Proving that $H_0(X)=\tilde{H_0}(X)\oplus\mathbb{Z}$ In the article about the Reduced Homology it's stated that $$H_0(X)=\tilde{H_0}(X)\oplus\mathbb{Z}$$
but I don't know how to prove that. I know $$H_0(X)=\bigoplus_{\alpha\in I}H_0(X_\alpha)=\bigoplus_{\alpha\in I} \mathbb{Z}$$where $X_\alpha$ are the connected components of $X$. That means we need to prove $$\tilde{H_0}(X)=\bigoplus_{\beta\neq\alpha\in I}\mathbb{Z}$$ for a specific $\beta\in I$. We know that $S_0(X)=\bigsqcup_{\alpha\in I}S_0(X_\alpha)$ where $S_0(X_\alpha)$ is the collection of 0-simplicies on each connectedness compoenent (which are simply points as as far as I understand). Also $$C_0(X)=\bigoplus_{\alpha\in I}C_0(X_\alpha).$$ This complex is mapped to $\bigoplus_{\alpha\in I}\mathbb{Z}$ by $C_0(X_\alpha)\ni 0\mapsto 1$. To conclude $$Z_0(X)\cong\bigoplus_{\alpha\in I}Z_0(X_\alpha)=\bigoplus_{\alpha\in I}\ker\partial_0(X_\alpha) \\ B_0(X)\cong\bigoplus_{\alpha\in I}B_0(X_\alpha)=\bigoplus_{\alpha\in I}\text{Im}\partial_1(X_\alpha)$$ but I cannot understand why from here it implies that $$\tilde{H_0}(X)=\bigoplus_{
\beta\neq\alpha\in I}\mathbb{Z}.$$ Am I doing correct? If so: How can I complete this proof?
 A: $\require{AMScd}$
Note that we have a commutative diagram
$$\begin{CD}
@.     0                @.         0      @.                  0            \\
@.     @VVV                        @VVV                       @VVV         \\
0 @>>> B_0(X)           @>>>       B_0(X) @>>>                0      @>>> 0\\
@.     @VVV                        @VVV                       @VVV         \\
0 @>>> \text{ker}(\varepsilon)@>i>>C_0(X) @>\varepsilon>>     \Bbb Z @>>> 0\\
@.     @VVV                        @VVV                       @|           \\
0 @>>> \tilde H_0(X)    @>\bar i>> H_0(X) @>\bar\varepsilon>> \Bbb Z @>>> 0\\
@.     @VVV                        @VVV                       @VVV         \\
@.     0                @.         0      @.                  0            \\
\end{CD}$$
where all columns are exact, as well as the top two rows. The nine lemma now implies that the lower row is exact, although that's maybe an overkill as a direct proof is also possible.
The top two rows split, i.e. there are maps $0\to B_0(X)$ and $s:\Bbb Z\to C_0(X)$ which are sections for $B_0(X)\to 0$ and $\varepsilon$, respectively, and both compositions $0\to C_0(X)$ are equal. We then have an induced map $\bar s:\Bbb Z\to H_0(X)$ such that $\bar\varepsilon\circ\bar s=\text{id}$. Hence the lower row splits and the map $\bar i\oplus\bar s:\tilde H_0(X)\oplus\Bbb Z\to H_0(X)$ is then an isomorphism.
For a more geometric idea what is actually going on here, note that $s$ is determined by $s(1)$ which is some $0$-simplex in $X$. Each $0$-chain can now be uniquely written as the sum of an element in $\text{ker}(\varepsilon)$ and $\eta\cdot s(1)$ for some integer $\eta$. This carries over to homology. The map $\bar s$ is determined by the choice of some path component $X_\beta$ of $X$. Each class $[c]\in H_0(X)$ is then sum of a class $[b]$ of an $\varepsilon$-cycle and that path component multiplied with some factor. Note that $[b]$ is determined by how often it meets every path component other than $X_\beta$ because of the restriction $\varepsilon(b)=0$, so we can indeed think of $\tilde H_0(X)$ as being generated by all path components $X_\alpha$ for $\alpha\ne\beta$.
