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Let $X$ be a space and $x_0\in X$ a base point. The Hurewicz map $\pi_k(X,x_0)\longrightarrow H_k(X)$ factors through oriented bordism $\pi_k(X,x_0)\longrightarrow MO_k(X)\longrightarrow H_k(X).$ Ordinary homology $H_k(X)$ can be defined as $MO_k(X)$, but taking oriented simplicial complexes with maximal faces of dimension $k$ instead of manifolds of dimension $k$: Define $S_k(X)=\{{\rm singular~oriented~simplicial~ complexes~in~}X~{\rm with~ maximal~ faces~ of ~dimension} ~k\}$ and $\partial_k\colon S_k(X)\rightarrow S_{k-1}(X)$ sends a simplicial complex $Y$ to its boundary (the singular sub-simplicial complex of $Y$ generated by those faces of $Y$ that are faces of only one (maximal) face). Then $H_k(X)\cong H_k(S_*(X))$. If we take the subcomplex $SM_*(X)$ generated by those oriented simplicial complexes that are oriented manifolds, then $H_k(SM_*(X))\cong MO_*(X)$.

Say that a homology theory $E_*$ is ''bordism-like'' if 1) $E_k(X) = H_k(SQ_*(X))$, for some subchain complex $SQ_*(X)\subset S_*(X)$ generated by the simplicial complexes that have some property $Q$ and 2) the Hurewicz map to ordinary homology factors throgh $E_*$.

Question: Is there a initial bordism-like homology theory? (might be $MO_*$).

Remarks: 1. I've tried restricting to the subcomplex generated by coproducts of spheres ${\mathbb S^k}$, discs $D^k$ and cylinders ${\mathbb S^{k-1}}\times I$, but it does not satisfy excision. 2. Probably I should be writing PL-bordism instead of $MO_*$.

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    This is my first question here. Should $I$ do something to mark this question as answered or even close it, since it was not a real question?2011-12-20

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(OK, let me try to move my answer from comments.)

Unfortunately, there is no (known reasonable) "(co)bodordism complex" (it's not that important for MO because $MO(X)=H(X;MO(pt))$; but, say, lifting $MU$ to a triangulated functor to complexes would have solved many problems -- but nothing like this is known). The problem is, bordisms are defined as quotient of some semigroup (as "homology of semigroup complex" in a sense), not of a group (one might try to make this semigroup a group by imposing relations like $\bar M=-M$ -- but homology of the resulting complex are not bordism groups).

Anyway, perhaps, "the right" answer to your question is stable homotopy groups aka framed cobordism (any Hurewicz map factors through $\pi^s$ since for any homology theory there is a suspension isomorphism).

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    @user17786 Th$a$t's true ($b$tw, ref.: Burdick, R. O., Conner, P. E., Floyd, E. E., 1968. Chain theories and their derived homology. Proceedings of the American Mathematical Society 19 (5), 1115-1118.), but the devil is in the details: there are different ways to make dream "obtained as homology of chained complexes" into precise statement -- see, for example, Neeman. Stable homotopy as a triangulated functor (Inventiones 109 (1), 17-40) for a kind of opposite answer.2011-12-22
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Also, any complex oriented cohomology theory will have the Hurewicz map factor through $MU_*X$, the "complex bordism" of X.