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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?

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    $\begingroup$ Does 0'th sheaf cohomology (with constant coefficients) count components or quasicomponents? $\endgroup$ – Grigory M Jul 3 '13 at 18:47
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    $\begingroup$ $H_{0}(PX)$, 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. $\endgroup$ – DBS Jul 7 '13 at 22:32
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    $\begingroup$ @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. $\endgroup$ – Giulio Bresciani Dec 22 '14 at 21:21
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    $\begingroup$ What exactly do you mean by « a broad sense»? Does the free abelian group on the set of connected components count? $\endgroup$ – Mariano Suárez-Álvarez Apr 30 '15 at 4:38
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    $\begingroup$ @rookie For general topological spaces there is a difference between path components and connected components. $\endgroup$ – Daniel Gerigk Sep 3 '17 at 4:58
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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.

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  • $\begingroup$ 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. $\endgroup$ – Pedro Tamaroff Mar 7 '18 at 22:14
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    $\begingroup$ +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). $\endgroup$ – Eric Wofsey Mar 7 '18 at 22:49
  • $\begingroup$ 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. $\endgroup$ – Paul Frost Jun 18 '18 at 22:57

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