In homology, when we operate the boundary twice we get zero, that is, $\partial^2=0$. Need help understanding proof.

Question

Proof for $S=\Delta_n=(v_0 ... \hat{v_i} ...v_n)=d_i$

$\partial=\displaystyle\sum_i^n(-1)^id_i.$

Thus,

$\partial^2=[\displaystyle\sum_i^n(-1)^id_i][\displaystyle\sum_j^n(-1)^jd_j] $ $=\displaystyle\sum_i\displaystyle\sum_j(-1)^{i+j}d_id_j $ $=\displaystyle\sum_{i\geq j}(d_id_j-d_id_j) $ $=\displaystyle\sum_{i\geq j}(-d_i^2+d_i^2) $ $=0.$

I'm completely new to homology and, in fact, algebraic topology. So i was wondering why does $\displaystyle\sum_{i\geq j}(-1)^{i+j}d_id_j=\displaystyle\sum_{i\geq j}(-1)^{j+i+1}d_jd_{i+1}$?

I completely understand when we show an example lets say for a tetrahedron simplex, say, $S=(v_0 \ v_1 \ v_2 \ v_3)$ then

$\partial(S)=(v_1 \ v_2 \ v_3) - (v_0 \ v_2 \ v_3) + (v_0 \ v_1 \ v_3) - (v_0 \ v_1 \ v_2)$

$\Rightarrow \partial(\partial(S))=\partial(v_1 \ v_2 \ v_3) - \partial(v_0 \ v_2 \ v_3) + \partial(v_0 \ v_1 \ v_3) - \partial(v_0 \ v_1 \ v_2)$

$=[(v_2 \ v_3) - (v_1 \ v_3) + (v_0 \ v_2)] - [(v_2 \ v_3) - (v_0 \ v_3) + (v_0 \ v_2)]+ [(v_1 \ v_3) - (v_0 \ v_3) + (v_0 \ v_1)] - [(v_1 \ v_2) - (v_0 \ v_2) + (v_0 \ v_1)] =0$.

Just need help understanding the proof.

Answer 1

You don't need homology to understand $\displaystyle\sum_{i\geq j}(-1)^{i+j}d_id_j=\displaystyle\sum_{i\geq j}(-1)^{j+i+1}d_jd_{i+1}$. Let us consider $$\displaystyle\sum_{i\geq j}(-1)^{i+j}d_id_j=\displaystyle\sum_{i\geq j}(-1)^{i+(j+1)}d_{i}d_{j+1}=\displaystyle\sum_{i\geq j}(-1)^{k+(l+1)}d_{k}d_{l+1}$$ Now let $k=j,l=i$. This gives $$\displaystyle\sum_{i\geq j}(-1)^{i+j}d_id_j=\displaystyle\sum_{i\geq j}(-1)^{j+i+1}d_jd_{i+1}$$