why 0-th sheaf cohomology is the global section

Question

Let $\mathcal{G}^*$ be some injective resolution for sheaf $\mathcal{F}$,hence we have the exact sequence $0\to \mathcal{F} \to \mathcal{G}^*$.

Then we have the chain complex after taking global section functor $\Gamma(X,-)$.Then we can define the sheaf cohomology to be $$H^i(X,\mathcal{F} ) = H^i(\Gamma(X,\mathcal{G}^i),d)$$

I need to prove that 0-th cohomology is the global section $\mathcal{F}(X) = H^0(X,\mathcal{F})$.

My attempt:first we can write out the chian complex of the global section $$0\to \Gamma(X,\mathcal{F})\to \Gamma(X,\mathcal{G}^0)\to \Gamma(X,\mathcal{G}^1)\to ...$$

Now we know global section funtor is left exact,we deduce the 0-th cohomology is 0?What's wrong with my understanding?

Answer 1

I remember getting tripped up on the same thing! In the definition of derived functors, you have to "chop off" the left-most object in the injective resolution (i.e., the object you started with) before applying the functor and taking homology.

So you should just be looking at $\ker \Gamma(X, \mathcal{G}^0) \to \Gamma(X, \mathcal{G}^1),$ which is indeed what you want.