Given two smooth manifolds with corners, let's say that a map $f:X\to Y$ is "transversally smooth" if it is smooth in the usual sense and if (in a local sense on $X$) for every open Whitney stratum $S'\subset Y,$ 

 - the preimage $f^{-1}(S')$ is an open stratum $S$ of $X$, 
 - Normal directions to $S$ map to normal directions to $S'$. In other words, the map $X\to Y$ is as close to being transversal near $S$ as it can be given that $S$ maps to $S'.$

In particular, any smooth map whose image lands in an open stratum (possibly of positive codimension) is transversally smooth. 

Obviously, restriction to a submanifold with corners $X'\subset X$ preserves this property. This means that if $\Delta^n$ is the $n$-simplex viewed as a manifold with corners and $f:\Delta^n\to X$ is a transversally smooth map, then restrictions of $f$ to subsimplices are transversally smooth. (In fact, precomposition with degeneracy maps will be transversally smooth as well.)

This means that one can define transversally smooth singular homology $H_*^{trans}(X)$ to be the homology of the chain complex built out of singular transversal maps $\Delta^n\to X$ in the usual way. The resulting complex will be a subcomplex of the full singular chain complex, giving a map $$H_*^{trans}(X) \to H_*(X).$$ My question is whether this map is is an isomorphism. There seem to be no obstructions for this to be true and of course the corresponding statement for manifolds without boundary and smooth chains is known. Is this a known result, or can it be proven easily by some comparison with the ordinary smooth cochain complex?