Vtyjkyuk

This refers to Spivak, Calculus on Manifolds pg. 98. Here he defines the (i,0)-face of $I^n$ as follows:

For $xin [0,1]^n-1$,
$$I^n_(i,0)(x)=I^n(x^1,...,x^i-1,0,x^i,...,x^n-1)=(x^1,...,x^i-1,0,x^i,...,x^n-1)$$

also the (i,1)-face is:

$$I^n_(i,1)(x)=I^n(x^1,...,x^i-1,1,x^i,...,x^n-1)=(x^1,...,x^i-1,1,x^i,...,x^n-1)$$

My first question is how this definition applies to 2-cubes, that is, when n=2.

I was thinking we have

For $xin [0,1]$ and $1leq i leq 2$

$$I^2_(i,0)=(0,x)$$

and

$$I^2_(i,1)=(1,x)$$

because regardless of which i, we pick, the fact that we only take some x in the interval $[0,1]$, we will have the same variable x leading the 0 or 1 before it.

My second question is how the boundary is defined when n=2. Here Spivak defines

$$partial c= sum^n_i=1 sum_alpha = 0,1 (-1)^i+alphac_(i,alpha)$$

where $c_(i,alpha)=c circ (I^n_(i,alpha))$

Below, I list some steps I took to trying to solve this:

$$partial c =sum^2_i=1 sum_alpha= 0,1 (-1)^i+alphac_(i,alpha)
=sum^2_i=1 (-1)^i c_(i,0) +sum^2_i=1 (-1)^(i+1) c_(i,1)
= -c_(1,0)+c_(2,0)+c_(1,1)-c_(2,1)$$

From here forth, I am having trouble concluding the rest due to my confusion on the $(i,alpha)$ face for a singular 2-cube...