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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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    What book/article did you read this in? References are always helpful for non-trivial questions :-)2012-08-18
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    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.2012-08-18
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    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 $i \geq 2$?2012-08-18
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    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.2012-08-18
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    But if I forget about all that can I simply take any "higher" sequence $I^2 \rightarrow I^3 \rightarrow I^4 \rightarrow \cdots$? would that sequence be exact? or $I^3 \rightarrow I^4 \rightarrow I^5 \rightarrow \cdots$? I thought that "higher" terms in sequences were unaffected if I took out "lower" terms, the terms $I^2$, $I^3$, $I^4$ and the maps $d^1$, $d^2$, $d^3$ are still the same no? or do they change if I remove "lower" terms? @t.b. Thanks, you're giving me great insight into this2012-08-18
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    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!2012-08-18
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    I'm confused, t.b. says the sequence $I^0 \rightarrow I^1 \rightarrow I^2 \rightarrow \cdots$ is exact, but @ZhenLin says it isn't. I read another definition and it says that "left exact functors are exact on injective objects", so if I take for example $I^1 \rightarrow I^2 \rightarrow I^3 \rightarrow \cdots$ and apply the left exact functor $F$ I get the exact sequence $F(I^1) \rightarrow F(I^2) \rightarrow F(I^3) \rightarrow \cdots$. But doesn't that imply that $R^iF(X) = 0$ for $i \geq 2$? Then what would be the point of homology? Wouldn't that make homology trivial from degree 2 onwards?2012-08-18
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    @Luis: I said the *complex* $\cdots \to 0 \to 0 \to I^0 \to I^1 \to \cdots$ and by a bad habit and general sloppiness I suppressed the zeros on the left, as we are talking about things that live in $\operatorname{Kom}^+(\mathcal{A})$, after all. Zhen Lin is right: if you consider the diagram $I^0 \to I^1 \to \cdots$ as a sequence in $\mathcal{A}$, it *is* exact since there's a zero missing on the left, however, that's not what we're talking about, really. Concerning your second question: it does not imply that. The adaptedness condition only applies to exact *complexes* (zeros on the left).2012-08-18
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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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