The coend of a parametric counit The following is true:

*

*The counit components $\epsilon : X^A\times A\to X$ of the cartesian closed structure of $Set$ are the components of the initial cowedge; in other words, $X$ is the coend $\int^A X^A\times A$. Size issues apart, the reason is that $\int^A X^A\times A$ is isomorphic to the value at $X$ of $\text{Lan}_11$, and the latter is the identity functor.

But what about the same statement in a generic cartesian closed category $\cal C$, with enough colimits if needed?


*The counit components $\epsilon : X^A\times A\to X$ are the components of the initial cowedge in $\cal C$; in other words, $X$ is the coend $\int^A X^A\times A$.

And what about the most general statement in this direction:


*Let $L_a\dashv R_a$ be a parametric adjunction -give by functors $L : A\times C\to D$ and $R : A^{op}\times D \to C$. Then the counit $\epsilon : L_aR_ad \to d$ is a cowedge, and $d\cong\int^aL_aR_ad $; dually, the unit $c\to R_aL_ac$ is a wedge, and $c\cong\int_a R_aL_ac$.

I think 3. is false in such a generality, but then

*

*What is an instructive counterexample?

*Under which conditions it is true, considering that it is sometimes true?

 A: Unfortunately statement 2 is not true in general. First, notice that
$$\hom\Bigl(\int^A X^A \times A,Y\Bigr) = \int_A \hom(X^A,Y^A) = \hom(X^{(-)},Y^{(-)}).$$
So the question is if the functor
$$\mathcal{C} \to [\mathcal{C}^{\mathrm{op}},\mathcal{C}], ~ X \mapsto X^{(-)}$$ is fully faithful. The canonical map
$$\hom(X,Y) \to \hom(X^{(-)},Y^{(-)})$$
has a retraction: we can evaluate a natural transformation at $1 \in \mathcal{C}$. So the composition $\hom(X,Y) \to \hom(X^{(-)},Y^{(-)}) \to \hom(X,Y)$ is always the identity. But I don't see a reason why the composition $\hom(X^{(-)},Y^{(-)}) \to \hom(X,Y) \to \hom(X^{(-)},Y^{(-)})$ should be the identity. It holds when $1$ is a separator of $\mathcal{C}$. I am too lazy to write down the proof, but it is an easy exercise.
But now consider the topos of $G$-sets $\mathcal{C} = {}_G \mathbf{Set}$ for some group $G$ with non-trivial center $Z(G)$ (maybe another argument works for all non-trivial groups). It is well-known that for two $G$-sets $X,A$ the $G$-set $X^A$ consists of all maps of the underlying sets $\alpha : U(A) \to U(X)$, and the $G$-action is $(g \cdot \alpha)(x) := \alpha(g^{-1} \cdot x)$. Any element $g \in Z(G)$ induces an element in the center $Z(\mathcal{C})$ and hence a natural transformation $X^{(-)} \to X^{(-)}$, namely $\alpha \mapsto g \cdot \alpha$. Evaluating this at $1 \in \mathcal{C}$ always gives the identity, so we get a counterexample when $g \neq 1$.
