Kan exquestions

Question

There seems to be two strange things in Borceux's treatment of Kan extensions.

1. Proposition 3.7.6 states that the following statements are equivalent for a functor between small categories $F : A \to B$:

   • $F$ has a right adjoint;
   • $\text{Lan}_FF \cong F \circ \text{Lan}_F1$;
   • $\text{Lan}_FL \cong L\circ \text{Lan}_F1$ for every functor $L : A \to C$.

   This seems to imply that $F$ preserves all colimits that exist in $A$ (which cannot be so many, since $A$ is small, but..); and yet I'm not able to see why the fact that it preserve that particular left Kan extension $\text{Lan}_F1$ gives that it preserves all colimits.

2. The statement of Exercise 3.9.5 makes no sense, as it is impossible to compose $\text{Lan}_GF\circ F$.

What's going on?

Answer 1

The first result is covered (in dual and enriched form) in Basic Concepts of Enriched Category Theory as Theorem 4.81 which is a corollary to Theorem 4.80 which I prove below (in the unenriched case), the easily proven fact (4.28) that $F\dashv U$ implies that $\mathsf{Lan}_FG\cong G\circ U$ and $\mathsf{Ran}_UK\cong K\circ F$, and the representation of Kan extensions as weighted (co)limits, e.g. $\mathsf{Ran}_UK(C)\cong\{\text{Hom}(C,U-),K\}$

A functor $H : \mathcal{C}\to\mathbf{Set}$ is representable iff the weighted limit $\{H,Id\}$ exists and is preserved by $H$ in which case $H\cong\text{Hom}(\{H,Id\},-)$.

Only if: If $H\cong\mathcal{C}(X,-)$, then $$H(\{H,Id\})\cong\mathcal{C}(X,\{H,Id\})\cong[\mathcal{C},\mathbf{Set}](H,\mathcal{C}(X,-))\cong\mathcal{C}(X,X)$$ by the universal property of weighted limits and Yoneda. So $X \cong \{H,Id\}$ by Yoneda. Preservation is obvious either because hom-functors preserve all limits or because $\{H,H\}\cong[\mathcal{C},\mathbf{Set}](H,H)$ and so preservation was implicit in the above.

If: Assume $H$ preserves $\{H,Id\}$, i.e. it takes limiting cylinders to limiting cylinders which is to say the universal element of the weighted limit $\{H,Id\}$, namely $\varepsilon : H \to \mathcal{C}(\{H,Id\},-)$, when followed by the action of $H$ on arrows produces the universal element of the weighted limit $\{H,H\}$, namely $\varepsilon' : H \to \mathcal{C}(\{H,H\},H-)$. That is, $$\require{AMScd} \begin{CD} H @>\varepsilon >> \mathcal{C}(\{H,Id\},-) \\ @V\varepsilon'VV @VVHV \\ \mathcal{C}(\{H,H\},H-)@>\cong>> \mathcal{C}(H\{H,Id\},H-) \end{CD}$$ The goal will be to show that $\varepsilon$ has an inverse. In this case, $\varepsilon'$ being a universal element means given a set $X$ and a natural transformation $\tau : H \to \mathbf{Set}(X, H-)$ there exists a unique function $f' : X \to \{H,H\}$ such that $\tau_C = \mathbf{Set}(f,HC)\circ\varepsilon_C'$. In particular, choosing $X=1$ and $\tau = id$ (identifying $H\cong\mathbf{Set}(1,H-)$) we get a unique $f'\in\{H,H\}$ such that $id = \mathbf{Set}(f',H-)\circ\varepsilon'$. That is, $\mathbf{Set}(f,-)\circ H\circ\varepsilon = H(\varepsilon(x))(f) = x$ where $f \in H\{H,Id\}$ corresponds to $f'$.

