$H_0(X)$ with coefficient in arbitrary PID ring It's a well know fact that if $X$ is a topological path connected space then $H_0(X) = \mathbb{Z}$ and $\tilde{H}_0(X) = 0$ as stated and linked here : $H_0 ( X )$ is free abelian on the path components of $X$ ., The zeroth homology group corresponds to path components, zero relative homology.
I was wondering if the statement or the proofs given in the homology with $\mathbb{Z}-$coefficient could be replaced word by word to obtain $H_0(X;R) = R$ and $\tilde{H}_0(X;R) = 0$ with $R$ arbitrary PID.
Is this a true extension ? Any help would be appreciated
 A: There are two ways to define the homology groups $H_n(X;R)$:

*

*Via the singular chain complex $C_*(X;R)$ with coefficients in $R$ whose $n$-th component $C_n(X;R)$ is the free $R$-module having as a basis the singular $n$-simplices in $X$. This is completely analogous to the "usual" singular chain complex $C_*(X)$ with coefficients in $\mathbb Z$ . In fact, $C_*(X;\mathbb Z) = C_*(X)$.


*For any abelian group $A$ one can define the singular chain complex with coefficients in $A$ by $C_*(X;A) = C_*(X) \otimes_{\mathbb Z} A$.
Clearly, if $A = R$, then 1. and 2. yield naturally isomorphic chain complexes. However, the perspective is somewhat different: In 1. we get a chain complex consisting of free $R$-modules, in 2. we get a chain complex consisting of abelian groups and ignore the $R$-module structure. Therefore in 1. the homology groups $H_n(X;R)$ have the structure of $R$-modules, but in 2. they are only abelian groups. As abelian groups both constructs are identical.
This  makes clear that for a  path connected space $X$ we have $H_0(X;R) = R$ and $\tilde{H}_0(X;R) = 0$. The proof is exactly the same as for "usual" homology groups.
