I have been trying to understand the statement of the title but seems that I am getting stuck on something in the very last of its proof.

To start with, assume that we have the adjunction $\mathsf{C} \overset{\mathcal{F}} {\underset{\mathcal{G}}\rightleftarrows} \mathsf{D}$, $\mathcal{F} \dashv \mathcal{G}$. Our statement, is that we have the following diagram \begin{array}{cc} \,\,\,\,\,\,\,\mathsf{C} \\ \\ \mathcal{F} \downarrow & \,\,\,\,\,\,\,\searrow \,{id_{\mathsf{C}}} \\ \\ \mathsf{D} & \xrightarrow{\mathcal{G} \cong Lan_{\mathcal{F}}id_{\mathsf{C}}} & \mathsf{C}, \end{array} In other words, that $\mathcal{G}$, is the left Kan Extension of the identity functor $id_{\mathsf{C}}$ along $\mathcal{F}$. In order to prove that, we have to prove that $\mathcal{G}$, fulfils the universality of left Kan extensions.

Where is my problem: Apparently, due to the adjunction we can define the natural transformation $\eta : id_{\mathsf{C}} \rightarrow \mathcal{G \circ F}$, to be the unit of the adjunction. Then we assume that another functor $\mathcal{H}: \mathsf{D} \rightarrow \mathsf{C}$ along with a natural transformation $\gamma : id_{\mathsf{C}} \rightarrow \mathcal{H \circ F}$, exists and then we have to find a unique natural transformation $\delta : \mathcal{G} \rightarrow \mathcal{H}$, such that $\delta_{\mathcal{F}} \circ \eta_{\mathcal{F}}= \gamma_{\mathcal{F}}.$ However, to find out $\delta$, it's kind of simple, since due to the adjunction again, we have the following adjunction too $\mathsf{C}^{\mathsf{C}} \overset{\mathcal{G}^{*}} {\underset{\mathcal{F}^{*}} \leftrightarrows} \mathsf{C}^{\mathsf{D}}$. Therefore the uniqueness of transformation follows. What I cannot prove is why we end up having the composition $\delta_{\mathcal{F}} \circ \eta_{\mathcal{F}}= \gamma_{\mathcal{F}}$.

Thank you.

• Technically, it's $\delta_\mathcal{F}\circ\eta=\gamma$; no subscripts on $\eta$ and $\gamma$. – Derek Elkins Jun 7 '17 at 22:21
• Yes you're right, I wrote it out to emphasize which diagram should be commutative due to universality of Kan extension. However, if you believe that creates more troubles than it solves I can edit it. – user321268 Jun 7 '17 at 22:26

You can do it directly using the unit-counit zig-zag identities. What you are trying to do is the argument for 1. implies 2. in the following lemma with $K=\mathrm{id}_{\mathcal A}$ (sorry, the names of my objects differ from yours).

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

1. There exists a $2$-morphism $FG\overset\epsilon\Rightarrow \mathrm{id}_{\mathcal B}$ satisfying the zig-zag identities

• $F\overset{F\eta}\Rightarrow FGF\overset{\epsilon F}\Rightarrow F$ is the identity $2$-morphism $F\overset{\mathrm{id}_F}\Rightarrow F$
• $G\overset{\eta G}\Rightarrow GFG\overset{G\epsilon}\Rightarrow G$ is the identity $2$-morphism $G\overset{\mathrm{id}_G}\Rightarrow G$
2. $G=\Lan_F^\eta\mathrm{id}_{\mathcal A}$ is an absolute left Kan extension, i.e. such that $KG=\Lan_F^{K\eta}K$ for any $1$-morphism $\mathcal A\xrightarrow{K}\mathcal C$.

3. $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF$ is a $2$-morphism such that

• The left extension of $\mathcal A\xrightarrow{F}\mathcal B$ along itself given by $F\overset{F\eta}\Rightarrow FGF$ is weakly initial among such extensions, i.e. has a morphism of extensions to any other extension.

• $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF$, considered as a left extension of $\mathcal A\xrightarrow{\mathrm{id}_{\mathcal A}}\mathcal A$ along $\mathcal A\xrightarrow{F}\mathcal B$, has no non-identity endomorphisms

Proof of 1. implies 2..

Consider a left extension of $\mathcal A\xrightarrow{K}\mathcal C$ along $\mathcal A\xrightarrow{F}\mathcal B$ given by a $2$-morphism $K\overset\tau\Rightarrow HF$. We claim that the unique morphism of extensions $KG\overset{\phi}\Rightarrow H$ is given by $KG\overset{\tau G}\Rightarrow HFG\overset{H\epsilon}\Rightarrow H$.