To show that $\mathbf{Set}(f,-)\circ H$ is an inverse we need to show that $\varepsilon \circ \mathbf{Set}(f,-)\circ H : \mathcal{C}(\{H,Id\},-)\to\mathcal{C}(\{H,Id\},-)$ is the identity. By Yoneda, this transformation is determined by its action at $id_{\{H,Id\}}$. We have $\varepsilon_{\{H,Id\}}(f) : \{H,Id\}\to\{H,Id\}$ and $$\begin{align} (\mathcal{C}(\varepsilon_{\{H,Id\}}(f),-)\circ\varepsilon)(x) & = \varepsilon(x)\circ\varepsilon_{\{H,Id\}}(f) \\ & = \varepsilon(H(\varepsilon(x))(f)) \\ & = \varepsilon(x) \end{align}$$ and since $\varepsilon$ is a universal element that means $\varepsilon_{\{H,Id\}}(f) = id$. $\square$

Answer 2

$\def\Lan{\text{Lan}}$ Borceux's proposition is simply the equivalence of the unit-counit formulation of adjoint functors and the absolute Kan extension formulation of adjoint functors. In particular, it is valid in any $2$-category. Below is my attempt at a brief self-contained exposition of the relevant results.

$$\require{AMScd}\begin{CD} \mathcal A @>Y>> \mathcal E @>G>> \mathcal C\\ @VXVV \overset\epsilon\Rightarrow @| \overset{\epsilon'}\Leftarrow @VFVV\\ \mathcal B @>Z>> \mathcal E @>J>>\mathcal D \end{CD}$$ (Since amscd is inflexbile, for clarity the natural transformations are supposed to be $ZX\overset\epsilon\Rightarrow Y$ and $FG\overset{\epsilon'}\Rightarrow J$).

Lemma 1. Suppose that

1. $JZ=\DeclareMathOperator{\Ran}{Ran}\Ran^{\epsilon_1}_XJY$, i.e. that $JZX\overset{\epsilon_1}\Rightarrow JY$ is a right Kan extension of $\mathcal A\xrightarrow{JY}\mathcal D$ along $\mathcal A\xrightarrow{X}\mathcal B$. In the diagram above, $JZ=\Ran_X^{J\epsilon}JY$, i.e. $J\epsilon=\epsilon_1$.
2. $GZ=\DeclareMathOperator{\Rift}{Rift}\Rift_{\epsilon_2}^FJZ$, i.e. that $FGZ\overset{\epsilon_2}\Rightarrow JZ$ a right Kan lift of $\mathcal B\xrightarrow{JZ}\mathcal D$ along $\mathcal C\xrightarrow{F}\mathcal D$. In the diagram above, $GZ=\Rift_{\epsilon'Z}^FJZ$, i.e. $\epsilon'Z=\epsilon_2$.
3. $GY=\Rift_{\epsilon_3}^FJY$, i.e. that $FGY\overset{\epsilon_3}\Rightarrow JY$ is a right Kan lift of $\mathcal A\xrightarrow{JY}\mathcal D$ along $\mathcal C\xrightarrow{F}\mathcal D$. In the diagram above, $GY=\Rift^F_{\epsilon Y}JY$, i.e. $\epsilon_3=\epsilon' Y$.

Then

• $GZ=\Ran_X^{\epsilon_4}GY$ where $GZX\overset{\epsilon_4}\Rightarrow GY$ is the unique $2$-morphism such that the right lift $FGZX\overset{\epsilon_2X}\Rightarrow JZX\overset{\epsilon_1}\Rightarrow JY$ of $\mathcal A\xrightarrow{GZX}\mathcal C$ of $\mathcal A\xrightarrow{JY}\mathcal D$ factors as $FGZX\overset{F\epsilon_4}\Rightarrow FGY\overset{\epsilon_3}\Rightarrow JY$ by the universal property of a right Kan lift. In the diagram above, $\epsilon_4=G\epsilon$.

Corollary 1. $G=\Rift_{\epsilon'}^FJ$ preserves a right Kan extension $Z=\Ran^\epsilon_XY$ in the sense that $GZ=\Ran^{G\epsilon}_XGY$ if both $\mathcal B\xrightarrow{Z}\mathcal C$ and $\mathcal A\xrightarrow{Y}\mathcal E$ respect the right Kan lift $G=\Rift_\epsilon^FJ$ in the sense that $GZ=\Rift_{\epsilon' Z}^FJZ$ and $GY=\Rift_{\epsilon' Y}^FJY$.

Definition 1. $\mathcal E\xrightarrow{G}\mathcal C$ is a $J$-relative right adjoint $F\dashv_JG$ to $\mathcal C\xrightarrow{F}\mathcal D$ if $G=\Rift_{\epsilon'}^FJ$ is an absolute right Kan lift in the sense that it is respected by every functor to $\mathcal E$.

Corollary 2.. A $J$-relative right adjoint $F\dashv_J G$ preserves all right Kan extensions that $J$ does.

