In algebraic topology we have the result $$H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}.$$ In Massey's book, this is a result that follows from the fact that the sequence $$0 \rightarrow \tilde{H}_0(X) \xrightarrow{\xi} H_0(X) \xrightarrow{\xi_*} \mathbb{Z} \rightarrow 0$$ is exact. So let me try and make sense of this:
The sequence is exact because $\text{im}(\xi) = \text{ker}(\xi_*)$. Where does this equality come from? We have that $\varepsilon \circ \partial_1 =0$ (where we have the sequence $\cdots \rightarrow C_1(X) \xrightarrow{\partial_1} C_0(X) \xrightarrow{\varepsilon} \mathbb{Z} \rightarrow 0$), hence $\varepsilon$ vanishes on $\partial_1$. We know that $\tilde{H}_0(X)=\text{ker}(\varepsilon)/\text{im}(\partial_1)$. Why is this the kernel of $\xi_* : H_0(X) \rightarrow \mathbb{Z}$ though? Can we show both inclusions?
Once that is understood, I need to understand where the isomorphism comes from. Do we have that for any map $\phi:A \to B$, then $A \cong \text{ker}(\phi) \oplus \text{im}(\phi)$? Or what's the deal? :)
I hope that you can help me clarify these issues.
$\require{AMScd}$
You have a commutative diagram $$\begin{CD} 0 @>>> ker(\varepsilon) @>i>> C_0(X) @>\varepsilon>> \Bbb Z @>>> 0\\ @. @VVV @VVV @| \\ 0 @>>> \tilde H_0(X) @>\bar i>> H_0(X) @>\bar\varepsilon>> \Bbb Z @>>> 0 \end{CD}$$ where the vertical maps send each element $\gamma\in C_0(X)$ to its homology class $[\gamma]$. Since $im(\partial_1)\subseteq ker(\varepsilon)$, there is an induced map $\bar\varepsilon$.
Using the exactness of the upper row, it is easy to show that the lower row is exact, as well. Moreover, since $\Bbb Z$ is free, we can construct a section $s$ for $\bar\varepsilon$, which is determined by some $[x_0]=s(1)$ An isomorphisms $H_0(X)\to\tilde H_0(X)\oplus\Bbb Z$ is given by $$[\gamma]\mapsto\left([\gamma-\varepsilon(\gamma)x_0],\varepsilon(\gamma)\right)$$