@ [Anindya](https://forum.azimuthproject.org/discussion/comment/19702/#Comment_19702)
> Michael's diagram suggests that \$$\textrm{Lan}_{G} (H) \circ G = H \$$ which turns out to be the case. Is it true for left Kan extensions in general?

I believe the answer, in general, is "No".

Let \$$I\_\mathcal{C}\$$ be the identity functor. We have the following:

$F \dashv \mathrm{Lan}\_{F}\, I\_\mathcal{C} \tag{A}$

In a [Cartesian Closed Category](https://en.wikipedia.org/wiki/Cartesian_closed_category#Definition), we have the following adjunction:

$A \times (-) \dashv (-)^A \tag{B}$

By (A) and (B), we have

$\mathrm{Lan}\_{A \times (-)}\, I\_\mathcal{C} \cong (-)^A$

This is because adjoints are unique up to isomorphism.

The composition \$$(-)^A \circ A \times (-) \$$ is a monad. In Haskell, it is commonly referred to as the [State Monad](https://wiki.haskell.org/State_Monad) - you can find details regarding this on [nLab](https://ncatlab.org/nlab/show/function+monad#relation_to_the_writer_comonad_and_state_monad).

While the identity functor is a monad too, it is distinct from the state monad.

So we have:

$(\mathrm{Lan}\_{A \times (-)}\, I\_\mathcal{C}) \circ (A \times (-)) \not\cong I_\mathcal{C}$

This does lead me to another curiosity...

**Puzzle MD1**. If \$$M\$$ is a monad and \$$\mathrm{Lan}\_{M}\, I\_\mathcal{C}\$$ exists, then is it the case that \$$(\mathrm{Lan}\_{M}\, I\_\mathcal{C}) \circ M \cong M\$$?