ F\eta_{U'F'X}\circ F\eta'_X&\text{by naturality of $\varepsilon$}\\ &=\varepsilon'_{F'X}\circ (\sigma_{U'F'X}\circ \underbrace{\varepsilon_{FU'F'X})\circ F\eta_{U'F'X}}_{=1_{U'F'X}}\circ F\eta'_X&\text{by naturality of $\varepsilon$}\\ &=\underbrace{\varepsilon'_{F'X}\circ F'\eta'_X}_{=1_{F'X}}\circ \sigma_X&\text{by naturality of $\sigma$}\\ &=\sigma_X \end{align*}$$