# Cobordism/bordism group based on orbifolds with corners

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 $C_{n-1}$ 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 $X$?

• Why does the boundary square to zero if one allows corners? You probably want to add compactness somewhere as well, otherwise every orbifold bords – Thomas Rot Apr 27 '17 at 9:08
• Does "oriented" condition avoid the issue you mentioned?Why does not it square zero, can you give an example? – Hao Yu Apr 27 '17 at 9:43
• 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 $x$ – Hao Yu Apr 27 '17 at 9:48
• 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. – Thomas Rot Apr 27 '17 at 13:19
• Also interesting to note is that a point is nullbordant if one admits orbifolds. – Thomas Rot Apr 27 '17 at 13:21