TITLE: Relationship between two universal properties of the category of elements?
QUESTION [10 upvotes]: Let $G: \mathcal{A} \to \mathsf{Set}$ be a functor. Recall that the category of elements $\mathsf{el}G$ is defined by the comma square
$\require{AMScd}$
\begin{CD}
    \mathsf{el}G @>!>> \ast \\
    @V U_G VV \Downarrow @VV \Delta_\ast V\\
    \mathcal{A} @>>G> \mathsf{Set}
    \end{CD}
where $\ast$ is used to label the terminal category and $\Delta_\ast$ is the functor that is constant at the terminal set.
But there is another key universal property of this diagram: it exhibits the left Kan extension $ G = \operatorname{Lan}_{U_G} (\Delta_\ast !)$. This fact is key in the reduction of weighted colimits to conical ones (as explained here).
It's easy to verify that this is a Kan extension directly, but it's a bit mysterious, because arbitrary comma squares are not Kan extensions (for example, the same square in $\mathsf{Cat}^{\mathrm{co}}$ doesn't seem to be a left Kan extension). So what is going on here? Why, conceptually (by which I suppose I mean: 2-categorically) is this particular comma square also a Kan extension triangle?

REPLY [7 votes]: The conceptual explanation is that a comma square exhibits the bottom morphhism as a left Kan extension if the right morphism is dense.
There is, of course, a problem what we mean by a "dense morphism" in an arbitrary 2-category. If we take the usual definition and say that a morphism is dense if its pointwise left Kan extension along itself exists and is the identity, then the above follows almost by the definition of density and the definition of "pointwiseness" in the sense of R. Street (i.e. stability under comma squares). 
In your example $\Delta_* \colon 1 \rightarrow \mathbf{Set}$ is obviously dense (in the classical sense) --- therefore its left Kan extension along itself is the identity:
$$\begin{CD}
    \ast @>\Delta_\ast>> \mathsf{Set} \\
    @V \Delta_\ast VV \Downarrow @| \\
    \mathsf{Set} @>>\mathit{id} = \mathit{Lan}_{\Delta_\ast}\Delta_\ast> \mathsf{Set}
\end{CD}$$
and when you draw the above left extension diagram on the right side of your comma square, then the fact that the bottom morphism is the left extension follows directly from the definition of pointwiseness.