Generaly $\tilde H_0 \oplus \mathbb Z =H_0$. (reduced homology and homology)
I'm interested in the specific case $H_0 =\mathbb Z \oplus \mathbb Z$ or a little more generally $H_0 =\bigoplus_{1\leq i\leq n} \mathbb Z$.
I obviously want to deduce $\tilde H_0 =\mathbb Z$ or in the general case $\tilde H_0 =\bigoplus_{1\leq i\leq n-1} \mathbb Z $.
Is this correct? How can it be done? All I can think of is goin to the definition of $H_0$, but it didn't work out well for me. Purely algebraic seems not so possible, but i might be very wrong ofc. If going the algebraic direction it is quite easy if we knew that $\tilde H_0$ is free, but I don't think that it is necessarily true.
Purely algebraically, you know that $H_0=\bigoplus\mathbb{Z}$ is free of finite rank $n$, and $\tilde{H}_0\subset H_0$ is a submodule. By the structure of modules over a principal domain, any submodule of a free module is again free, hence $\tilde{H}_0$ is free and it will be of rank $n-1$.
In fact, you don't need to know the structure of modules over a principal ring, you can explicitly give a basis of $\tilde{H}_0$.
First, you know a basis of $H_0$, namely the class of a point in each path-component. In other words, $H_0=\bigoplus_{i=1}^n\mathbb{Z}x_i$.
For $\tilde{H}_0$, choose a point, say $x_n$. Then the differences $x_i-x_n$ for $1\leq i\leq n-1$ form a basis of $\tilde{H}_0$. Indeed, they are of degree 0, they form a linearly independent family, and they generate $\tilde{H}_0$ since if $c=\sum_{i=1}^n a_i x_i$ is of degree 0, then $$ c = \sum_{i=1}^{n-1} a_i (x_i-x_n) + (\sum_{i=1}^n a_i)x_n = \sum_{i=1}^{n-1} a_i (x_i-x_n)$$ Hence, $\tilde{H}_0$ is free of rank $n-1$.