Is there a homology theory that counts connected components of a space?

It is well-known that the generators of the zeroth singular homology group H0(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 ˇH0(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?

Answer

There is no homology theory which satisfies the following conditions:

  1. H0(X) is the free abelian group generated by the connected
    components of X.

  2. The homomorphism f:H0(X)H0(Y) induced by a continuous map
    f:XY, maps a generator [x]H0(X) to the generator [f(x)]H0(Y).

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

Proof. Let
X={(0,1)}([0,1]×{0})n1({1n}×[0,1])R2
with the usual subspace topology. Let A=X{(0,1)} and U={(x,y)X | y<x}X. Then U and A are open and
¯Uint(A)=A. By Excision H0(XU,AU) is isomorphic to H0(X,A). Since A and X are
connected, H0(A)H0(X) is the identity by 2. Since A is
contractible, by Homotopy it has the homology of a point, so by Dimension
H1(A)=0. Then by Exactness H0(X,A)=0. On the other hand
{(0,1)} is a connected component of XU and
(0,1)AU. Therefore H0(AU)H0(XU) is not surjective (by 1 and 2). By Exactness,
H0(XU)H0(XU,AU) is
not the trivial homomorphism. Then H0(XU,AU)0, a contradiction.

Attribution
Source : Link , Question Author : Dejan Govc , Answer Author : Jon Barmak

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