Why integrate on cubes that's not injective?

Question

Again, this is a conceptual(soft) problem I had while reading Spivak's calculus on manifold. There, to develop the theory of integration, Spivak chose to integrate k-forms on singular cubes. However, as pointed out here, singular cubes can collapse dimension, so it seems theoretically possible that one pulls back a differential form onto a higher dimension cubical region for evaluating the integral, and I wonder when could that ever be useful? Why don't we just add the assumption that singular cubes are injective? Also, without this injectivity I find the picture of chains even less geometrically intuitive...

Answer 1

For the purpose of Spivak's book, injective cubes will suffice (in few cases, one would need to subdivide cubes to achieve injectivity, say, when you parameterize circle by an interval). However, if you want to move further, restricting to injective cubes will leads to unnecessary complications. While I cannot read Spivak's mind, it is not unreasonable to suppose that he wanted to write a follow-up to his book, connecting differential forms to algebraic topology (he never wrote such a book, but Bott and Tu did, it is a very nice book called "Differential forms in algebraic topology"). In singular homology theory one should not restrict to injective maps of cubes/simplices (in cubical singular homology theory, "degenerate cubes," i.e. the ones which lower the dimension, play critical role), so in order to pair forms and singular chains via integration, it is natural to integrate over arbitrary singular cubes, including degenerate ones.