# Suffices to compute Cech cohomology on an arbitrarily fine cover?

For a topological space $$X$$ and a sheaf $$\mathscr{F}$$, define the Cech cohomology groups in the usual way $$\check{H}^{q}(X, \mathscr{F}) = \varinjlim H^q(\mathfrak{U}, \mathscr{F})$$ where the limit is taken over classes of open covers $$\mathfrak{U}$$, ordered by refinement. To be clear, for open covers $$\mathfrak{U} = \{U_i\}_{i \in I }$$ and $$\mathfrak{V} = \{V_j\}_{j \in J}$$ we say $$\mathfrak{U} \prec \mathfrak{V}$$ if there is a function $$\tau \colon I \to J$$ such that for each $$i \in I$$ we have $$U_i \subseteq V_{\tau(j)}$$.

I would like to do some computations of cohomology, and the paper I am working out of does so by computing cohomology on open covers that we will call 'nice' and asserting that 'nice open covers are arbitrarily fine'. I need to make sure I understand the logic, and part of that is making sure I understand what arbitrarily fine means.

Here is what I think arbitrarily fine should mean: considering we have defined cohomology as a direct limit, I would say that the 'nice' open covers are 'arbitrarily fine' if given any open cover $$\mathfrak{V}$$, there is a 'nice' open cover $$\mathfrak{U}$$ such that $$\mathfrak{U} \prec \mathfrak{V}$$. Then if you showed that $$H^{q}(\mathfrak{U}, \mathscr{F}) = 0$$ for every 'nice' open cover, it would follow that the direct limit will also be trivial because this would mean everything must become 0 eventually.

This is in the paper:

1. The collection of all 'nice' subsets of $$X$$ form an open base of the topology of $$X$$.

2. The coverings of $$X$$ consisting of 'nice' subsets are arbitrarily fine.
proof: Note that if $$\mathfrak{U} = \{ U_i \}_{i \in I}$$ is such a cover, the $$U_{i_0} \cap \dots \cap U_{i_p}$$ are also 'nice'.

He then defines a special type of 'nice' open cover, computes the cohomology to be trivial, then notes the the 'nice' open cover was arbitrarily fine by the above. Can someone help me solidify what is going on here?

• arbitarily fine: a cofinal subset of the poset we take the inverse limit over. And it's a standard fact that the inverse limit of a cofinal subsystem is the same as that of the whole system. So yes, I agree with your interpretation. – Henno Brandsma Feb 27 at 21:34
• Ok I will look into that, I guess I just didn't see why his facts (1. and 2.) above were needed to justify this – Prince M Feb 27 at 21:36
• Fact 1 and 2 justify the cofinalness and the fact that the basic covers form a directed subsystem (which is what 2 says), so we can take a limit over just these. – Henno Brandsma Feb 27 at 21:39
• Oh I see, hes not just using those facts to guarantee these covers are 'suitably fine', but we also need to make sure that the collection of them satisfies the set up in order to take a direct limit – Prince M Feb 27 at 21:42
• That's what I think yes. – Henno Brandsma Feb 27 at 21:42

The crucial point is this. Let $$\mathcal{B}$$ be any base for the topology of $$X$$. A $$\mathcal{B}$$-cover of $$X$$ is one whose members belong to $$\mathcal{B}$$. Then for any open cover $$\mathfrak{V}$$ of $$X$$ there exists a $$\mathcal{B}$$-cover $$\mathfrak{U}$$ of $$X$$ such that $$\mathfrak{U} \prec \mathfrak{V}$$.
Take $$I = X$$. For any $$x \in I$$ choose $$\tau(x) \in J$$ such that $$x \in V_{\tau(x)}$$. Then choose $$B(x) \in \mathcal{B}$$ such that $$x \in B(x) \subset V_{\tau(x)}$$. Now $$\mathfrak{U} = \{ B(x) \}_{x \in I}$$ is finer than $$\mathfrak{V}$$.