Usually we define singular homology via the complex of singular chains: $$C_\ast(X)=\bigoplus_{n\geq 0}\mathbb Z\langle\sigma:\Delta^n\to X\rangle$$ where the right hand side denotes the free abelian group on the set of maps $\sigma:\Delta^n\to X$.

Suppose that we want the flexibility to use more general *manifolds with corners* in place of standard simplices. So, we define an alternative chain complex:
$$C_\ast'(X)=\bigoplus_{n\geq 0}\mathbb Z\langle f:M^n\to X\rangle/\sim$$
where the right hand side denotes the free abelian group on the set of *isomorphism classes of* maps $f:M^n\to X$ where $M^n$ is a smooth oriented manifold with corners, modulo the relation that reversing orientation negates the generator (and also restricting to maps $f:M^n\to X$ for which reversing the orientation on $M^n$ gives a non-isomorphic map, just so we don't have a bunch of $\mathbb Z/2$'s floating around). The boundary map is clear: take the sum of all the "faces" of $M^n$ (this is ok, even though some of the faces of $M^n$ may not be embedded in $M^n$).

There is an obvious map $C_\ast(X)\to C_\ast'(X)$. Is it a quasi-isomorphism? My guess would be that it is, but I also found I was unable to prove it after thinking for quite a while. Surprising as it may sound, I don't even know how to, given a cycle in $C_\ast'(X)$, construct the homology class in $H_\ast(X)$ it is "supposed" to represent!

There's an obvious strategy of proof, namely "triangulate the manifolds with corners". However doing this compatibly with the boundary map, even for a specific cycle in $C_\ast'(X)$, seems hard.

I've tagged this "reference request" since it seems that a positive answer to my question is a "folk theorem". For instance, some quick searching revealed a paper http://arxiv.org/abs/math/0509532 which claims in Theorem 5 that (a slightly different version of) $C_\ast'(X)$ calculates singular homology. However all they say for the proof is that manifolds with corners can be triangulated (which, on its own, does not seem to me to be sufficient).

Another related construction in the literature is that of Max Lipyanskiy (http://www.math.columbia.edu/~mlipyan/SingHom.pdf), who shows that if we are only interested in the case $X$ is a *smooth* manifold, then (something very close to) the analogous complex built out of isomorphism classes of *smooth* maps from smooth manifolds with corners to $X$ does indeed calculate singular homology. Unfortunately, his proof of this fact uses smoothness in an essential way, so I do not see how to extend it to the case I am interested in.