I am currently trying to compute the first homology group of the sphere, using the open tetrahedron (not solid).
Let $\gamma, \beta, \alpha, \rho, \delta, \nu$ represent the six edges (1-simplexes) making up the open tetrahedron.
Then $$C_1= < \gamma, \beta, \alpha, \rho, \delta, \nu>$$ the free abelian group generated by these edges.
Thus every arbitrary element in $C_1$ can be represented as,
$a_1 \gamma + a_2 \beta + a_3 \alpha + a_4 \rho + a_5 \delta + a_6 \nu$ with $a_i \in \mathbb{Z}$.
Then, $$im(d_1)= a_1 (b-a) + a_2 ( a-d) + a_3 (d-b) + a_4 (d-c) + a_5 ( c-a) + a_6 ( b-c)$$
I found the $ker(d_1)$ to be spanned by $a_1 + a_2 + a_3$, $a_2 + a_3 + a_5$, and $-a_2 - a_3 + a_6$.
I am not sure how to interpret this result. Is the above summation of the $a_i's$ a cycle? A video I watched interpreted it this way but I can't really make sense of it.