[proofplan] The proof is a direct unpacking of the defining universal properties of left and right Kan extensions. For the left Kan extension, the universal arrow $\eta: F \Rightarrow L K$ makes every natural transformation $F \Rightarrow G K$ factor uniquely through a natural transformation $L \Rightarrow G$, which gives the stated bijection. Naturality in $G$ follows by comparing the two possible factorizations after composing with a natural transformation $G \Rightarrow G'$. The proof for right Kan extensions is the dual argument, with the counit $\varepsilon: R K \Rightarrow F$ and precomposition by natural transformations into $R$. [/proofplan] [step:Extract the left Kan extension bijection from its defining factorization property] Let $(L,\eta)$ be a left Kan extension of $F$ along $K$, with $L: \mathcal D \to \mathcal E$ and $\eta: F \Rightarrow L\circ K$. Fix a functor $G: \mathcal D \to \mathcal E$. For every natural transformation $\bar{\alpha}: L \Rightarrow G$, define \begin{align*} \Phi_G(\bar{\alpha}) := (\bar{\alpha}K)\circ \eta, \end{align*} where $\bar{\alpha}K: L\circ K \Rightarrow G\circ K$ is the natural transformation obtained by whiskering $\bar{\alpha}$ with $K$. By the defining universal property of the left Kan extension $(L,\eta)$, for every natural transformation $\alpha: F \Rightarrow G\circ K$ there exists a unique natural transformation $\bar{\alpha}: L \Rightarrow G$ such that \begin{align*} \alpha = (\bar{\alpha}K)\circ \eta. \end{align*} This says exactly that $\Phi_G$ is surjective and injective. Hence \begin{align*} \Phi_G: [\mathcal D,\mathcal E](L,G) \to [\mathcal C,\mathcal E](F,G\circ K) \end{align*} is a bijection. [guided] We first translate the definition of a left Kan extension into the notation of the statement. A left Kan extension of $F$ along $K$ is a pair $(L,\eta)$ consisting of a functor $L: \mathcal D \to \mathcal E$ and a natural transformation \begin{align*} \eta: F \Rightarrow L\circ K \end{align*} with the following universal property: whenever $G: \mathcal D \to \mathcal E$ is a functor and $\alpha: F \Rightarrow G\circ K$ is a natural transformation, there is a unique natural transformation $\bar{\alpha}: L \Rightarrow G$ satisfying \begin{align*} \alpha = (\bar{\alpha}K)\circ \eta. \end{align*} Now fix $G: \mathcal D \to \mathcal E$. The proposed map \begin{align*} \Phi_G: [\mathcal D,\mathcal E](L,G) \to [\mathcal C,\mathcal E](F,G\circ K) \end{align*} sends a natural transformation $\bar{\alpha}: L \Rightarrow G$ to the composite \begin{align*} F \xRightarrow{\eta} L\circ K \xRightarrow{\bar{\alpha}K} G\circ K. \end{align*} That composite is precisely $(\bar{\alpha}K)\circ \eta$. The universal property says that every natural transformation $\alpha: F \Rightarrow G\circ K$ is obtained in this way from some $\bar{\alpha}: L \Rightarrow G$, so $\Phi_G$ is surjective. It also says that this $\bar{\alpha}$ is unique, so two natural transformations $L \Rightarrow G$ cannot have the same image under $\Phi_G$; hence $\Phi_G$ is injective. Therefore $\Phi_G$ is a bijection. [/guided] [/step] [step:Verify naturality of the left Kan extension bijection in the target functor] Let $G,G': \mathcal D \to \mathcal E$ be functors, and let $\theta: G \Rightarrow G'$ be a natural transformation. The functor $[\mathcal D,\mathcal E](L,-)$ sends $\bar{\alpha}: L \Rightarrow G$ to $\theta\circ \bar{\alpha}: L \Rightarrow G'$, while the functor $[\mathcal C,\mathcal E](F,-\circ K)$ sends $\alpha: F \Rightarrow G\circ K$ to $(\theta K)\circ \alpha: F \Rightarrow G'\circ K$. For every $\bar{\alpha}: L \Rightarrow G$, associativity of [vertical composition of natural transformations](/theorems/3961) gives \begin{align*} \Phi_{G'}(\theta\circ \bar{\alpha}) &= (((\theta\circ \bar{\alpha})K)\circ \eta) \\ &= ((\theta K)\circ(\bar{\alpha}K))\circ \eta \\ &= (\theta K)\circ ((\bar{\alpha}K)\circ \eta) \\ &= (\theta K)\circ \Phi_G(\bar{\alpha}). \end{align*} Thus the family $(\Phi_G)_G$ is natural in $G$. [/step] [step:Extract the right Kan extension bijection from its defining factorization property] Let $(R,\varepsilon)$ be a right Kan extension of $F$ along $K$, with $R: \mathcal D \to \mathcal E$ and $\varepsilon: R\circ K \Rightarrow F$. Fix a functor $G: \mathcal D \to \mathcal E$. For every natural transformation $\bar{\beta}: G \Rightarrow R$, define \begin{align*} \Psi_G(\bar{\beta}) := \varepsilon\circ(\bar{\beta}K), \end{align*} where $\bar{\beta}K: G\circ K \Rightarrow R\circ K$ is the natural transformation obtained by whiskering $\bar{\beta}$ with $K$. By the defining universal property of the right Kan extension $(R,\varepsilon)$, for every natural transformation $\beta: G\circ K \Rightarrow F$ there exists a unique natural transformation $\bar{\beta}: G \Rightarrow R$ such that \begin{align*} \beta = \varepsilon\circ(\bar{\beta}K). \end{align*} This says exactly that $\Psi_G$ is surjective and injective. Hence \begin{align*} \Psi_G: [\mathcal D,\mathcal E](G,R) \to [\mathcal C,\mathcal E](G\circ K,F) \end{align*} is a bijection. [/step] [step:Verify naturality of the right Kan extension bijection in the source functor] Let $G',G: \mathcal D \to \mathcal E$ be functors, and let $\theta: G' \Rightarrow G$ be a natural transformation. The functor $[\mathcal D,\mathcal E](-,R)$ sends $\bar{\beta}: G \Rightarrow R$ to $\bar{\beta}\circ \theta: G' \Rightarrow R$, while the functor $[\mathcal C,\mathcal E](-\circ K,F)$ sends $\beta: G\circ K \Rightarrow F$ to $\beta\circ(\theta K): G'\circ K \Rightarrow F$. For every $\bar{\beta}: G \Rightarrow R$, associativity of vertical composition of natural transformations gives \begin{align*} \Psi_{G'}(\bar{\beta}\circ \theta) &= \varepsilon\circ((\bar{\beta}\circ \theta)K) \\ &= \varepsilon\circ((\bar{\beta}K)\circ(\theta K)) \\ &= (\varepsilon\circ(\bar{\beta}K))\circ(\theta K) \\ &= \Psi_G(\bar{\beta})\circ(\theta K). \end{align*} Thus the family $(\Psi_G)_G$ is natural in $G$. Combining this with the previous steps proves both universal-property isomorphisms. [/step]