# Adapted classes of objects and left (right) exact functors

I had a question about adapted classes of objects, I was confused by the definition and how it relates to left exact functors. Let $\mathcal{A}$ be an abelian category with enough injectives, let $F: \mathcal{A} \rightarrow \mathcal{B}$ be a left exact functor to an abelian category $\mathcal{B}$, to construct the right derived functor $RF$ of $F$ we take an object $X$ of $\mathcal{A}$ and embed it in an injective resolution

$0 \rightarrow X \rightarrow I^0 \rightarrow I^1 \rightarrow I^2 \cdots$,

and $R^iF(X)$ is the homology at the i-th spot of the complex

$0 \rightarrow F(I^0) \rightarrow F(I^1) \rightarrow F(I^2) \rightarrow \cdots$

right?

Let $F:\mathcal{A} \rightarrow \mathcal{B}$ be a left exact functor. A class of objects $\mathcal{R}$ in $\mathcal{A}$ is an adapted class of objects for $F$ if the following conditions are satisfied:

1 - $F$ maps any acyclic complex from $Kom^+(\mathcal{R}$) into an acyclic complex.

...

Among other conditions, it also says that the class of injective objects is adapted to all left exact functors, so now I'm like if that's the case, if I take this part of the sequence

$I^0 \rightarrow I^1 \rightarrow I^2 \rightarrow \cdots$,

this is exact therefore acyclic no? if I apply the left exact functor $F$, I get

$F(I^0) \rightarrow F(I^1) \rightarrow F(I^2) \rightarrow \cdots$,

which would also be acyclic by the above definition? But wouldn't that make the right derived functor groups $R^iF(X) = 0$ for $i \geq 2$?

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What book/article did you read this in? References are always helpful for non-trivial questions :-) – Peter LeFanu Lumsdaine Aug 18 '12 at 17:57
The complex $I^0 \to I^1 \to \cdots$ is not exact (it fails to be exact at the $0$th place). @Peter: If memory serves, the notation of this question is that of Gelfand-Manin (who talk about adapted classes), I think that's a pretty safe bet. – t.b. Aug 18 '12 at 17:59
It's Methods of homological algebra by Gelfand-Manin. @t.b. I thought the complex $X \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots$ was the one that wasn't exact but acyclic? In any case going by that definition aren't the groups $R^iF(X) = 0$ for $i \geq 2$? – Mario Carrasco Aug 18 '12 at 18:10
No, if you take an injective (or adapted) resolution of $X$ then by definition the complex $\cdots 0 \to 0 \to X \to I^0 \to I^1 \to \cdots$ is exact. If you throw away $X$ then the resulting complex $\cdots \to 0 \to 0 \to I^0 \to I^1 \to \cdots$ isn't exact anymore (the first morphism of an exact complex should be injective) and this can result in cohomology in all degrees. A good example of an adapted class are the flat modules, which are (by definition) those adapted to the tensor product. You can compute Tor by taking flat resolution of a module, not only by a projective resolution. – t.b. Aug 18 '12 at 18:30
The sequence $I^0 \to I^1 \to I^2 \to \cdots$ is exact, but the sequence $\cdots \to 0 \to I^0 \to I^1 \to \cdots$ is not. The $0$ in front makes a difference! – Zhen Lin Aug 18 '12 at 18:51