$\newcommand{\cSet}{\mathsf{Set}}$ I'd like a verification of my computation of the Left Kan Extension of a functor along the identity functor.
Consider the group $G \equiv \mathbb Z/2\mathbb Z \equiv \{+1, -1 \}$ as a one-object category with single object $(\star)$, morphisms $+1, -1 \in \hom(\star, \star)$. Let $\cSet$ be the category of sets. Let $\phi: G \to \cSet$ be the action of $G$ on the set $A \equiv \{ (a, +), (a, -) \} \in \cSet$ be given by swapping $(a, +)$ and $(a, -)$. Formally:
$$ \begin{align*} &\phi: G \to \cSet \\ &\phi_{\texttt{obj}}(\star) \equiv A \\ &\phi_{\texttt{mor}}(+1) \equiv id_A \\ &\phi_{\texttt{mor}}(-1) \equiv (a, +) \mapsto (a, -); (a -) \mapsto (a, +) \end{align*} $$
Let's now compute the Left Kan Extension of $\phi: G \to \cSet$ along the identity $id_G: G\to G$, denoted by $L : G \to \cSet$ In particular, let us compute $L(\star)$. Recall that this computation is given by the colimit construction:
$$ L(\star) \equiv (\phi \downarrow \star) \xrightarrow{\Pi_{\star}} G \xrightarrow{\phi} \cSet $$
I see that the combined map will look like:
But this colimit asks us to glue $(a, +)$ with $(a, -)$, which collapses $A$ into a single element set!
Is this computation correct? I find it very puzzling, since it seems to imply that "the best extension of $\phi$ along $id_G$" somehow doesn't even act on $A$! What's the intuition here?
On the other hand, if the computation is wrong, which step went wrong?
Screenshot of colimit construction from Riehl's "Category theory in Context":
First of all, it is $(1 \downarrow \star)$, not $(\phi \downarrow \star).$
Now, this should look very suspicious because the indexing category $(1 \downarrow c)$ has a terminal object $1_c$, so taking colimit over it should not collapse anyone anywhere. Another way to see it is that the adjunction $\operatorname{Nat}(\operatorname{Lan}_KF, G) = \operatorname{Nat}(F, G \circ K)$ for $K = 1$ means that we should have $\operatorname{Lan}_KF \simeq F$.
In what you wrote everything looks fine until the moment of taking the colimit. The way you tried to take it is not quite correct, as you should, informally, factor by some relation a union of copies of $A$ indexed by the indexing diagram, not factor the image of the diagram.
So the moral is that the colimit over $\bullet \to \bullet$ has to be the image of the terminal object. You can take your favorite approach to colimit computation and accurately ensure that it agrees with this statement.