Proof of Left adjoint preserves Left Kan extension

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My intuitive understanding is the following:

Kan extensions are the closest "approximation" of a given functor, that factors along a given functor. What's happening is that left adjoints preserve colimits, so this guarantees the existence of the Kan extension of the composition. So since $\mathcal{L}$ is the 'closest' functor to $\mathcal{L}$, we should have, $$\text{Lan}_\mathcal{F} (\mathcal{L}\circ \mathcal{G})\cong\mathcal{L}\circ \text{Lan}_\mathcal{F} \mathcal{G}.$$ So I was expecting this.

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Now for the proof, the idea I felt was to apply Yoneda in some way. So, $$\text{Hom}_{\mathcal{D}^{\mathcal{B}}}(\mathcal{L}\text{Lan}_{\mathcal{F}}\mathcal{G},\mathcal{K}) \cong \text{Hom}_{\mathcal{C}^{\mathcal{B}}}(\text{Lan}_{\mathcal{F}}\mathcal{G},\mathcal{R}\mathcal{K})\cong\text{Hom}_{\mathcal{C}^{\mathcal{A}}}(\mathcal{G},\mathcal{R}\mathcal{K}\circ \mathcal{F})\cong \text{Hom}_{\mathcal{D}^{\mathcal{A}}}(\mathcal{L}\mathcal{G},\mathcal{K}\mathcal{F})$$ Doesn't this already say, by precomp-left Kan adjointness, that $\text{Lan}_{\mathcal{F}}(\mathcal{L}\circ \mathcal{G})\cong \mathcal{L}\circ \text{Lan}_\mathcal{F} \mathcal{G}$? In the usual books they assume $\mathcal{K}=\mathcal{L}\text{Lan}_{\mathcal{F}}\mathcal{G}$, and make some arguments I don't understand.

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The fact that left adjoints preserve left Kan extensions can be motivated in many ways according to the generality you are forced to work in.

Your argument is a viable solution provided there exists a "left Kan extension along $F$" functor, which by uniqueness of adjoints must be a left adjoint to precomposition with $F$. This is not always the case; it is the case when the left extensions you are considering are pointwise, and strictly speaking this is a less general concept than the bare definition of Kan extension with its 2-dimensional universal property.

Left adjoint 1-cells of any 2-category, however, preserve left Kan extensions, and this can be deduced quite swiftly from some basic 2-category theory. If you feel you might need this more general argument, I'll update my answer.

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Let $\mathcal A, \mathcal C, \mathcal D, \mathcal E$ be categories, $L:\mathcal D \to \mathcal E$ be a left adjoint to $R:\mathcal E\to \mathcal D$, $F:\mathcal C$ and $K:\mathcal C\to \mathcal A$ such that the left Kan extension $\mathtt{Lan}_K F$ exists.

Diagram of the situation

Recall that the adjunction $L\dashv R$ induces a natural isomorphism $\mathcal E^\mathcal A(L-, =)\simeq \mathcal D^\mathcal A(-, R=)$, and that the left Kan extension along $K$ can be defined as a left adjoint to $-\circ K$.

Let $G$ be any functor $\mathcal A\to \mathcal E$. Then we have the following chain of isomorphisms:

$\begin{aligned} \mathcal E^\mathcal A(L(\mathtt{Lan}_K F), G) & \simeq \mathcal D^\mathcal A(\mathtt{Lan}_K F, RG) \\ & \simeq \mathcal D^\mathcal C(F, RGK) \\ & \simeq \mathcal E^\mathcal C(LF, GK) \\ & \simeq \mathcal E^\mathcal A(\mathtt{Lan}_K (LF), G) \end{aligned}$

Hence, by the Yoneda Lemma, we have $L(\mathtt{Lan}_K F)\simeq \mathtt{Lan}_K (LF)$.