Adjoints are Kan extensions

Question

How do I prove that if $F:\mathcal{C}\leftrightarrows\mathcal{D}:G$ is an adjunction with unit $\eta:id_{\mathcal{C}}\Rightarrow GF$ and counit $\epsilon: FG\Rightarrow id_\mathcal{D}$, then $(G,\eta)$ is a left Kan extension of the identity functor at $\mathcal{C}$ along $F$ and $(F,\epsilon)$ is a right Kan extension of the identity functor at $\mathcal{D}$ along $G$.

Answer 1

We have the following diagram:

The right Kan extension of $T$ along $K$ is a pair $(R,\epsilon)$ where $Ran_KT:=R$ is the functor in the diagram and $\epsilon:RK\to T$ is a natural transformation. The UMP of this construction is: if $(H,\delta)$ is another pair, where $H$ is the functor in the diagram and $\delta:HK\to T$ is a natural transformation, then there is a $unique$ natural transformation $\sigma:H\to R$ such that $\delta = \epsilon\circ\sigma K$.

Uniqueness establishes a natural bijection $\sigma\mapsto \epsilon\circ\sigma K$

$[H, Ran_KT]\cong [HK, T]\tag1 $ ($K$ and $T$ are fixed here)

Now, suppose $(F,\eta)\dashv (G,\epsilon'):\mathcal X\leftrightharpoons \mathcal A$ is an adjoint pair. Then, if $S:\mathcal X\to \mathcal C$ is any functor, for all $H$ as in the diagram, we have the following bijection $\Phi$, natural in $S$:

$[S, HF]\cong [SG, H]\tag2$

with $\Phi(\sigma')=H\epsilon'\cdot \sigma' G.$ Show that $\Phi^{-1}$ is given by $\tau\mapsto \tau F\cdot S\eta$ by taking the naturality square

\begin{array}{ccc} S & \overset{\sigma'}\rightarrow & HF \\ \downarrow S\eta& & \downarrow HF\eta \\ SGF & \overset{\sigma' GF}\rightarrow & HFGF \end{array}

and adjoining to it on the right one of the triangular identities. This will give $\Phi^{-1}\circ \Phi=id.$ A similar analysis gives $\Phi\circ \Phi^{-1}=id.$

Now let $\mathcal C=\mathcal A$ and $H=1_{\mathcal A}$, so that $(2)$ becomes

$[S,F]\cong [SG,1]\tag3 $

and now, we have, by defintion of $\Phi$ and comparison of $(3)$ with $(1)$ along with the UMP of the Kan extension, that $F=Ran_G 1_{\mathcal A}.$