Descent of singular cohomology

When proving that singular cohomology of an appropriate space $X$ equals sheaf cohomology of $X$ with "values" (does one say that?) in the sheaf $\mathbb{Z}_X$ of locally constant functions, the author of the proof I have read observes that the sheafifications $\mathcal{C}^n$ of the singular cochain complexes $C^n(-)=\hom_{Ab}(\mathbb{Z}\hom_{Top}(\Delta^n,-),\mathbb{Z})$ form an injective resolution $$0\to\mathbb{Z_X}\to\mathcal{C}^0\to\mathcal{C}^1\to\ldots$$ of $\mathbb{Z}_X$.

Why must one sheafify the singular cochain complexes? Aren't they sheaves since they satisfy descent (= have the excision property) and "sheaf=presheaf+descent"?

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They aren't sheaves. Consider a covering $(U_i)$ of $X$ and take a $k$-cochain $f_i$ on each $U_i$. If one has a singular $k$-chain $\sigma:\Delta^k\to X$ whose image is not in any of the $U_i$, then its image under the putative cochain on $X$ isn't determined by the $f_i$. – Robin Chapman Jul 4 '10 at 11:31
I see. The excision theorem (or "theorem of small chains") implies that $C^\bullet(U)$ is chain homotopic to $\mathcal{C}^\bullet(U)$ but with U fixed, right? – user7316 Jul 4 '10 at 13:04