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In singular homology one of the first calculations you can make is $H_0(X)=H_0(pt)$ for path-connected $X$. This seems to be a property which does not follow from the axioms for a generalized homology theory. This raises the following question:

Assume $H_* : Top^2 \to Ab^\mathbb{N}$ is a homology theory. Thus we impose homotopy invariance, excision, the long exact sequence, the dimension axiom and if you wish also the disjoint union axiom. Is then $H_0(X)=H_0(pt)$ for every path-connected space $X$?

I believe that there is a counterexample. Of course this can't be homotopy equivalent to a CW-complex. And probably this is the reason why this question is not reasonable at all. It's just my curiosity.

Harry Gindi
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Martin Brandenburg
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    So really your question is to produce such an H and a space X to evaluate it on. You'll of course need an H which is not invariant under weak homotopy equivalences, so singular homology is out. – Chris Schommer-Pries Oct 06 '10 at 18:13
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    Maybe you shouldn't say "generalized homology theory" if you want the dimension axiom. – Tom Goodwillie Oct 06 '10 at 19:16
  • What is the Cech homology of the topologist's sine curve? – Charles Rezk Oct 06 '10 at 23:55
  • @Charles It may depend which exact version of that space (TSC) you are thinking of.Essentially the usual TSC has the same Cech invariants as if the sin 1/x bit was omitted. Cech homology is a shape invariant and the usual variants of the TSC (thought of as being in the plane) have contractible open neighbourhoods. – Tim Porter Oct 07 '10 at 06:06
  • As your question isn't quite clear as stated (is Chris's interpretation correct? and you should correct it in line with Tom's comment) I can't be sure of this, but it looks as though you're asking essentially the same thing as is asked in this question: http://mathoverflow.net/questions/1750 – Andrew Stacey Oct 07 '10 at 09:05
  • @Andrew: Of course Chris' interpretation is correct. The linked question is another one. – Martin Brandenburg Oct 07 '10 at 10:27
  • It is not hard to come up with a space whose Cech homology differs from its singular homology. But can you make an example where the $0$th homology is bigger (not smaller) than $0$th singular? Isn't that basically the question? – Tom Goodwillie Oct 07 '10 at 10:55
  • Hm, does Cech cohomology satisfy the axioms of a homology theory (without disjoint union)? If yes, what is a reference for this fact? – Martin Brandenburg Oct 07 '10 at 13:34
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    @Martin: Cech homology has at least three definitions: one is the inverse limit of homology groups over open covers (Aleksandrov–Cech homology), one takes values in pro-abelian groups, and one applies a derived version of the inverse limit functor to the chain complexes in the system of open covers (the version I was using). I'm not sure of a reference and I would have to check closely to see if it satisfies the other axioms. One has to define the relative Cech homology using a mapping cylinder so that the long exact sequence becomes a tautology. – Tyler Lawson Oct 07 '10 at 14:08

2 Answers2

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Here is a candidate space for Cech homology. (which doesn't satisfy the disjoint union axiom, and so strictly speaking it's not a homology theory)

For any natural number n, let $X_n$ be the circular arc of radius 1 + 1/n centered at $( 1+1/n,0)$ and extending between angles $-\pi \leq \theta \leq (\pi - 1/n)$. Let $X$ be the union of these circles with the subspace topology, which is path connected.

Covering this space with sufficiently small disks (each $X_n$ needing to be covered by progressively smaller disks) gives a Cech nerve homotopy equivalent to a similar space where all but finitely many of the circular arcs have been closed up to circles. The zero'th homology of this cover is $\mathbb{Z}$ and the first homology is an infinite direct sum $\oplus_{n \geq N} \mathbb{Z}$. All other homology groups are zero.

As you decrease the size of the cover, you get a cofinal sequence of open covers inducing a decreasing sequence of abelian groups as N grows. There is a resulting exact sequence $$ 0 \to lim^1(\oplus_{n \geq N} \mathbb{Z}) \to \check{H}_0(X) \to lim^0(\mathbb{Z}) \to 0 $$ and the left-hand side is $(\prod \mathbb{Z}) / (\oplus \mathbb{Z}) \neq 0$.

Tyler Lawson
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I'm not sure exactly what Cech homology is, but I'll assume that something which may or may not be called Cech homology has the following properties:

It's a generalized homology theory.

It vanishes in negative dimensions.

It satisfies the dimension axiom (so it's a homology theory).

There's a natural map from singular to Cech that is part of a natural triangle, an exact sequence

$\dots \to Cech_{n+1}\to ?_n\to Sing_n\to Cech_n\to \dots$.

The map $Sing_0\to Cech_0$ is always surjective, so that $?_n$ vanishes for $n<0$.

The map $Sing_0(TSC)\to Cech_0(TSC)$ is not injective if TSC is the topologist's sine curve.

(End of list of assumed properties.)

Then $?$ is a generalized homology theory vanishing in negative dimensions and vanishing on a point (therefore on CW complexes) but not vanishing on the path-connected space that you get by attaching a suitable $1$-cell to TSC.

The direct sum of ? and singular will then do the job, because $?_0(TSC\cup cell)=?_0(TSC)\ne 0$

Tom Goodwillie
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