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An answer to the following question would clarify my understanding of what a cohomology theory is. I know it's something that satisfies the Eilenberg-Steenrod axioms, and I know that those axioms allow you to work out quite a lot. But what sort of thing is not determined by the axioms? In particular, can someone give me a simple example of a space that has different cohomology groups with respect to two different theories? Obviously a trivial answer would be to take coefficients in different rings, so let me add the requirement that the coefficient rings should be the same. And if there's some other condition needed to make the question non-trivial, then add that in too.

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7 Answers 7

For any space that has the homotopy type of a CW complex, its cohomology is determined purely formally by the Eilenberg-Steenrod axioms, so a counterexample is necessarily some reasonably nasty space. Here's an example you can see with your bare hands: consider the space $X=\{1,1/2,1/3,1/4,...,0\}$. Now 0th singular cohomology is exactly the group of $\mathbb{Z}$-values functions on your space which are constant on path-components, so $H^0(X)=X^\mathbb{Z}$ (an uncountable group) naturally for singular cohomology. On the other hand, 0th Cech cohomology computes global sections of the constant $\mathbb{Z}$ sheaf, i.e. locally constant $\mathbb{Z}$-valued functions on your space. These must be constant in a neighborhood of 0, so the Cech cohomology $H^0(X)$ is actually free of countable rank, generated (for example) by the functions $f_n$ that are $1$ on $1/n$, $-1$ on $1/(n+1)$, and $0$ elsewhere, plus the constant function $1$.

I should add that topologists don't actually care about such examples. The point of the Eilenberg-Steenrod axioms is to show that cohomology of reasonable spaces is determined by purely formal properties, and these formal properties are actually much more useful than any specific definition you could give (the only point of a definition is to show that the formal properties are consistent!). What is of interest is when you remove the dimension axiom to get "extraordinary" cohomology theories, which Oscar talks about in his answer.

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Here's a question I don't know the answer to (perhaps I should post it, hmm): What is the class of spaces for which the ES axioms determine ordinary cohomology? We know it includes all spaces with the homotopy type of a CW complex, but is that all, or are there others? –  Charles Rezk Oct 21 '09 at 23:14
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@Charles: I think this indeed deserves its own question-page; please post it! –  Benoit Jubin Oct 22 '09 at 0:33
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I unfortunately don't have anything of value to add, but I couldn't resist pointing out that "an example you can see with your bare hands" is a wonderful (and apparently original!) mixed metaphor. :) –  Harrison Brown Oct 22 '09 at 1:07
    
Oh, and @Charles: I'd like to second Benoit here, please ask it! –  Harrison Brown Oct 22 '09 at 1:08

An example I recently came across while learning about shape theory is the closed topologist's sine curve. The zero-th singular cohomology (integer coefficients) is Z^2, while the zero-th Cech cohomology (again integer coefficients) is Z.

I can't give a great reason why this is so as I'm just learning about the technicalities myself, but it seems to be a standard example.

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Does Cech cohomology in this setting satisfy all the Eilenberg-Steenrod axioms? –  Andreas Holmstrom Oct 21 '09 at 23:01
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Yes. In fact, it actually satisfies a stronger form of excision than singular cohomology. However Cech cohomology fails to satisfy that a weak equivalence of spaces induces an isomorphism on (ohomology. –  Eric Wofsey Oct 21 '09 at 23:09
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And, to close the circle: if your cohomology theory does take weak equivalences to isomorphisms, then it is uniquely determined (up to isomorphism). –  Charles Rezk Oct 21 '09 at 23:35

It might be interesting to note that the examples given by Eric and Justin also show why it is necessary to use Čech cohomology when treating Alexander duality for arbitrary closed subsets of a closed (and orientable) manifold. If one considers A={1/n | n>=1} as a subset of R/Z=S^1 then its complement is a countable disjoint union of open intervals and its 0-th homology (for any theory satisfying the axioms) is free with a countable basis. So Alexander duality would demand that the 0-th cohomology of A also have a countable basis, which is true for Čech cohomology but not for singular cohomology.

Similarly for for the closed topologist's sine curve as a subset of the two-dimensional sphere. Its complement has the homology of a point, so it itself should have the cohomology of a point, which again is true for Čech cohomology but not for singular cohomology.

This has indeed lead to a remark in the classical textbook by Seifert and Threllfall that while it is possible to define singular homology for arbitrary topological spaces and not just complexes, it is not very useful to do so. But that was in 1934.

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The sheaf cohomology with coefficients in the constant sheaf $Z$ coincides with the singular cohomology or the Cech cohomology for all homologically locally connected hausdorff paracompact spaces (this is the case for finite CW-complexes or locally contractible spaces).

