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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?

But then I was reading about adapted classes of objects, and the definition says:

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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    I was curious, if the complex $0 \rightarrow A \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots$ is exact, can't I just filter through $A$ to get $0 \rightarrow I^0 \rightarrow I^1 \rightarrow I^2 \rightarrow \cdots$ exact? where the map from $0$ to $I^0$ is the composition of the map from $0$ to $A$ and the map from $A$ to $I^0$?2012-09-01

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