Is the sheafified Cech complex for an étale cover a resolution?

Question

Suppose I have an étale cover of a scheme $X$ by etale $X$-schemes $U_{i}$, let $U:=\coprod_{i=1}^{n} U_{i}$ and write $f:U\rightarrow X$ for the induced map. Then for every quasicoherent sheaf $\mathcal{F}$ it is possible to define a sheafified Cech complex with generic term $\mathcal{C}^{n}(U,\mathcal{F})=f_{n*}f_{n}^{*}\mathcal{F}$, where $f_{n}:U\times_{X}\times \cdots \times_{X}U \rightarrow X$ (n+1 copies). When we have an open cover instead of an etale cover we can take the complex without repetitions which is concentrated in degrees $\leq n$, which moreover is a resolution of $\mathcal{F}$. Does something similar hold here, in the sense that we get a resolution and that at some point some differential is zero so that we can divide the complex in two?

Answer 1

Yes the complex $\mathcal{C}^\bullet(U,\mathcal{F})$ is a resolution of $\mathcal{F}$. To prove it, it is enough to work locally : more precisely, if $g:V\to X$ is a cover, then $\mathcal{F}\to \mathcal{C}^\bullet$ is a resolution iff $g^*\mathcal{F}\to g^*\mathcal{C}^\bullet$ is a resolution.

There is an obvious cover that we can use : the cover $f:U\to X$ itself. It is easy to see that the complex you get is given by $g_n:U\times_X U\times_X ...\times_X\ U\to U$ ($n+2$ copies). But now, there are obvious sections. It follows that $f^*\mathcal{F}\to f^*\mathcal{C}^\bullet$ has an homotopy inverse. It follows that $\mathcal{C}^\bullet(U,\mathcal{F})$ is a resolution of $\mathcal{F}$ (though of course not an homotopy equivalence). This is trick is standard but extremely powerful, you should write the details to see what is going on.

---

I am not sure I understand what you mean by divide the complex in two. If you ask whether you can replace it by a bounded one, the answer is no. Even fields may non bounded étale cohomology.