Unclear step on the calculation of a homology group.

Question

let $K=\{p_0,p_1,(p_0 p_1)\}$ is a simplicial complex. We want to calculate the homology group $H_0(K)$.

$Z_0(K)=\{ip_0+jp_1:i,j\in\mathbb{Z}\}$ and $B_0(K)=\{ip_0-ip_1:i\in\mathbb{Z}\}$ so we want to learn $$H_0(K)=Z_0(K)/B_0(K)$$ that would be $\mathbb{Z}\oplus\mathbb{Z}/\mathbb{Z}$. However this is calculated different in the book I study. It follows like this. Let $f$ is a surjective homomorphism $f:Z_0(K)\rightarrow\mathbb{Z}$ by $$f(ip_0+jp_j)=i+j$$ then we have $$\ker f=f^{-1}(0)=B_0(K) \ \ \ \ \ \ \ (@)$$ then $$Z_0(K)/\ker f\cong \text{im}f=\mathbb{Z}$$ thus $$H_0(K)=Z_0(K)/B_0(K)\cong\mathbb{Z}$$

In this derivation I have two questions:

1. How the equivalence at step (@) is true?
2. How one could calculate $\mathbb{Z}\oplus\mathbb{Z}/\mathbb{Z}$ instead of the alternative approach?

Answer 1

1. Step $(@)$ follows from $$f(ip_0+jp_1)=0\Longleftrightarrow i+j=0\Longleftrightarrow j=-i$$

2. As was said in this answer, writing $(\mathbb{Z}\oplus \mathbb{Z})/\mathbb{Z}$ makes no sense by itself. In this particular framework you quotient by the subgroup of $\mathbb{Z}^2$ generated by $(1,-1)$ (which is isomorphic to $\mathbb{Z}$, but there are other subgroups isomorphic to $\mathbb{Z}$ that would yield other results).

An alternative proof would be to check that $\mathbb{Z}^2$ is the internal direct product of the subgroups generated by the elements $(1, -1)$ and $(1,0)$; therefore when you quotient by the subgroup generated by $(1, -1)$, the quotient is isomorphic to the subgroup generated by $(1, 0)$.

In a more advanced language the elements $(1, -1)$ and $(1,0)$ form a basis of $\mathbb{Z}^2$ seen as a $\mathbb{Z}$-module, and the key-property to prove this is that the matrix $\left(\matrix{1& -1\\ 1&0}\right)$ has determinant $\pm 1$ (hence is invertible in $\mathcal{M}_{2\times 2}(\mathbb{Z})$).