I am reading the chapter on Kan extenions in the Handbook of Categorical Algebra by Francis Borceux and I am a bit confused about some of the steps he takes.
My main question is a step in his proposition 3.7.4:
Let $\mathcal{A},\mathcal{B},\mathcal{C},\mathcal{D}$ be categories, with $\mathcal{A}$ and $\mathcal{B}$ small. Let $F:\mathcal{A}\rightarrow\mathcal{B}$ and $G:\mathcal{A}\rightarrow\mathcal{C}$ such that $\operatorname{Lan}_F{G}$ exists. Let $L:\mathcal{C}\rightarrow \mathcal{D}$ be the left adjoint of a functor $R:\mathcal{D}\rightarrow\mathcal{C}$. Then $$\textbf{Nat}(\operatorname{Lan}_FG,RH)\cong\textbf{Nat}(G,RHF)$$
$\operatorname{Lan}_FG$ denotes the left Kan extension of $G$ along $F$. I have no clue why this should hold. At first I thought that for $\alpha\in \textbf{Nat}(\operatorname{Lan}_FG,RH)$, you would take $\alpha F$, the natural transformation which acts on objects as $(\alpha F)_A = \alpha_{F(A)}$ and then compose with the natural transformation $G\Rightarrow \operatorname{Lan}_FG\circ F$given by the Kan extension. But going from $\alpha$ to $\alpha F$ does not give a bijection, since for example $F$ is not necessarily surjective on the objects. What am I missing here?
The definition of the left Kan extension used in Borceux's book is
It's a useful skill to be able to convert between the presentation of something as a universal property and as saying that a certain functor is representable. Often this second presentation manifests as an isomorphism between two homsets, or in this case, two sets of natural transformations.
There's a map from $\textbf{Nat}(K, H)$ to $\textbf{Nat}(G, H \circ F)$ given by $\gamma \mapsto (\gamma * F) \circ \alpha$. By the universal property, this is an isomorphism. For any $\beta$ in $\textbf{Nat}(G, H \circ F)$, there's a unique $\gamma$ in $\textbf{Nat}(K, H)$ that maps to it.
Polishing up the notation, we have that for any functors $F : \mathcal A \to \mathcal B$, $G : \mathcal A \to \mathcal C$ and $H : \mathcal B \to \mathcal C$, if $\operatorname{Lan}_F G$ exists, then $\textbf{Nat}(\operatorname{Lan}_F G, H) \cong \textbf{Nat}(G, H F)$. (Note that $\operatorname{Lan}_F G$ is the $K$ from above).
So in particular, with $R H$ as $H$, we get $\textbf{Nat}(\operatorname{Lan}_F G, R H) \cong \textbf{Nat}(G, R H F)$.