Elementary proof of the exactness of Čech complex associated to a hypercovering ("Illusie's Conjecture")

Question

Let $\mathcal{E}$ be a sheaf of abelian groups on a topological space (or a site). For an open covering $\mathfrak{U} = (U_i)_i$, it is well known that the augmented Čech complex $0 \to \mathcal{E} \to \mathcal{\check C}^\bullet(\mathfrak{U}, \mathcal{E})$ is a resolution of $\mathcal{E}$. (Depending on the sheaf cohomology of $\mathcal{E}$ on the intersections of the members of $\mathfrak{U}$, this resolution might compute $\mathbb{R}\Gamma(\mathcal{E})$, but this need not concern us here.)

A simple proof runs as follows: Let $s \in \mathcal{\check C}^{n+1}(\mathfrak{U}, \mathcal{E})(V)$ such that $ds = 0$. Then $t = (s_{i_\text{fix},i_0,\ldots,i_n})_{i_0,\ldots,i_n} \in \mathcal{\check C}^{n}(\mathfrak{U}, \mathcal{E})(V \cap U_{i_\text{fix}})$ is a local preimage of $s$ under $d$, as $(dt)_{i_0, \ldots, i_{n+1}} = s_{i_0, \ldots, i_{n+1}} - (ds)_{i_\text{fix}, i_0, \ldots, i_{n+1}} = s_{i_0, \ldots, i_{n+1}}$. As $V = \bigcup_{i_\text{fix}} (V \cap U_\text{fix})$, that's all what's needed.

I'm looking for a similarly elementary proof in the case that $\mathfrak{U}$ is a hypercovering. Apparently this fact is sometimes referred to as Illusie's Conjecture. A proof involving trivial Kan fibrations and an induction reducing to the case of ordinary covers is in Section 01GA of the Stacks Project, but I'd prefer either a more direct proof or some insight as to why a much simpler argument isn't likely to exist.

Answer 1

If you are working on a topological space (or more generally a site with enough points), then it's enough to show that $E \to \check C(\mathfrak U,E)$ is a quasi-isomorphism on stalks. This reduces the claim to the case that the topological space is a point. But a hypercover of a point is (after some unpacking) a simplicial set $X$ such that every map $\partial \Delta_n \to X$ extends to $\Delta_n \to X$, i.e. a contractible Kan complex. So the claim is that such an $X$ has the cohomology of a point. Not sure if this is the type of argument you want - it is more "simplicial" than "chainy" - but on the other hand, so is the definition of a hypercover.

In general the "meat" of Illusie's conjecture lies in the case when the site does not have enough points.