# Nilpotent elements of cohomology ring

It is well known that every element of positive degree in the Cech cohomology ring of a compact space is nilpotent. How can ı prove that?

The Cech cohomology ring of a space $$X$$ is the direct limit of the cohomology rings of nerves of open covers of $$X$$. If $$X$$ is compact, then finite open covers are cofinal in all open covers, so every Cech cohomology class on $$X$$ comes from the nerve of a finite open cover. But the nerve of a finite open cover of $$X$$ is just a finite simplicial complex, and in particular has nontrivial cohomology in only finitely many degrees. Thus every positive-degree cohomology class on the nerve of a finite open cover is nilpotent, and remains nilpotent when mapped into the Cech cohomology of $$X$$.
(Incidentally, this result is very false for singular cohomology. By a theorem of Przeździecki (see the references and comment thread here), every topological space of cardinality less than the least measurable cardinal is weak equivalent to a compact Hausdorff space. In particular, if you take any familiar example of a space $$X$$ with a non-nilpotent cohomology class in positive degree (e.g., $$X=\mathbb{C}P^\infty$$), there is a compact Hausdorff space $$Y$$ which is weak equivalent to $$X$$ and in particular has an isomorphic singular cohomology ring.)
• Weird. What's the Cech cohomology of $Y$ then? Jul 6 '17 at 7:19
• @QiaochuYuan: My guess would be that Cech cohomology thinks that $Y$ is something like a monstrously huge wedge of circles. The construction of $Y$ is roughly by adding lots of paths to a compactification of $X$; these paths end up giving you something equivalent to $X$ as far as singular cohomology is concerned but will create loops as far as Cech cohomology is concerned (similar to how a Warsaw circle is weakly contractible but Cech cohomology thinks it's a circle). Jul 6 '17 at 7:47
• What I don't have any understanding of is to what extent, if any, the cohomology of $X$ survives to the Cech cohomology of $Y$. For instance, in the case $X=\mathbb{C}P^\infty$, no nontrivial cohomology classes of $X$ can lift to the Cech cohomology of $Y$, since $Y$ can't have any non-nilpotent classes in Cech cohomology. But I don't have a clear understanding of how the relevant open covers of $X$ fail to extend to $Y$, or what would happen for an arbitrary infinite CW-complex $X$. Jul 6 '17 at 7:49
• (To give some idea of what I mean by "monstrously huge", I would guess that $\check{H}^1(Y)$ has cardinality $\beth_\omega$ when $X$ is a countably infinite CW-complex.) Jul 6 '17 at 8:13