There is something very weird with the way Hartshorne defines right derived functors.

Hartshorne p 204 Let $\mathfrak A$ be an abelian category with enough injectives, and let $F \colon \mathfrak A \to \mathfrak B$ be a covariant left exact functor. Then we construct the right derived functors $R^iF, i ≥ 0$, of $F$ as follows. For each object $A$ of $\mathfrak A$, choose once and for all an injective resolution $I^.$ of $A$. Then we define $R^iF(A) = h^i(F(I^.))$

With Prop 1.2A. Hartshorne claims that one can also compute the right exact functor $R^iF$ with a resolution $J^i$ that is acyclic for F.

But then, in prop. 2.6 (and similarly in prop. 8.4) Hartshorne states what was already true by definition: that the derived functors of $\Gamma(X,^.)$ coincide with the cohomology functors $H^i(X,^.)$. The proof goes around a circle to show the obvious fact that an injective resolution is equivalent to itself. Namely, that every injective is flasque (2.4), and flasques are acyclic (2.5), so by 1.2A. (above) we can use this very injective resolution(!).

What do we need to go around like that?

up vote 16 down vote accepted

If you look closely, Hartshorne defines the cohomology functors as the right-derived functors of

$\Gamma(X, -) : \mathfrak{Ab}(X) \to \mathfrak{Ab}$

and then claims in Prop 2.6 that these agree with the right-derived functors of

$\Gamma(X, -) : \mathfrak{Mod}(X) \to \mathfrak{Ab}$

There's some abuse of notation here -- in the first definition $X$ is just a topological space, whereas in the second definition $X$ is a ringed space, and of course we're also using $\Gamma$ for two different functors.

So Proposition 2.6 says that the derived functors of $\Gamma(X, -) : \mathfrak{Mod}(X) \to \mathfrak{Ab}$, which a priori depend on the structure sheaf of $X$, don't really. What's being checked in the proof is that an injective resolution in the category $\mathfrak{Mod}(X)$ is still at least an acyclic resolution when we forget the $\mathcal{O}_X$-module structure and regard it as a chain complex in $\mathfrak{Ab}(X)$.

You might want to try finding (or asking for) an example of an injective $\mathcal{O}_X$-module which fails to be injective as a sheaf of abelian groups if this doesn't make sense.

  • Very nice answer. I also remember I found these parts of the wonderful Hartshorne's book to be extremely tricky... – Piotr Pstrągowski Jul 19 '13 at 9:56
  • 3
    To be honest, when I read the question originally I was as confused as the OP. Hartshorne could really stand a revised edition cleaning up all these things. – Daniel McLaury Jul 19 '13 at 17:14
  • @DanielMcLaury from the bottom of my heart: Clean up all the things!! – guest Dec 30 '14 at 6:24

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