Boundary operator of $n$-chain composed with itself is zero

Question

Suppose that $X$ is a topological space, and $S_n(X)$ is the singular $n$-chain. The $i$th face of $s$, $\partial_i(s)$, where $s$ is a singular $n$-simplex is defined by $\partial_i(s)(t_0,...,t_{n-1})=s(t_0,...,t_{i-1},0,t_{i},...,t_{n-1})$. The boundary operator is defined to be $\partial:=\partial_0-\partial_1+...+(-1)^n\partial_n$. I want to show that the composition $S_n(X)\xrightarrow{\partial}S_{n-1}(X)\xrightarrow{\partial}S_{n-2}(X)$ yields $0$.

I obtained a sum and decomposed it into three sums, indexes being equal and unequal. But then the terms are not canceled out. How should I proceed?

Answer 1

I'm going to write a somewhat standard $\sigma \mid_{[v_0, \dots \hat{v_i}, \dots v_n]}$ for what is your $\sigma(v_0, \dots,0, \dots v_n)$ just so it is easier to see the $i$.

Note that

\begin{align*}\partial_{n-1} \partial_{n}(\sigma)&=\sum_{i=0}^n(-1)^i \cdot\partial(\sigma \mid_{[v_0, \dots \hat{v_i}, \dots v_n]})\\ &=\sum_{i=0}^n(-1)^i\left(\sum_{j<i}(-1)^j\sigma \mid_{[v_0, \dots,\hat{v_j}, \dots ,\hat{v_i}, \dots v_n]}+\sum_{j>i}(-1)^{j-1} \sigma \mid_{[v_0, \dots,\hat{v_i}, \dots ,\hat{v_j}, \dots v_n]} \right)\\ \end{align*}

Question 1: why is the first power of $(-1)$ written with "$j$" and the second one "$j-1$"?

Question 2: Why is this enough to conclude the claim? (hint: how many times does map occur in the sum, and what are the parities of $i+j-1$ vs. $i+j$?)

bonus question: who do mathematicians insist on using the similar looking symbols $i,j$ for indices, even in tiny little subscripts?

Answer 2

$\partial_{n-1}\partial_n (\sigma)=\sum_{i=0}^n (-1)^i.\partial_{n-1}\sigma_{|[v_0,...,\bar v_i,..., v_n]}$. Is it correct?

The answer to Question 1: In stepping from the $(i-1)$th slot in the first sum to the $(i+1)$th slot in the second sum, the alternating signature must not change.

The answer to Question 2: There are $n(n+1)$ sums which is an even number. The number of $(i+j-1)$ and that of $(i+j)$ are both the same, $0+1+2+...+n$.

An answer to Bonus Question: I think an answer is that $i$ is a standard symbol for electric current, so it runs, and it is customary to use such mathematical expressions as the sum as $i$ runs over a certain set.