Sheaf cohomology is better behaved than singular cohomology when it comes to dimension. Recall that the cohomological dimension of a compact topological space X is the greatest integer $n$ (or $\infty$) such that $H^n(X)\neq 0$.

Theorem: If $F$ is a closed subset of a compact topological space $X$, then its sheaf cohomological dimension is less than the sheaf cohomological dimension of $X$ (see e.g. Bredon "sheaf theory", II.16).

Surprisingly, this result is false for singular cohomology. The following example is due to Barratt and Milnor (1962). Let X be the union in $R^3$ of countably many spheres of radius 1/n, all tangent to the xy-plane at the origin. The sheaf cohomological dimension of this compact space is 2. But the singular cohomology of this space is non-zero in arbitrarily large degrees. In other words, its singular cohomological dimension is infinite.

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This is really cool! –  Lennart Meier May 30 '10 at 8:20

The examples mentioned so far, made the point that Cech cohomology differs from singular cohomology in many examples which are, to the eyes of most topologists, quite pathological. I want to concentrate on an example of an ordinary homology theory, which is probably much less well known than Cech cohomology, but quite interesting in geometric contexts: stratifold homology (see Kreck's Differential Algebraic Topology). It is defined as bordism classes of certain stratified spaces (including manifolds, but much more).

An example, where stratifold homology and singular homology differ is the one-point compactification of the surface of infinite genus (i.e. the infinite connected sum of tori), denoted by $F_\infty^+$. This can be given the structure of a stratifold and therefore possesses a fundamental class, which is non-zero, even after applying the induced map of $F_\infty^+\to F_g$ to the surface of genus g. Such a class does not exist in singular homology (see Differential Algebraic Topology, chapter 20.2). The space $F_\infty^+$ is surely not the first thing one considers, but it has still some geometric appeal (and is a compact set of $\mathbb{R}^3$). I think, there are probably similar phenomena for one-point-compactifications of certain other open manifolds.

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An interesting cohomology theory is the continuous singular cohomology theory with real coefficients introduced, as far as I know, in the seventies by R.Bott, G.D.Mostow and others.

Let a singular cochain be called continuous if it restricts to a real continuous map on the space of singular simplices, endowed with the compact-open topology. It is readily seen that continuous cochains provide a subcomplex of the complex of singular cochains. The (co)homology of the complex of continuous cochains is called continuous cohomology.

As one can expect, continuous cohomology is canonically isometric to singular cohomology for reasonable spaces (for example, for locally contractible metrizable spaces). However, one can construct spaces with non-isomorphic continuous and singular cohomology just by looking at subspaces of the Euclidean plane.

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(Some inaccurate info appeared in the Answers above).

Different co/homology theories may behave differently on infinite CW-complexes. If two E-S theories are the same for the singleton then they are equivalent for all spaces homotopically dominated by finite polyhedra. It seems to me that for any other topological space one can construct two E-S theories (equal for singletons) which will give different groups (let's consider completely regular $T_1$-spaces only).

Surely dimension of the Hilbert cube   $C$   is infinite, while   $H^n(C)=0$   for every   $n > 0$. In the case of any Hausdorff compact space   $X$   the (topological) covering dimension, say   $n = \dim(X)$,   is equal to the cohomological dimension, meaning that there exists a closed subset   $A$   of   $X$   such that   $H^n(X\ A) \ne 0$,   but there is no such closed subset   $B$   that   $H^m(X\ B)\ne 0$   for any   $m > n$   (a few more words should be devoted to the infinite-dimensional case).

Intuitively the singular and the Cech co/homology view topological spaces in a drastically different way. What is connected for the cruder Cech theory might be not connected for the more sensitive singular theory; the former one is much more tolerant. One can say that the Cech theory applies the shape theory as its foundation rather than just homotopy theory. More precisely, all co/homology functors split into a composition of the homotopy functor and the rest; while the Cech functor does more, it splits into composition of the shape functor and the rest (while the shape functor is a composition of the homotopy functor and the rest).

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In the grand scheme of things, it is probably better not to tweak the spacing like you did here. Consistently across the site is, in the end, better than getting one's own personal posts to look perfectly to one's own eye. –  Mariano Suárez-Alvarez Jul 3 '13 at 6:37
    
@MarianoSuárez-Alvarez I disagree - I don't think the spacing here is a big enough deal to be discouraged or altered from the author's own intentions, and I don't have any desire to find some editing of spacing bumping this answer unnecessarily three weeks from now –  Yemon Choi Jul 3 '13 at 6:51
    
You will not find, Yemon, in what I wrote any suggestion whatsoever to edit this answer... It was expressing my preference for MO not becoming the math-social-site analogue of a kidnapper's letter. –  Mariano Suárez-Alvarez Jul 3 '13 at 6:55

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