$H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}$

I can't seem to figure out why $H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}$. In my notes it says:

''...Since $\varepsilon \partial_1 = 0$, $\varepsilon$ vanishes on $im \partial_1$ and hence induces a map $H_0(X) \rightarrow \mathbb{Z}$ with kernel $\tilde{H}_0(X)$...''

How is this map induced? Why is $\mathbb{Z}$ relevant?. Many thanks for your help.

The map $H_0(X)\to\mathbb{Z}$ is induced by the map on chains $C_0(X)\to\mathbb{Z}$ sending each singular simplex to $1$. Since $\mathbb{Z}$ is a projective $\mathbb{Z}$-module (even free), you can always pick a spliting $\mathbb{Z}\to H_0(X)$. Therefore, $H_0(X)\simeq \mathbb{Z}\oplus\tilde{H}_0(X)$.