Question about Grothendieck's vanishing theorem

Question

In the proof of Grothendieck's vanishing theorem, one of the steps says that if $X$ is irreducible of dimension o, then $H^{i}(X,\mathcal{F})=0$ for all $i>0$. This is because functor $\Gamma(X, .)$ induces an equivalence of categories, and $\Gamma(X, .)$ is an exact functor. But this is where my question arises.

Accroding to the definition, functor $H^{i}(X,.)$ is the right derived functor of $\Gamma(X, .)$, and the injective resolution of a sheaf $\mathcal{F}$ is an exact sequence. Since $\Gamma(X, .)$ is exact, $H^{i}(X,\mathcal{F})$ is always zero, which is impossible. What do I miss? Thanks!

Answer 1

Edit: Seems like I misunderstood the question. Cohomology doesn't vanish for all $X$ and all $\mathcal F$ simply because $\Gamma(X,\cdot)$ is left exact, but not exact in general.

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Resolution itself is not exact, it is only exact with augmentation map. What I mean is this: let $0\to \mathcal F\to \mathcal I^\bullet$ be injective resolution of $\mathcal F$. Then the chain complex

$$0\to\mathcal F\to \mathcal I^0\to \mathcal I^1\to\ldots$$

is exact, but you don't apply $\Gamma(X,\cdot)$ to that complex, but to the complex

$$0\to \mathcal I^0\to\mathcal I^1\to \mathcal I^2\to\ldots$$

which has trivial cohomology except at degree $0$: $H^0(\mathcal I^\bullet)\cong \mathcal F.$

So, if $\Gamma(X,\cdot)$ is exact, $H^i(\Gamma(X,\mathcal I^\bullet)) = 0,\ i > 0,$ and $H^0(\Gamma(X,\mathcal I^\bullet))\cong \Gamma(X,\mathcal F)$.

The same reasoning applies to any derived functor: if the original functor is already exact, then all the higher derived functors vanish.

Answer 2

I think I see where the confusion arises. Your statements

"functor $\Gamma(X, \cdot)$ induces an equivalence of categories, and $\Gamma(X, \cdot)$ is an exact functor"

appear to be quoting from Step 2 of the proof (p.210). But look at the first sentence of Step 2:

"Suppose $X$ is irreducible of dimension $0$."

The proof is just saying that under that assumption, the global section functor is exact (so higher cohomology of any sheaf on $X$ vanishes, as the theorem claims). But of course exactness is not true for a general $X$.