What is the point of pointwise Kan extensions?

A functor $F: A \to B$ is dense if and only if its left Kan extension along itself exists, is pointwise, and is isomorphic to the identity. This is because the defining universal property of a pointwise extension is

$B(\mathrm{Lan}_F F (b), b') \cong [A^\mathrm{op}, \mathsf{Set}](B(F-,b),B(F-,b'))$

To say that $\mathrm{Lan}_F F$ exists, is pointwise, and is isomorphic to the identity is to say that the left hand side is $B(b,b')$, while to say that $F$ is dense is to say that the right hand side is $B(b,b')$.

Here's an example of a functor $F$ which is not dense, but such that $\mathrm{Lan}_F F$ exists and is isomorphic to the identity (in particular, the extension is not pointwise). Take $A=B$ to be the monoid of natural numbers under addition, viewed as a 1-object category $\mathbf{B}\mathbb{N}$. Let $F: \mathbf{B}\mathbb{N} \to \mathbf{B}\mathbb{N}$ be the "multiply by 2" functor. Then $F$ is not dense, but $\mathrm{Lan}_F F$ exists and is the identity $\mathrm{Id}$. For, any functor $H: \mathbf{B}\mathbb{N} \to \mathbf{B}\mathbb{N}$ is given by multiplication by some $m$, and there is a natural transformation from $\mathrm{Id} \implies H$ if and only if $H= \mathrm{Id}$. Whereas a natural transformation $F \implies HF$ exists if and only if $2n = H(2n)$ for all $n$, which implies that $H = \mathrm{Id}$. In both cases, if a natural transformation exists, then any $\alpha \in \mathbb{N}$ is natural.

The argument that $F$ is not dense goes as follows: $\mathbf{B}\mathbb{N}(F-,\bullet)$ is the set $\mathbb{N}$ but the $\mathbb{N}$-action is addition by 2 rather than 1. The image of $\mathbf{B}\mathbb{N}$ under $\mathbf{B}{N}(F-,1)$ includes all addition-by-$n$ actions on this $\mathbb{N}$-set, but there are more $\mathbb{N}$-set endomorphisms: namely, the odd numbers and even numbers can be shifted independently.

I have always thought that pointwise Kan extensions are better than normal Kan extensions because you can actually compute them using a (co)end

In a Kan extension $\mathrm{Lan}_F G$, $\require{AMScd}$ \begin{CD} A @>G>> C\\ @VFVV \nearrow \mathrm{Lan}_F G \\ B \end{CD}

If $F$ is fully faithful and the extension is pointwise, then the comparison 2-cell is an isomorphism. This is because, if the extension is pointwise, we have

$C(\mathrm{Lan}_G F(Fa),c) \cong [A^\mathrm{op},\mathsf{Set}]( B(F-,Fa),C(G-,c))$

$\qquad \qquad \qquad \, \cong [A^\mathrm{op},\mathsf{Set}]( A(-,a),C(G-,c))$

$\qquad \qquad \qquad \, \cong C(Ga,c)$

and we can take $c = Fa'$.

In order to get an example of a fully faithful $F$ such that $\mathrm{Lan}_F G$ does not have an isomorphism for a comparison cell, we need to move beyond monoids, because a fully faithful functor between monoids is an isomorphism. So the simplest sort of fully faithful functor which is not an equivalence would be the inclusion $F: \mathbf{B}M \to \mathbf{B}M_+$ of a monoid into the category which freely adjoins either an initial or terminal object to it. If we adjoin a terminal object, then left Kan extensions will all be pointwise (with the terminal object being sent to the colimit of the original diagram), so we adjoin an initial object instead. And we might as well take $M$ to be the simplest possible monoid $\mathbb{N}$:

$\require{AMScd}$ \begin{CD} \mathbf{B}\mathbb{N} @>G>> C\\ @VFVV \nearrow L \\ \mathbf{B}\mathbb{N}_+ \end{CD}

It turns out there is a "minimal" $C$ admitting such a left Kan extension $L$, which looks like this:

$G\bullet \overset{(\eta_n)_{n \in \mathbb{N}}}{\overset{\to}{\to}} L\bullet \overset{L!}{\leftarrow} L\emptyset$

Here $G\bullet$ is the fully faithful image of $G$ (so it's a copy of $\mathbf{B}\mathbb{N}$), and $L\emptyset \overset{L!}{\to} L\bullet$ is the fully faithful image of $L$ (so it's a copy of $\mathbf{B}\mathbb{N}_+$ where $L\emptyset$ is the initial object). $C(G\bullet, L\bullet)$ is generated by an arrow $\eta = \eta_0$ under the equation $\eta_n = \eta \circ Gn = Ln \circ \eta$. So $\eta$ constitutes a comparison natural transformation $G \implies LF$.

An exhaustive analysis of the functors $H: \mathbf{B}\mathbb{N}_+ \to C$ (there are 5 families of them, depending on where the objects are sent) reveals that the only one which admits a natural transformation $G \implies HF$ or $L \implies H$ is $L$ itself, and this diagram is in fact a left Kan extension $L = \mathrm{Lan}_F G$ with $F$ fully faithful, but the comparison 2-cell $\eta$ is not invertible.