Isomorphism of coends

Question

This is a follow-up to this question: Reduction to graph subgroups for Bredon homology when the $G_1\times G_2$ is $G_2$-free

In his (very nice) answer Gregory Arone stated the following fact. Let $Q:\mathcal{C}_0\to\mathcal{C}$ be a functor between locally small categories. Then there is adjunction between functor categories $$ Q_!\colon[\mathcal{C}_0,\mathcal{D}]\leftrightarrows[\mathcal{C},\mathcal{D}]\colon Q^\ast, $$ where $Q^\ast$ is a precomposition with $Q$ and $Q_!$ is the left Kan extension (I hope I have the correct notation). Now if we have functors $F\colon\mathcal{C}_0\to\mathcal D$ and $G\colon\mathcal{C}^{op}\to\mathcal{D}$, then there is an isomorphism of coends: $$ F\otimes_{\mathcal{C_0}}Q^\ast G\cong Q_!F\otimes_{\mathcal{C}}G. $$

So my question is: how to prove this isomorphism? And is it something coming from having just an djoint pair of functors, or it applies only to this particular setting - i.e., we have a functor and its left Kan extension?

P.S. If this is a simple fact, reference will also do.

Answer 1

Knowing that $Q_!F : y\mapsto \int^z \hom(Qz,y)\otimes Fz$ (this is often called "pointwise formula" for Kan extensions) it is easy to derive the isomorphism in question: $$ \begin{align} Q_!F\;\otimes_{\cal C_0} G& := \int^y Q_!F(y)\otimes Gy \\ &\cong \int^y \int^z \hom(Qz,y)\otimes Fz \otimes Gy \\ &\cong \int^z Fz\otimes \left( \int^y \hom(Qz,y)\otimes Gy\right)\\ &\cong \int^z Fz\otimes GQ(z) \\ & = F \;\otimes_{\cal C_0} GQ = F \;\otimes_{\cal C_0} Q^*G \end{align} $$