I am lost in a bit of a haze of abstraction and am wondering if someone can set me straight.
I'm reading the chapter in Categories for the Working Mathematician, and Mac Lane points out on page 238 that the argument for vanilla categories (that the Kan extension can be given by a coend formula) may fail for Abelian categories and additive functors unless one uses the proper notion of "cotensor." So I am trying to make sure I understand at what point the usual argument fails. This is my attempt to work through the proof:
Let $T: M\to A$, let $K: M\to C$ be additive functors between Abelian categories. We want to define $R = \int_m T(m)^{Hom_C(c, K(m))} : C\to A$ and demonstrate a correspondence, for functors $S: C\to A$, $Nat(S,R)\cong Nat(SK,T)$ natural in $S$.
Now
- $Nat(S, R)\cong \int_c Hom_A(S(c), R(c))$
- $\cong\int_c Hom_A(S(c), \int_m T(m)^{Hom_C(c, K(m))})$
- $\cong\int_c\int_m Hom_A(S(c),T(m)^{Hom_C(c, K(m))})$
And here is where the divergence arises, as far as I understand it, in the proofs. If we interpret $T(m)^{Hom_C(c,K(m))}$ as being given by $\prod_{f\in Hom_C(c,K(m))}T(m)$, then there is a natural correpondence between morphisms $S(c)\to \prod T(m)$ and $|Hom_C(c,K(m))$|-indexed tuples of morphisms $S(c)\to T(m)$; that is, the next line should be
- $\cong\int_c\int_m Hom_{\mathbf{Sets}}(Hom_C(c, K(m)),Hom_A(S(c),T(m)))$ and indeed this is the next line in Mac Lane.
But working in an Abelian category, i.e. take $R$-mod, it perhaps makes more sense to define $T(m)^{Hom_C(c,K(m))}$ as the Abelian group $Hom_{R-mod}(Hom_C(c,K(m)),T(m))$ as one can then, in the next line, get the isomorphism
- $\cong\int_c\int_m Hom_{\mathbf{Ab}}(Hom_C(c, K(m)),Hom_A(S(c),T(m)))$ by simple manipulations with tensor products, etc.
Now I am trying to understand what the difference is between these when we take the end with respect to $c$. As I understand it, the end of the second version should be $\int_c Hom_{\mathbf{Ab}}(Hom_C(c,K(m)),Hom_A(S(c),T(m)))$ should be $Nat(Hom_C(-,K(m)),Hom_A(S(-),Tm))$ as taken in $Ab$; whereas the end of the first one should be more like $\int_c Hom_{\mathbf{Sets}}(...)=Nat(U(Hom_C(-,K(m))),U(Hom_A(S(-),Tm))$ where $U$ is the forgetful functor $Ab\to Sets$.
But aren't these sets the same? In both cases the Yoneda lemma - abelian cat version or normal version - allows us to pull back the natural transformations to morphisms between the original objects, i.e. a map in $A$ from $SK(m)\to T(m)$. So why does it matter which version of the "cotensor" we use?
It's not true that the sets are the same for an arbitrary functor $S$. This is what Mac Lane is trying to point out when he stresses that the universal property of the constructed $R$ holds for additive functors $S$.
It's obviously true that the additive natural transformations $\operatorname{Hom}_C(-,Km) \to \operatorname{Hom}_C(S(-),Tm)$ are a subset of all natural transformations. Let $\sigma : S(K(M))\to Tm$ be a morphism which by Yoneda represents a natural transformation; the natural transformation it is associated to gives for each $c$ a map $\operatorname{Hom}_C(c,Km) \to \operatorname{Hom}_C(Sc,Tm)$ sending $g : c\to Km$ to $\sigma\circ S(g)$. This is not additive in $g$ unless $S$ is itself an additive functor, and the proof gets stuck. So we shouldn't expect the cotensor formula for the right Kan extension to give the right Kan extension where all functors are regarded as enriched over Sets.