It is mentioned here in 4x how for a torus $\mathbb{T}^2$ one has the homology groups of $H_0 \cong \mathbb{Z}$, $H_1 \cong \mathbb{Z}\oplus \mathbb{Z}$ WITH generators $\{a+b+c\}$ and $\{d+e+f\}$, as well as $H_2 \cong \mathbb{Z}$. I understand the basic idea behind homology, but I struggle understanding why these sets specifically form generators. Could someone explain this to me? Here's also the triangulation of the torus.
Best Answer
I find the text a bit confusing. Usually the torus is presented as the quotient of the square by identifying parallel edges in a way to preserve orientations. This old question contains more material on such construction $CW$ complex structure and intuitive construction of spaces
Here instead from what I see you have a triangulation on the starting square and you want to present the generators using the edges of the triangulation. The ending result is correct but the method is more convoluted of what it needs to be. I suppose such triangulation will be used later for exercises or applications.
I would suggest first to understand the situation without the presented triangulation, like in the question I linked. To see the boundary edges give the generators of $H_1(\mathbb{T}^2)$ you can use either Seifert-Van Kampen theorem and then the fact that $H_1(X)$ is the abelianization of $\pi_1(X)$ for $X$ a path connected space or use the Mayer-Vietoris sequence.
If you manage to do that it should be easy to understand the generators presented via this triangulation. It is basically like presenting the circle $S^1$ as the quotient $\mathbb{R}/3\mathbb{Z}$ instead of the usual $\mathbb{R}/\mathbb{Z}$.