Zeroth homology is free on set of path components

Statement

With coefficients in integers (default choice)

Suppose $X$ is a topological space. The zeroth homology group $H_{0} (X)$ (with coefficients in $Z$ ) is a free abelian group whose rank is equal to the number of path components of $X$ , i.e., the cardinality of the Set of path components (?) $π_{0} (X)$ . More specifically, $H_{0} (X)$ is equipped with a natural choice of freely generating set which is in canonical bijection with $π_{0} (X)$ .

With coefficients in an arbitrary abelian group or module

Suppose $X$ is a topological space and $M$ is an abelian group (which may be additionally interpreted as a module over a commutative unital ring $R$ ). The zeroth homology group $H_{0} (X; M)$ , with coefficients in $M$ , can be canonically identified with $M^{π_{0} (X)}$ , i.e., a direct sum of copies of $M$ , one copy for each path component of $X$ . Here, $π_{0} (X)$ denotes the set of path components of $X$ .

Proof

Proof for integers

Given: A topological space $X$ .

To prove: $H_{0} (X)$ is a free abelian group with freely generating set corresponding to the elements of $π_{0} (X)$ .

Proof: Recall that $H_{n} (X) = Z_{n} (X) / B_{n} (X)$ where $Z_{n} (X)$ and $B_{n} (X)$ are subgroups of $C_{n} (X)$ comprising the cycles and boundaries respectively, and $C_{n} (X)$ is the free abelian group on the set of singular n-simplices, which we sometimes denote by $S_{n} (X)$ . Compute $H_{0} (X)$ thus boils down to unraveling the meaning of $S_{0} (X), C_{0} (X), Z_{0} (X), B_{0} (X)$ .

Step no.	We're trying to determine...	It turns out to be ...	Explanation
1	$S_{0} (X)$ , i.e., the singular 0-simplices in $X$	canonically identified with the underlying set of $X$ .	A point $x \in X$ corresponds to the singular 0-simplex that sends the standard 0-simplex (which is a one-point space) to $x$ .
2	$C_{0} (X)$ , i.e., the singular 0-chains in $X$	canonically identified with the free abelian group on the underlying set of $X$ , i.e., formal linear combinations of points in $X$ .	None needed.
3	$Z_{0} (X)$ , i.e., the singular 0-cycles in $X$	the same as $C_{0} (X)$ . Thus, $Z_{0} (X)$ is canonically identified with the free abelian group on the underlying set of $X$ , i.e., formal linear combinations of points in $X$ .	The boundary map from degree $0$ to degree $- 1$ is the zero map.
4	$B_{0} (X)$ , i.e., the singular 0-boundaries in $X$	the subgroup of $C_{0} (X)$ generated by elements of the form $a - b$ where $a$ and $b$ are in the same path component of $X$ :	[SHOW MORE] An element of $B_{0} (X)$ arises as the boundary of an element of $C_{1} (X)$ . Since $C_{1} (X)$ is generated by the singular 1-simplices $S_{1} (X)$ , $B_{0} (X)$ is generated by the boundaries of singular 1-simplices. A singular 1-simplex is a path, and its boundary is the formal difference of its endpoints. A given difference of points occurs as the boundary of a singular 1-simplex if and only if the two points can be joined via a path.
5	$H_{0} (X) = Z_{0} (X) / B_{0} (X)$	free abelian group on the set of path components	[SHOW MORE] We construct a homomorphism from $Z_{0} (X)$ to the free abelian group on the set of path components which is surjective and has kernel $B_{0} (X)$ , that is sufficient. The homomorphism is as follows: consider the set map $X \to π_{0} (X)$ that sends each point to its path component. This induces a corresponding group homomorphism from the free abelian groups on these, so we get a map from $Z_{0} (X)$ to the free abelian group on the set of path components. The kernel of this map is precisely those singular 1-chains whose sum of coefficients of points is zero in each path component. Asingular 1-chain satisfies this condition iff it can be written as a formal sum of differences $a_{i} - b_{i}$ where $a_{i}$ and $b_{i}$ are in the same path component, and hence is in $B_{0} (X)$ .