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 $\mathbb {Z}$) 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 (?) $\pi _{0}(X)$. More specifically, $H_{0}(X)$ is equipped with a natural choice of freely generating set which is in canonical bijection with $\pi _{0}(X)$.

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 $M^{\pi _{0}(X)}$, i.e., a direct sum of copies of $M$, one copy for each path component of $X$. Here, $\pi _{0}(X)$ 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 $\pi _{0}(X)$.

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 $X\to \pi _{0}(X)$ 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 $B_{0}(X)$.