6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.)

$\endgroup$
7
  • 2
    $\begingroup$ Could you elaborate on the last paragraph? I got the first paragraph like you, just abused notation and wrote $\eta$ for $\eta_{U'}$. But the $D^2=Id$ gave me the condition I stated which I couldn't prove. $\endgroup$ May 5 '14 at 22:08
  • $\begingroup$ Ah, I see now your problem. There is a twist in the operations 'paste $\varepsilon$ to the left' and '$\eta$ to the right' while taking the opposite category. I will update my answer. $\endgroup$
    – Berci
    May 5 '14 at 23:43
  • 1
    $\begingroup$ Thanks for your work, but I still don't get it. Why is what you wrote for $D^2(\sigma)$ equal to $\sigma$ or in your diagrams how does vertical composition enter. I only see one type of composition in the formulae. $\endgroup$ May 6 '14 at 7:57
  • $\begingroup$ Unfortunately \xymatrix is not supported here, so I could not label the arrows, but I think you can find them, most of them are identity. Now e.g. the upper left part of the diagram, consisting of squares $\eta',\ \varepsilon',\ \sigma$ corresponds to the composite $\varepsilon F'\circ \sigma U'F'\circ F\eta'$. Here we evaluated horizontally first, and I claim that if we instead evaluate this part vertically first, then -- using naturality of transformations -- we get that this is equal to $\varepsilon'F'\circ F'\eta'\circ\sigma$. $\endgroup$
    – Berci
    May 6 '14 at 11:21
  • $\begingroup$ Perhaps it is easier to state and prove the more general form: whenever $\pmatrix{\alpha&\beta\\ \gamma&\delta}$ are pastable squares, i.e. $\alpha:F_2U_1\Rightarrow V_1F_1$ pictured with $U_1$ on top and $V_1$ on bottom, $\beta:F_3U_2\Rightarrow V_2F_2$, etc. Then the horizontal compositions $\alpha|\beta$, $\ \gamma|\delta$ and vertical compositions $\frac\alpha\gamma$, $\ \frac\beta\delta$ are defined in a straightforward way, and these operations commute: $$\frac{\alpha|\beta}{\gamma|\delta}\ =\ \frac\alpha\gamma \,\vert\,\frac\beta\delta\,.$$ $\endgroup$
    – Berci
    May 6 '14 at 11:31
1
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ So, for natural transformations $\alpha:F_1\to G_1$ and $\beta:F_2\to G_2$, using naturality, we get the lemma that $G_1\beta\circ \alpha F_2\ =\ \alpha G_2\circ F_1\beta$. Now define the squares to be quintets $\langle T,B,L,R,\phi\rangle$ such that $\phi:RT\Rightarrow BL$, and try to define the horizontal and vertical pasting of these squares (both using the vertical composition of nat.transformations), and prove the interchangability condition for general four pastable squares, using the lemma for two squares in diagonal. $\endgroup$
    – Berci
    May 8 '14 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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