I am working on calculating the first homology group $H_1(X)$ for a simplicial complex X, and I have found the kernel of the boundary operator $\partial_1: C_1 \rightarrow C_0$ (chain complex) to be the set:
\begin{equation} \{\alpha \cdot [p_0p_1p_2] + \lambda \cdot [p_0p_3p_4]| \alpha, \beta \in \mathbb{Z}\} \end{equation}
I understand that to find $H_1(X)=\frac{Ker(\partial_1)}{Im((\partial_2)}$, I need to find the image of $\partial_2$. This is not the problem. However, I want to demonstrate that the kernel I have computed is isomorphic to $\mathbb{Z}^2$.
Could someone guide me on how to properly show that this kernel is indeed isomorphic to $\mathbb{Z}^2$? Specifically, I am thinking of proving that the elements of the set form a basis for the kernel and are linearly independent.
Any help would be appreciated!
With the issues pointed out in the comments assumed fixed, let $\newcommand{\Z}{\mathbb{Z}}\varphi\colon \Z^2 \to \ker(\partial_1)$ be the map $\varphi(a, b) = a \cdot ([p_1 p_2] - [p_0 p_2] + [p_0 p_1]) + b \cdot ([p_3 p_4] - [p_0 p_4] + [p_0 p_3])$. It is surjective by definition and injective since $$ a \cdot ([p_1 p_2] - [p_0 p_2] + [p_0 p_1]) = b \cdot ([p_3 p_4] - [p_0 p_4] + [p_0 p_3]) $$ is to say that $$ a \cdot [p_1 p_2] = b \cdot ([p_3 p_4] - [p_0 p_4] + [p_0 p_3]) + a \cdot ([p_0 p_2] - [p_0 p_1]) $$ the only solution to which is $a = b = 0$ since by definition the 1-simplices form a basis of $C_1$ and are therefore in particular linearly independent.
In other words, $\varphi$ is the (/an) isomorphism you want.