Understand Cech Cohomology - MathOverflow

Understand Cech Cohomology

2010-06-27T22:29:42Z

I am currently trying to understand Cech cohomology. Five questions arised and I would be glad for help. In what follows $X$ is a topological space.

I really like Dugger's and Isaksen's paper "Topological Hypercovers and ...". They prove for an arbitrary open cover $U=(U_a)_{a\in A}$ of $X$ the weak equivalence \[ hocolim ~C(U)\to X \] with $C(U)$ is the usual simplicial space of Cech namely \[ ...\to\coprod_{(b,c)\in A\times A}U_b\cap U_c\to\coprod_{a\in A}U_a \] Related statements are due to Segal. If $U$ is now a cover with every iterated intersection contractible we have a weak equivalence \[ hocolim ~C(U)\to hocolim ~\pi_0C(U) \] The space $\pi_0C(U)$ is a simplicial set. You can find such a cover for every locally contractible space for instance. Dugger and Isaksen build the category $OpCov(X)$. This has open covers of $X$ as objects and morphisms are mappings between the index sets $f:A\to B$ and for every $a$ in $A$ a mapping $U_a\to V_{f(a)}$. One can also build the category $Cov(X)$ with same objects but morphisms only strict containments of covers. Is then \[ hocolim ~C(U)\sim lim_{U\in Cov(X)}~hocolim ~\pi_0C(U)\sim lim_{U\in OpCov(X)}~hocolim ~\pi_0C(U) \] all weakly equivalent for locally contractible $X$?
Why does one define Cech cohomology as $H^n(X)=colim H^n(hocolim C(U))$ with colimit over $Cov(X)^{op}$? Is this the right definition? Why not the limit over $Cov(X)$ instead of colimit over the opposite category?
What is the problem with Cech Homology? One defines $\hat H^n(X)=colim ~H^n(C(U))$ as Cech cohomology. Then for a locally contractible $X$ it coincides with singular cohomology. Why not define $\hat H_n(X)=lim ~H_n(C(U))$. I have read that it is because the limit functor does not respect exact sequences. Does it mean $\hat H_n(X)$ coincides not with singular homology for locally contractible $X$?
Can one see directly without going through sheaf cohomology that singular cohomology and Cech cohomology are the same for locally contractible $X$? I have written down complexes for the boundary of the two simplex with a nice cover and yes the cohomology is the same. Is there a kind of MayerVietoris argument?
Here is my last question. $X$ is a good space now. Where is the mistake in the following equality. I consider Top as a topologgical model category and the crucial step is perhaps the one which looks like a smallness argument on the homotopy category. \[ \begin{array}{rcl} \pi_0(X)&=&[S^0, hocolim ~C(U)]\\ &=& hocolim[S^0, C(U)]\\ &=& hocolim \pi_0(C(U))\\&=&X \end{array} \] Here $U$ is a cover with everything contractible as above.

Thank endurance reading and for help.

Answer by GPadilla for Understand Cech Cohomology

2010-08-19T23:09:17Z

For Question #2: Look at chapter 2 of Bott, R. & Lu, T. "Differential forms in Algebraic Topology", Springer GTM vol 82. There you will find a nice introduction to Cech cohomology through a Mayer-Vietoris general point of view.

For Question #4: Yes. Look at Bredon, G. "Topology and Geometry", Springer GTM vol 139, p. 289. There is a Mayer-Vietoris set-construction which is a nice way to avoid presheaf calculations.

Answer by Zoran Škoda for Understand Cech Cohomology

2010-10-01T14:56:47Z

On 3rd question: Čech homology is not a homology theory in the sense of Eilenberg-Steenrod: the exactness axiom (long exact sequence in homology) does not hold, exactly because of the problem with limits. There is however a sophisticated method, discovered by Sibe Mardešić, to correct this, by modifying slighly the Čech definition. The resulting "strong homology theory" agrees with singular homology on the spaces having homotopy type of CW complexes, and does give long exact sequence of pairs $(X,A)$ where $X$ is paracompact and $A$ closed; moreover for metric compacta it satisfies not only all the axioms of Eilenberg-Steenrod, but also the relative homeomorphism axiom and the wedge axiom. The only homology theory on the metric compacta satisfying not only the Eilenberg-Steenrod but also the wedge axiom is the Steenrod-Sitnikov homology theory, hence the strong homology agrees with it.