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In Chapter 4's section on Geometric Preliminaries, I am confused by the definition of "boundary."

The standard $n$-cube $I^n$ is defined to be $I^n(x^1,...,x^n) = (x^1,...,x^n)$, and two associated $n-1$-cubes are defined as $$I^n_{(i,0)}(x^1,...,x^{n-1}) = I^n(x^1,...,x^{i-1},0,x^i,...,x^{n-1})$$ $$I^n_{(i,1)}(x^1,...,x^{n-1}) = I^n(x^1,...,x^{i-1},1,x^i,...,x^{n-1})$$

With these definitions, the boundary of the standard $n$-cube is defined as: $$\partial I^n = \sum_{i=1}^n \sum_{a=0,1} (-1)^{i+a} I^n_{(i,a)}$$

I'm having trouble interpreting these definitions in the case of $n=2$. The image of $I^2$ in $R^2$ is just the unit square with endpoints at $(0,0)$ and $(1,1)$. The image of $\partial I^2$ is thus the outline of the square (as depicted in the book). However, when I work out the definition of boundary given above, I get:

\begin{align*} \partial I^2(x) &= \sum_{i=1}^2 \sum_{a=0,1} (-1)^{i+a} I^2_{(i,a)}(x)\\ &= (-1)^1 I^2_{(1,0)}(x) + (-1)^2 I^2_{(1,1)}(x) + (-1)^2 I^2_{(2,0)}(x) + (-1)^3 I^2_{(2,1)}(x)\\ &= -(0,x) + (1,x) + (x,0) - (x,1)\\ &= (1,-1) \end{align*}

Spivak's provided equation suggests that the image of $\partial I^2$ is a single point, which doesn't make any sense.

Does this equation contain a typo, and if not, where am I going wrong in my interpretation of it?

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    $\begingroup$ For that matter, what does the notation $I^n(x^1,...,x^n) = (x^1,...,x^n)$ mean? Surely the RHS doesn't simply mean a single point. Does this mean the collection of all such points where the coordinates vary with $0\leq x^i \leq 1$ for each $i$? And what does the $\sum$ notation mean? It can't mean coordinate wise addition as you suggest. $\endgroup$
    – MPW
    Mar 6 '20 at 21:26
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    $\begingroup$ You don't do the algebraic sum of the functions. You take the singular chain, whose image is the union of the respective images. $\endgroup$ Mar 6 '20 at 23:09
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It's indeed not the best choice of notation.

Geometrically it's clear what should happen (modulo the alternating sum): The boundary of an $n$ dimensional box is the 'collection' of its faces, each of which is an ($n$-$1$)-cube.

Now, instead of taking their collection we take their (alternating) formal sum (which can be rigorously achieved by introducing a free Abelian group generated by the set of all $n$-cubes).

The notation $I^n(x^1,\dots,x^n)=(x^1,\dots, x^n)$ defines the embedding of the standard $n$-cube $[0,1]^n$ into $\Bbb R^n$.

Then $I^n_{i,0}$ is the embedding $[0,1]^{n-1}\to\Bbb R^n$ that chooses the face on the hyperplane $x_i=0$.

Note also that the alternating sum is in accordance with the orientation of the (embeddings of the) faces.


Specifically for $n=2$, if we denote the (signed) segments as $\def\segment#1#2#3#4{\big[(#1,#2)\multimap(#3,#4)\big]} \segment{x_1}{y_1}{x_2}{y_2}$ we get: $$ \partial I^2\ =\ -I^2_{(1,0)}+I^2_{(1,1)}+I^2_{(2,0)}-I^2_{(2,1)}\ =\\ -\segment0001 + \segment1011 + \segment0010 -\segment0111 \\ =\ \segment0010 + \segment1011 + \segment1101 + \segment0100\,. $$

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