I was reading Fosco Loregian's paper This is the co/end, my only co/friend, and here's something that I don't understand in an exercise.
The exercise is to prove that given $F: C\to D, U: D\to C$ two functors, then $F\dashv U$ if and only if for all $G: D^{op}\times C\to E$ such that $\int^c G(Fc, c)$ or $\int^d G(d, Ud)$ makes sense, then both do and they are isomorphic, naturally in $G$.
Assuming $F\dashv U$ one can easily prove this; it's the other direction that's bugging me for the following reason. Take $G= (\hom_C(-, U(-))\circ \tau)^{op} : D^{op}\times C\to \mathbf{Set}^{op}$, where $\tau : D\times C^{op}\to C^{op}\times D$ is the obvious functor. $\mathbf{Set}^{op}$ being cocomplete, these coends always make sense, and we have
$$\int^c G(Fc,c) = \int^c \hom_C^{op}(c, UFc) = \int_c \hom_C(c,UFc) \cong \mathrm{Nat}(id_C, UF)$$
and also
$$\int^d G(c,Ud) = \int_d \hom_C(Ud, Ud) \cong \mathrm{Nat}(U,U)$$
so if both coends are indeed isomorphic, $\mathrm{Nat}(U,U) \cong \mathrm{Nat}(id_C,UF)$. But this is odd because in an adjunction, what we actually get is something like $\mathrm{Nat}(id_C, UF) \cong \mathrm{Nat}(F,F)$, not $\mathrm{Nat}(U,U)$, which will rather be isomorphic to $\mathrm{Nat}(FU, id_D)$.
Now I don't know if I made a mistake in my calculation, or simply I just found out something I didn't know about adjunctions. Which is it ? (I think I made a mistake at some point, probably when going from $\int^d$ to $\int_d$ but I don't see how: if I'm not mistaken, $\int^c T(c,c) = \int_c T^{op}(c,c)$ where, if $T: A^{op}\times A\to C$, $T^{op} : (A^{op})^{op}\times A^{op} = A\times A^{op}\to C^{op}$)