Boundary of boundary of singular cube is zero (Spivak) At the bottom of page 99 of M. Spivak's Calculus on Manifolds he arrives at the formula
$$\partial (\partial c)=\sum_{i=1}^n \sum_{\alpha=0,1} \sum_{j=1}^{n-1} \sum_{\beta=0,1} (-1)^{i+\alpha+j+\beta} (c_{(i,\alpha)})_{(j,\beta)} $$
Here $c$ is a singular $n$-cube in $A \subseteq \mathbb{R}^n$; that is, a continuous function $[0,1]^n \to A$, and $$c_{(i,\alpha)}=c \circ (I^n_{(i,\alpha)}),$$ where $I^n(i,\alpha)$ is the identity function on $[0,1]^n$ applied to $x \in [0,1]^n$ with its i'th coordinate replaced by $\alpha \in \{0,1\}$.
The author then claims:

In this sum $(c_{(i,\alpha)})_{(j,\beta)}$ and $(c_{(j+1,\beta)})_{(i,\alpha)}$ occur with opposite signs. Therefore all terms cancel out in pairs and $\partial (\partial c) = 0$.

I can't see how to pair up the terms exactly: If $i=n$ for example, the term $(c_{(n,\alpha)})_{(j,\beta)}$ seems to be paired with $(c_{(j+1,\beta)})_{(n,\alpha)}$. But doesn't that third subscript take values only in $\{1,2,\dots,n-1\}$?
Can you please help me understand this?
Thank you!
 A: Let's try a small value of $n$ first to see how it works. The first nontrivial case is $n=2$, so take $c : [0,1]^2 \to A$ a singular $2$-cube. Its boundary is
$$\partial c = \sum_{i=1}^2 \sum_{\alpha \in \{0,1\}} (-1)^{i+\alpha} c_{(i,\alpha)} = -c_{(1,0)} + c_{(1,1)} + c_{(2,0)} - c_{(2,1)}.$$
If $u : [0,1] \to A$ is a singular $1$-cube, its boundary is:
$$\partial u = -u_{(1,0)} + u_{(1,1)} = -u(0) + u(1)$$
And so we get that:
$$\begin{align}
\partial^2(c) & = - \bigl(-(c_{(1,0)})_{(1,0)} + (c_{(1,0)})_{(1,1)}\bigr) + \bigl(-(c_{(1,1)})_{(1,0)} + (c_{(1,1)})_{(1,1)}\bigr) \\
& \mathrel{\hphantom{=}} + \bigl(-(c_{(2,0)})_{(1,0)} + (c_{(2,0)})_{(1,1)}\bigr) - \bigl(-(c_{(2,1)})_{(1,0)} + (c_{(2,1)})_{(1,1)}\bigr)
\end{align}$$
Now how can we pair this?


*

*The terms of the form $(c_{(2,0)})_{(1,i)}$ are those that look like $c(i,0)$, and so will be paired with the terms of the form $(c_{(1,i)})_{(1,0)}$. But the sign of $(c_{(2,0)})_{(1,i)}$ is $(-1)^{2+0+1+i} = -(-1)^i$, while the sign of $(c_{(1,i)})_{(1,0)}$ is $(-1)^{1+i+1+0} = (-1)^i$, so they are opposite and cancel.

*Similarly, the terms of the form $(c_{(2,1)})_{(1,i)}$ are paired with those of the form $(c_{(1,i)})_{(1,1)}$, and the signs cancel again.



So hopefully we now see how it works. What happened above is that in $(c_{(i,\alpha)})_{(j,\beta)}$, if $i > j$, then we first "fix" the $i$th coordinate in $c(t_1, \dots, t_n)$ and then the $j$th; but if $i \le j$, when we fix the the $i$th coordinate to get $c_{(i,\alpha)}$, the new "$j$th" coordinate in $(c_{(i,\alpha)})_{(j,\beta)}$ is really the $(j+1)$st coordinate of $c$! Let's try to formalize that.
We have the set of indices:
$$J_n = \{(i,\alpha,j,\beta) | 1 \le i \le n, \alpha \in \{0,1\}, 1 \le j \le n-1, \beta \in \{0,1\} \},$$
and $\partial^2(c) = \sum_{(i,\alpha,j,\beta) \in J_n} (-1)^{i+\alpha+j+\beta} (c_{(i,\alpha)})_{(j,\beta)}$. We separate $J_n$ in two parts: the set $J_n''$ of indices $(i,\alpha,j,\beta)$ such that $i > j$, and the set $J_n'$ such that $i \le j$. Then there is a bijection between $J_n''$ and $J_n'$ given by:
$$\begin{align}
\theta : J_n' & \to J_n'' \\
(i,\alpha,j,\beta) & \mapsto (j+1,\beta,i,\alpha)
\end{align}$$
Of course, the bijection isn't chosen at random. If $(i,\alpha,j,\beta) \in J_n'$, ie $i \le j$, then:
$$(c_{(i,\alpha)})_{(j,\beta)}(t_1, \dots, t_{n-2}) = c(t_1, \dots, \underbrace{\alpha}_{i\text{th position}}, \dots, \underbrace{\beta}_{(j+1)\text{st position}}, \dots, t_{n-2}),$$
and you see that it is equal to $(c_{(j+1,\beta)})_{(i,\alpha)}$. (At this point I should advise you to check what I've written, if you're not comfortable with all this. Try to see how it works for $n=2$ above, and maybe try $n=3$ if you're courageous. No amount of explanation can replace writing it out for yourself.)
But the signs in front of each are different because of the $+1$ in $j+1$, and so when you pair $(i,\alpha,j,\beta)$ with $\theta(i,\alpha,j,\beta)$ they cancel each other and in the end the sum is zero.

PS: If you're interested in this, this is essentially the combination of the proof that the singular cubes of a space form a (semi-)simplicial abelian group (I didn't mention degeneracies), and that the differential of the Moore complex of a simplicial abelian group has square zero.
