Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 18 / 4 / i / Solution

Let $T:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$. An end of a functor is an object $E$ with maps $e_C:E\to T(C,C)$ satisfying $T(1_C,f)e_C=T(f,1_D)e_D$ for every $f:C\to D$, universal among all such families. Thus every compatible family with vertex $A$ factors uniquely through $E$. Write it as ${\int }_{C}T(C,C)$. These are the wedge equations for a dinatural transformation from a constant functor to $T$.

Dually, a coend of a functor is an object $Q$ with maps $i_C:T(C,C)\to Q$ satisfying $i_{C}T(f,1_C)=i_{D}T(1_D,f)$ as maps from $T(D,C)$, universal among compatible families with a common target. Write it as ${\int }^{C}T(C,C)$. These are cowedge equations, and every cowedge to $A$ factors uniquely through $Q$.

For small $\mathcal{C}$, if the needed products and equalizers exist, the end is the equalizer of ${\prod }_{C}T(C,C)\rightrightarrows {\prod }_{f:C\to D}T(C,D)$. If the needed coproducts and coequalizers exist, the coend is the coequalizer of ${\coprod }_{f:C\to D}T(D,C)\rightrightarrows {\coprod }_{C}T(C,C)$. The two arrows in each construction are precisely the two sides of the compatibility equation. This explains both the mixed variance and the universal property, rather than treating the integral notation as an ordinary numerical integral.