To show that the above defines a factorization, observe that the $2$-moprhism $KG\overset{\tau G}\Rightarrow HFG\overset{H\epsilon}\Rightarrow H$ is in fact an extension in two steps. First, $K\overset{K\eta}\Rightarrow KGF\overset{\tau GF}\Rightarrow HFGF\overset{H\epsilon F}\Rightarrow HF$ is, by naturality/horizontal composition/interchange law, the same as $K\overset{\tau}\Rightarrow HF\overset{HF\eta}\Rightarrow HFGF\overset{H\epsilon F}\Rightarrow HF$. Second, use the first zig-zag identity to conclude that this is precisely $K\overset\tau\Rightarrow HF$.

To show uniqueness of the factorization, suppose $K\overset\tau\Rightarrow HF$ factors as $K\overset{K\eta}\Rightarrow KGF\overset{\phi F}\Rightarrow HF$ for a $2$-morphism $KG\overset\phi\Rightarrow H$. We again proceed in two steps. First, the composite $KG\overset{K\eta G}\Rightarrow KGFG\overset{\phi FG}\Rightarrow HFG\overset{H\epsilon}\Rightarrow G$ is, by naturality/horizotnal composition/interchange law, the same as $KG\overset{K\eta G}\Rightarrow KGFG\overset{KG\epsilon}\Rightarrow KG\overset{\phi}\Rightarrow H$. Second, the latter is the same as $KG\overset{K\mathrm{id}_G}\Rightarrow KG\overset{\phi}\Rightarrow H$ by th second zig-zag identity.

Proof of 3. implies 1. Begin by observing that the first zig-zag identity says precisely that $F\overset{\mathrm{id}_F}\Rightarrow F$, interpreted as an extension $\mathrm{id}_{\mathcal B}F\Rightarrow F$ of $\mathcal A\xrightarrow{F}\mathcal A$ along itself, has a morphism of extensions $FG\overset\epsilon\Rightarrow\mathrm{id_{\mathcal B}}$ from the extension $F\overset{F\eta}\Rightarrow FGF$. But the existence of this morphism is precisely the weakly initial hypothesis on the extension $F\overset{F\eta}\Rightarrow FGF$.

Next, observe that, given the first zig-zag identity, the $2$-endomorphism $G\overset{\eta G}\Rightarrow GFG\overset{G\epsilon}\Rightarrow G$ is in fact an endomorphism of the left extension $\mathrm{id_{\mathcal A}}\overset\eta\Rightarrow GF$. Indeed, the composite $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF\overset{\eta GF}\Rightarrow GFGF\overset{G\epsilon F}\Rightarrow GF$ is by naturality/horizontal composition/interchange law the same as $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF\overset{GF\eta}\Rightarrow GFGF\overset{G\epsilon F}\Rightarrow GF$, and the first zig-zag identity says that this is precisely $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF\overset{\mathrm{id}_GF}\Rightarrow GF$. But by hypothesis, $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF$ has no non-identity endomorphisms, hence the second zig-zag identity holds.

Corollary. Because the zig-zag identities are $1$- and $2$-dual, the following conditions are equivalent: $\DeclareMathOperator{\Lift}{Lift} \DeclareMathOperator{\Rift}{Rift} \DeclareMathOperator{\Ran}{Ran}$

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
• Thank you very much Vladimir, very nice proof (probably a couple of typos if I understand correctly) and explanation! I posted a comment, for something I didn't understand in the first place , but I did eventually. – user321268 Jun 8 '17 at 12:55
• One question, when you write naturality/horizontal composition/interchange law, could you please be more precise for which diagram exactly you mean? – user321268 Jun 8 '17 at 20:39
• I'm referring to the following fact: given $2$-morphisms $H\overset\beta\Rightarrow K\colon\mathcal C\to\mathcal D$ and $F\overset\alpha\Rightarrow G\colon\mathcal B\to\mathcal C$, the two composites $HF\overset{\beta F}\Rightarrow KF\overset{K \alpha}\Rightarrow KG$ and $HF\overset{H\alpha}\Rightarrow HG\overset{\beta G}\Rightarrow KG$ are equal. In the case of categories, functors, and natural transformations, this holds precisely because of the commutative squares that naturality requires. – Vladimir Sotirov Jun 9 '17 at 6:11
• Thank you for your reply again. So if I perceive correctly, in our case we have $\mathcal{B}=\mathcal{D}=\mathcal{C}$? And the 2-cells are what? (Sorry for nitpicking I do want really to understand that). – user321268 Jun 9 '17 at 9:15
• Yes; the $2$-morphisms that are equal are $K\overset{K\eta}\Rightarrow KGF\overset{\tau GF}\Rightarrow HFGF$ and $K\overset{\tau}\Rightarrow HF\overset{HF\eta}\Rightarrow HFGF$ in the proof of existence, and $KGFG\overset{\phi FG}\Rightarrow HFG\overset{H\epsilon}\Rightarrow G$ and $KGFG\overset{KG\epsilon}\Rightarrow KG\overset{\phi}\Rightarrow H$ in the proof of uniqueness. – Vladimir Sotirov Jun 10 '17 at 0:48