Why does $\partial^2 = 0$ in simplicial homology? Let $\sigma:\Delta^n\rightarrow{X}$ be a map from the standard n-simplex to a topological space $X$. Now let us define the boundary operator as:
$$\partial(\sigma)=\sum_{j=0}^{n}{(-1)^{j}\sigma}i_j,$$
where $i_j:\Delta^{n}\rightarrow\Delta^{n-1}$ defined as:
$$i_j([e_0,...,e_n]) = [e_0,...,\hat{e_j},...,e_n],$$
i.e. $i_j$ is a restriction to the $j^{th}$ side.
I am trying to prove that $\partial\partial=0.$ I am reading Hatchers text on algebraic topology and his proof looks like this:
Firstly, does he split the summation into cases where $k<j$ and $k>j$ since when $k=j$ this doesn't make sense since you are removing the $j^{th}$ element twice which is undefined.
   Secondly, I don't understand why the power of $-1$ is $j-1$ for the second sum?
 A: The boundary (in a topological, not algebraic sense) of an $n$-simplex is a union of $(n-1)$-simplicies. However, when doing homology, we throw in negatives to make things work out. This is done by putting a negative in front of every other $(n-1)$-simplex we encounter. 
Setting aside the negatives for the moment, imagine an equilateral triangle (like, say $\Delta$). By successively removing vertices, we can arrive at the top point of the triangle in two ways: 1. remove the bottom right vertex, then the left 2. remove the left first, then the right. 
This holds more generally: given any $n$-simplex, you can find each of its sub-$(n-2)-simplicies by removing two points.
Now, when we apply the boundary twice, we obtain a sum of $(n-2)$-simplicies. In fact, we obtain two "copies" of every sub-$(n-2)$-simplex, one for each way to "arrive" at the $(n-2)$-simplex by removing two points. (do it explicitly for the triangle).
Now, if we want to show the entire sum is 0, it suffices to show that the two sub-$(n-2)$-simplicies have different sign once we throw in the negatives. 
Say we remove vertices $i$ and $j$, ($i < j$). We would like to check that the simplex obtained by removing $i$ then $j$ has the opposite sign as the simplex obtained by removing $j$ then $i$.
Removing $j$ then $i$ gives us a factor of $(-1)^i(-1)^j$. However, if we remove $i$ first, $j$ will "move down" in the list of vertices, so it contributes a factor of $(-1)^{j-1}$. Then, removing $i$ then $j$ gives us a factor of $ (-1)^i(-1)^{j-1}$. 
$(-1)^i(-1)^j$ and $ (-1)^i(-1)^{j-1}$ obviously have opposite signs, so we're done.

If that was at all confusing, do it explicitly for the triangle. It will become clear.
I wouldn't get too caught up on Hatcher's proof. He's just saying what this answer says in far fewer words. 
A: You have to compute $\partial_{n-1}(\sigma \mid [w_0,\ldots,w_{n-1}])$ where $[w_0,\ldots,w_{n-1}] = [v_0,\ldots,\hat v_i,\ldots,v_n])$. The result is $\sum_{j=0}^{n-1}(-1)^j \sigma \mid[w_0,\ldots,\hat w_j,\ldots,w_{n-1}])$. The $j$-th summand is $(-1)^j \sigma \mid[v_0,\ldots,\hat v_j,\ldots,\hat v_i,\ldots, v_{n}])$ if $0 \le j < i$ and $(-1)^j \sigma \mid[v_0,\ldots,\hat v_i,\ldots,\hat v_{j+1},\ldots, v_{n}])$ if $i \le j \le n-1$ because in $[v_0,\ldots,\hat v_i,\ldots,v_n]$ the vertex $v_{j+1}$ is at position $j$. Now write $k = j+1$. We get $(-1)^j \sigma \mid[v_0,\ldots,\hat v_i,\ldots,\hat v_{j+1},\ldots, v_{n}]) = (-1)^{k-1} \sigma \mid[v_0,\ldots,\hat v_i,\ldots,\hat v_k,\ldots, v_{n}])$. Thus
$$\partial_{n-1}(\sigma \mid [w_0,\ldots,w_{n-1}]) \\=\sum_{j < i} (-1)^j \sigma \mid[v_0,\ldots,\hat v_j,\ldots,\hat v_i,\ldots, v_{n}]) + \sum_{k > i}(-1)^{k-1} \sigma \mid[v_0,\ldots,\hat v_i,\ldots,\hat v_k,\ldots, v_{n}]) .$$
In the second sum we may replace the summation index $k$ by $j$.
