It is wellknown 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 EilenbergSteenrod 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:

H0(X) is the free abelian group generated by the connected
components of X. 
The homomorphism f∗:H0(X)→H0(Y) induced by a continuous map
f:X→Y, maps a generator [x]∈H0(X) to the generator [f(x)]∈H0(Y). 
The theory satisfies the Homotopy, Exactness, Excision and Dimension
axioms.
Proof. Let
X={(0,1)}∪([0,1]×{0})∪⋃n≥1({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
¯U⊆int(A)=A. By Excision H0(X∖U,A∖U) 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
H−1(A)=0. Then by Exactness H0(X,A)=0. On the other hand
{(0,1)} is a connected component of X∖U and
(0,1)∉A∖U. Therefore H0(A∖U)→H0(X∖U) is not surjective (by 1 and 2). By Exactness,
H0(X∖U)→H0(X∖U,A∖U) is
not the trivial homomorphism. Then H0(X∖U,A∖U)≠0, a contradiction.
Attribution
Source : Link , Question Author : Dejan Govc , Answer Author : Jon Barmak