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.



singular homology of a differential manifold
This depends on what exactly is a smooth mapping from a simplex to the manifold. The standard definition is that the mapping of a non-open subset $X$ of $\mathbf{R}^n$ to a manifold is smooth iff it can be extended to a smooth mapping of an open neighborhood of $X$. With this definition the comparison theorem is true and a very detailed proof can be found e.g. in the book "Introduction to smooth manifolds" by Lee, chapter 16.


