# Kan extension “commutes” with a certain left adjoint

Let $$\mathcal{A},\mathcal{B}$$ be small categories, $$\mathcal{C}$$ a cocomplete category and $$\mathcal{D}$$ an arbitrary category. Consider functors $$F:\mathcal{A}\rightarrow\mathcal{B}$$, $$G:\mathcal{A}\rightarrow\mathcal{C}$$, $$R:\mathcal{D}\rightarrow\mathcal{C}$$ and $$L:\mathcal{C}\rightarrow\mathcal{D}$$, where $$L$$ is left adjoint to $$R$$. We want to show that $$L\circ\text{Lan}_F(G)=\text{Lan}_F(L\circ G).$$

The author provides the following proof:

For every functor $$H:\mathcal{B}\rightarrow\mathcal{D}$$, we get the following bijections:

\begin{align} \text{Nat}\left(L\circ\text{Lan}_F(G),H\right) & \cong \text{Nat}\left(\text{Lan}_F(G),R\circ H\right) \\ & \cong \text{Nat}\left(G,R\circ H\circ F\right) \\ & \cong \text{Nat}\left(L\circ G,H\circ F\right)\\ & \cong \text{Nat}\left(\text{Lan}_F(L\circ G),H\right) .\end{align}

What is the justification for these isomorphisms; i.e. how does one deduce them? On the other hand, are these classes of natural transformations sets?

Edit:

Let $$R_*:[\mathcal{B},\mathcal{D}]\rightarrow[\mathcal{B},\mathcal{C}]$$ be the functor defined by $$R_*(H)=R\circ H$$ and $$R_*(\alpha)=R*\alpha$$. Then $$R_*$$ is left adjoint to $$L_*$$. But I am not sure how this helps. If $$\mu:L\circ\text{Lan}_F(G)\Rightarrow H$$, then $$R_*(\mu):R\circ L\circ\text{Lan}_F(G)\Rightarrow R\circ H$$ –but this doesn't give a natural transformation from $$\text{Lan}_F(G)$$ to $$R\circ H$$.

## 1 Answer

If $$L\dashv R$$, then $$L_*\dashv R_*$$, not the other way around (the places are switched if you take precomposition, $$L^*, R^*$$)

A good way to see that $$L_*\dashv R_*$$ is via the formulation of adjoints with a unit and a counit satisfying the triangle identities.

You'll want $$\epsilon_* : L_*R_*\to id$$, which is simply given by $$\epsilon G : LRG\to G$$ for any $$G$$, and $$\eta_* : id\to R_*L_*$$ which is also given by $$\eta G : G\to RLG$$. That they satisfy the triangle identities is essentially obvious because $$\epsilon,\eta$$ do (if you're not convinced, write it down !)

Therefore all the isomorphisms in the proof are justified.

(with regards to your edit, maybe a more concrete way to see it : suppose you have $$\theta : LT\to S$$, then you get $$R\theta : RLT\to RS$$, and then you can precompose by $$\eta T : T\to RLT$$ to get $$(R\theta)\circ \eta T : T\to RS$$)

• Thank you. Just a quick question: Are these classes of natural transformations sets? I ask because I would like to know if any link can be drawn between this result and Yoneda. – alf262 Jul 30 '20 at 17:59
• Since $\mathcal{A,B}$ are assumed to be small, they are sets indeed (for any small category $C$, and any category $D$ - which is assumed to be locally small, as is done most of the time -, $Fun(C,D)$ is locally small, that is, $Nat(F,G)$ is a set). But even if they were classes, you could use Yoneda, you just have to be careful in a sense – Maxime Ramzi Jul 30 '20 at 18:05