The composition of derived functors - commutation fails hazardly?

Question

Hello,

When we have left exact functors $F: A \to B , G: B \to C$ (between abelian categories), we would like sometimes to state that $D(GF)=D(G)D(F)$ (functors between bounded below derived categories). It holds when $F$ sends some adopted to $F$ class into an adopted to $G$ class.

My question is whether this is only technical issue, or there is something meaningful behind. For example, is there a situation when this does not hold, and somehow thinking that it does hold give wrong answers. Or, is there a more general context in which it always holds.

Sasha

Answer 1

This is far from being a technical issue, there are many examples when it fails. Suppose that A is the category of $\mathbb F_p$-vector spaces, $B = C$ the category of abelian groups, $F$ the embedding, $G = \mathrm{Hom}(\mathbb Z/p\mathbb Z, -)$. Then it is easy to see that $DF = F$, $D(GF) = GF$, but $DG\circ F \neq G \circ F$, and the equality does not hold.

Answer 2

My favorite (counter)example is this: let $A$, $B$, $C$ be the categories of left modules over some rings $R$, $S$, $T$ (respectively), and let $F$ and $G$ be the functors of tensor product with some bimodules, that is $F(M)=K\otimes_RM$ and $G(N)=L\otimes_SN$, where $K$ is an $S$-$R$-bimodule and $L$ is a $T$-$S$-bimodule. Then the left derived functor $\mathbb D (GF)$ is the functor of derived tensor product with the underived tensor product $L\otimes_SK$, while the composition of derived functors $\mathbb D (G)\mathbb D(F)$ is the functor of derived tensor product with the derived tensor product $L\otimes_S^{\mathbb D}K$, that is $\mathbb D(GF)(M)=(L\otimes_SK)\otimes_R^{\mathbb D}M$ and $\mathbb D(G)\mathbb D(F)(M)=L\otimes_S^{\mathbb D}K\otimes_R^{\mathbb D}M$.