It is well-known that the generators of the zeroth singular homology group $H_0(X)$ of a space $X$ correspond to the path components of $X$.

I have recently learned that for Čech homology the corresponding statement would be that $\check{H}_0(X)$ is generated by the quasicomponents of $X$. This leads me to my question:

Are there any homology theories (in a broad sense; i.e. not necessarily satisfying all of Eilenberg-Steenrod axioms) being used such that the zeroth homology of a space is generated by its connected components?

Dan Rust
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Dejan Govc
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  • (my answer is incorrect, I was forgetting my basic point-set nonsense: components don't have to be clopen... Sorry!) – Dylan Wilson Mar 02 '13 at 06:46
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    Does 0'th sheaf cohomology (with constant coefficients) count components or quasicomponents? – Grigory M Jul 03 '13 at 18:47
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    $H_{0}(PX)$, where $PX =$ the path space of X with compact open topology. I would have rather liked to put it as a comment but I do not have enough points to do so. – DBS Jul 07 '13 at 22:32
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    @GrigoryM quasicomponents: if $p,q\in X$ are in the same quasicomponent, they can't be divided by a global section of a locally constant sheaf. – Giulio Bresciani Dec 22 '14 at 21:21
  • I would comment this, but don't have points to do so. Does Alexander-Spanier cohomology work? A reference would be Massey's book on homology and cohomology. – Kyle Dec 31 '14 at 17:05
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    What exactly do you mean by « a broad sense»? Does the free abelian group on the set of connected components count? – Mariano Suárez-Álvarez Apr 30 '15 at 04:38
  • @MarianoSuárez-Alvarez OP states "... [homology theories] being used". So I think a reasonable interpretation of "broad sense" here could mean any homology theory that is actually used (more or less commonly). – Ittay Weiss Apr 30 '15 at 06:25
  • @MarianoSuárez-Alvarez: I would be happy with any homology theory that is actually used. My main reason for using the phrasing "broad sense" was to allow things like Čech homology theory, which is not exact. (I hear Steenrod-Sitnikov homology corrects this defect, but I'm not sure what it counts.) So, ideally, this homology theory should satisfy all of Eilenberg-Steenrod axioms, but if e.g. one of them fails (or, preferably, holds in some weaker form), I'm still interested. – Dejan Govc Apr 30 '15 at 12:32
  • @MarianoSuárez-Alvarez: Answering your second question, yes, $H_0$ in this theory should probably be the free abelian group on the set of connected components. But this should be just the first term in a sequence $H_n$ of functors defined in some "natural and uniform manner" (i.e. in the same way for each $n$; for example, but not necessarily, via a chain complex). I hope this is not too vague. – Dejan Govc Apr 30 '15 at 12:53
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    That is not what I mean. *Define* $H_0(X)$ to be the free abelian group on the components of $X$ and let $H_0(X,Y)$ and $H_1(X,Y)$ be the cokernel and kernel of the map $X_0(Y)\to H_0(X)$. Let $H_p=0$ for all $p>1$. This has long exact sequences for pairs, is additive and satisfies the dimension axiom. I have no idea about excision and homotopy, but I guess they are not satisfied. I don't know how reasonable is to ask for these for a theory having components in degree zero, though — maybe one can prove these two cannot be satisfied? – Mariano Suárez-Álvarez Apr 30 '15 at 17:37
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    @Kyle : From the [Wikipedia](https://en.wikipedia.org/wiki/Alexander%E2%80%93Spanier_cohomology): "The Alexander–Spanier cohomology groups coincide with Čech cohomology groups for compact Hausdorff spaces, and coincide with singular cohomology groups for locally finite complexes." So Alexander-Spanier cohomology appears to capture the two examples given by the OP, not connected components. – Eric Towers May 27 '16 at 19:08
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    https://en.wikipedia.org/wiki/Betti_number: Isn't $0^{\text{th}}$-Betti number counting the number of connected components of a space? – rookie Feb 15 '17 at 13:33
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    @rookie For general topological spaces there is a difference between [path components](https://en.wikipedia.org/wiki/Connected_space#Path_connectedness) and [connected components](https://en.wikipedia.org/wiki/Connected_space#Connected_components). – Daniel Gerigk Sep 03 '17 at 04:58

1 Answers1


There is no homology theory which satisfies the following conditions:

  1. $H_0(X)$ is the free abelian group generated by the connected components of $X$.

  2. The homomorphism $f_*:H_0(X)\to H_0(Y)$ induced by a continuous map $f:X\to Y$, maps a generator $[x]\in H_0(X)$ to the generator $[f(x)]\in H_0(Y)$.

  3. The theory satisfies the Homotopy, Exactness, Excision and Dimension axioms.

Proof. Let $$X=\{(0,1)\} \cup ([0,1]\times \{0\}) \cup \bigcup\limits_{n\ge 1} (\{\frac{1}{n}\}\times [0,1]) \subseteq \mathbb{R}^2$$ with the usual subspace topology. Let $A=X\smallsetminus \{(0,1)\}$ and $U=\{(x,y)\in X \ | \ y<x\} \subseteq X$. Then $U$ and $A$ are open and $\overline{U}\subseteq int(A)=A$. By Excision $H_0(X\smallsetminus U, A\smallsetminus U)$ is isomorphic to $H_0(X,A)$. Since $A$ and $X$ are connected, $H_0(A)\to H_0(X)$ is the identity by 2. Since $A$ is contractible, by Homotopy it has the homology of a point, so by Dimension $H_{-1}(A)=0$. Then by Exactness $H_0(X,A)=0$. On the other hand $\{(0,1)\}$ is a connected component of $X\smallsetminus U$ and $(0,1)\notin A\smallsetminus U$. Therefore $H_0(A\smallsetminus U)\to H_0(X\smallsetminus U)$ is not surjective (by 1 and 2). By Exactness, $H_0(X\smallsetminus U)\to H_0(X\smallsetminus U, A\smallsetminus U)$ is not the trivial homomorphism. Then $H_0(X\smallsetminus U, A\smallsetminus U)\neq 0$, a contradiction.

Jon Barmak
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  • As someone pointed out to me just now, I suppose the intuition to gather here is that the comb (with the point in the $y$-axis) is weakly homotopy equivalent to the space with two points, but certainly not homotopy equivalent to it. – Pedro Mar 07 '18 at 22:14
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    +1, though I'm not convinced that this definitively answers the question. In particular, Cech cohomology reaches the correct answer in your example by violating (1), in this case keeping track not just of the _set_ of (quasi)components but also their topology. It seems plausible that some other theory could work by similarly violating (1) (but unlike Cech cohomology, it would need to somehow still separate components that are in the same quasicomponent). – Eric Wofsey Mar 07 '18 at 22:49
  • The answer also depends on the class of pairs $(X,A)$ on which the homology theory $H_\ast$ is defined. For example, if we allow only compact pairs, then the above example no longer works. In fact, in compact spaces components and quasi-components agree so that Cech homology will do. With suitable coefficients (e.g. $\mathbb{Q}$) Cech homology is even exact. – Paul Frost Jun 18 '18 at 22:57