Definition 2. A right adjoint to $\mathcal A\xrightarrow{F}\mathcal B$ is $\mathrm{id_B}$-relative right adjoint, i.e. an absolute right Kan lift $G=\Rift^F_\epsilon\mathrm{id_B}\colon\mathcal B\to\mathcal A$.

Corollary 3. Since a right adjoint is the adjoint relative to the identity functor, and the identity functor preserves all right Kan extensions, right adjoints preserve all right Kan extensions (in particular all limits).

Lemma 2. Consider $1$-morphisms $\mathcal A\xrightarrow{F}\mathcal B$, $\mathcal A\xleftarrow{G}\mathcal B$, and a $2$-morphism $FG\overset\epsilon\Rightarrow\mathrm{id}_B$. Then the following conditions are equivalent

1. $G=\Rift^F_\epsilon\mathrm{id}_B$ and $GF=\Rift^F_{\epsilon F}F$

2. There exists a unique $2$-morphism $\mathrm{id}_A\overset\eta\Rightarrow GF$ satisfying the zig-zag identities

   • $F\overset{F\eta}\Rightarrow FGF\overset{\epsilon F}\Rightarrow F$ is the identity $2$-morphism
   • $G\overset{\eta G}\Rightarrow GFG\overset{G\epsilon}\Rightarrow G$ is the identity $2$-morphism

3. $G=\Rift^F_\epsilon\mathrm{id}_B$ is an absolute right Kan lift.

Proof. The existence and uniqueness of $\eta$ satisfying the first zig-zag identity follows from interpreting $F\mathrm{id_{A}}\overset{\mathrm{id}_F}\Rightarrow F$ as a right lift of $F$ along itself, and applying the universal property of $GF=\Rift^F_{F\epsilon}F$. The second zig-zag identity follows from the uniqueness property of $G=\Rift^F_\epsilon\mathrm{id}_B$ since $FG\overset{F\eta G}\Rightarrow FGFG\overset{FG\epsilon}\Rightarrow FG\overset{\epsilon}\Rightarrow\mathrm{id}_B$ is the same as $FG\overset{F\eta G}\Rightarrow FGFG\overset{\epsilon FG}\Rightarrow FG\overset{\epsilon}\Rightarrow\mathrm{id}_B$ by horizontal composition, and then the same as $FG\overset{\mathrm{id}_FG}\Rightarrow FG\overset\epsilon\Rightarrow$ by the first zig-zag identity.

That the zig-zag identities imply $G=\Rift^F_{\epsilon}\mathrm{id}_B$ is an absolute right Kan lift, i.e. that each $LG=\Rift^F_{\epsilon L}L$ is a right Kan lift is easy to see. If $FH\overset\tau\Rightarrow L$ is a natural transformation, then it factors as $FH\overset{F\eta H}\Rightarrow FGFH\overset{FG\tau}\Rightarrow FGL\overset{\epsilon L}\Rightarrow L$ using the first-zag identity and horizontal composition. Uniqueness follows from an application of the second zig-zag identity.

Corollary. Because the zig-zag identities are $1$- and $2$-dual, the following conditions are equivalent:

1. $F\dashv G$ with unit $\mathrm{id}_A\overset\eta\Rightarrow GF$ and counit $FG\overset\epsilon\Rightarrow\mathrm{id}_A$
2. $G=\Rift_\epsilon^F\mathrm{id}_B$ is an absolute right Kan lift
3. $G=\Rift_\epsilon^F\mathrm{id}_B$ and $GF=\Rift_{\epsilon F}^FF$ are right Kan lifts
4. $F=\DeclareMathOperator{\Lift}{Lift}\Lift_\eta^G\mathrm{id}_A$ is an absolute left Kan lift
5. $F=\Lift_\eta^G\mathrm{id}_A$ and $FG=\Lift_{\eta G}^G G$ are left Kan lifts
6. $G=\Lan^\eta_F\mathrm{id}_A$ is an absolute left Kan extension
7. $G=\Lan^\eta_F\mathrm{id}_A$ and $FG=\Lan^{F\eta}_FF$ are left Kan extensions
8. $F=\Ran^\epsilon_G\mathrm{id}_B$ is an absolute right Kan extension
9. $F=\Ran^\epsilon_G\mathrm{id}_B$ and $GF=\Ran^{G\epsilon}_GG$ are right Kan extensions