TITLE: singular homology of a differential manifold
QUESTION [13 upvotes]: Let $M$ be a differentiable manifold, $\Delta$ the closed simplex $[p_0, p_1,...,p_k]$. A differential singular $k$-simplex $\sigma$ of $M$ is a smooth mapping $\sigma:\Delta \to M$.
And we construct a chain complex in the same way we construct the chain complex of singular homology, we gain its homology group.
My question is why this homology group equals the singular homology group? I have tried finding in many books but there is no answer of this question.

REPLY [5 votes]: ons can find a proof for my quesion in "algebraic topology: a first course" by W. Fulton. The proof bases on the fact that in some simple sample like a ball, these two are identical, and the general case would be implied by applying the Mayer-Vertoris principle (in fact, ones can see a manifold as a set of balls that overlap each other, and that principle allows to compute homology groups by reducing on each ball).