We define a geometric homology group of a topological space $X$ as follows: the chain complex $C_{\bullet}$is freely generated by the maps $f$ from a compact oriented orbifold with corners $P$ to $X$, graded by the dimension of $P$, the boundary map from $C_n$ to $C_{n1}$ is defined to be the restriction of $f$ to the (oriented)sum of its codimension $\ge$ 1 faces. If $X$ is itself a compact oriented orbifold with corners, is it true that the homology $H_{\bullet}(C, Q)$ over $Q$ is isomorphic to the rational singular homology group of $X$?
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1$\begingroup$ Why does the boundary square to zero if one allows corners? You probably want to add compactness somewhere as well, otherwise every orbifold bords $\endgroup$ – Thomas Rot Apr 27 '17 at 9:08

$\begingroup$ Does "oriented" condition avoid the issue you mentioned?Why does not it square zero, can you give an example? $\endgroup$ – Hao Yu Apr 27 '17 at 9:43

$\begingroup$ The orbifold with corners is well formed,similar to the simplexes. Any point $x$ lies in exactly $n$ faces, where $n$ is the codimension of $x$, where "codimension" is the number of zero coordinates in any chart containing $x$ $\endgroup$ – Hao Yu Apr 27 '17 at 9:48

$\begingroup$ The problem i was thinking of is that the boundary of a manifold with corners is not a manifold with corners. This is already not true for the model of the positive quadrant in $R^n$. Or think of a space homeomorphic to a disc, but with one corner. I don't see how orbifolds solve this. $\endgroup$ – Thomas Rot Apr 27 '17 at 13:19

$\begingroup$ Also interesting to note is that a point is nullbordant if one admits orbifolds. $\endgroup$ – Thomas Rot Apr 27 '17 at 13:21
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This seems closely related to BaasSullivan theory of bordism with singularities. Look up Nils Baas's 1973 Math. Scand. paper on MathSciNet, and then look at the various papers that cite this (including very recent ones).