Given an injective resolution for an abelian sheaf $\mathcal{F}$ on $X$, that is an exact sequence $$ 0 \to \mathcal{F} \xrightarrow{\epsilon} \mathcal{A}^0 \xrightarrow{\delta^0} \mathcal{A^1} \xrightarrow{\delta^1} \mathcal{A^2} \xrightarrow{\delta^2} \ldots$$ of abelian sheafs with all $\mathcal{A}^i$ injective, why is the sequence $$ 0 \to \mathcal{A}^0(X) \xrightarrow{\delta^0_X} \mathcal{A}^1(X) \xrightarrow{\delta^1_X} \mathcal{A}^2(X) \xrightarrow{\delta^2_X} \ldots$$ a cochain complex?
I know that the global section functor is left exact, but I can't convince myself that this implies $\text{im}(\delta^q_X) \subseteq \text{ker}(\delta_X^{q+1})$.
Okay, it's actually easier than I thought. Since taking global sections is a functor, we get as MooS pointed out $$\delta_X^{q+1} \circ \delta_X^q = (\delta^{q+1} \circ \delta^q)_X = 0,$$ so the left-exactness of the global sections functor is not needed to get a cochain complex.