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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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Probably my notation is confusing: I'm denoting by $S_k(pt)$ the set of simplicial complexes all of whose maximal faces are of dimension $k$. It is \emph{not} the singular complex on a point, which as you say, is $1$-dimensional. It is the set of finite simplicial complexes (in $\mathbb R^\infty$, for that to be a set), each of them endowed with the unique map to a point (this is a ''singular simplicial complex'' on the space $\{pt\}$) It is much larger than the singular chain complex but has the same homology. – user17786 Dec 20 '11 at 17:39
In short, what I denote by $S_*(X)$ is like the bordism complex, but instead of taking maps from a oriented manifold of dimension * to $X$, I take maps from finite oriented simplicial complexes of "pure dimension" * to $X$. – user17786 Dec 20 '11 at 17:44
I see. But the problem is, there is no "bordism complex" -- bordisms are not defined as homology of some complex of abelian groups (well, MO(X)=H(X;MO(pt)), so it can be resolved, but, say, for MU there is no known way to construct a reasonable complex). – Grigory M Dec 20 '11 at 18:19
Thanks! I got it. The bordism group $MO_*(X)$ has as underlying set the quotient of set of singular manifolds in $X$ by the relation given by singular bordisms in $X$. This quotient has a group structure, but the set of all singular manifolds does not have it (the group structure I put in the set of all singular manifolds was supposed to be "disjoint union", but clearly it is only defined up to bordism). The same applies to my description of ordinary homology. – user17786 Dec 20 '11 at 20:38
My idea was "homotopy groups are like pointed spheres up to pointed cylinders. If I forget the basepoint, how many generators and relators do I have to add to get a homology theory?" Bordism was an example, as well as ordinary homology, as I described it. – user17786 Dec 20 '11 at 20:39
up vote 3 down vote accepted

(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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For completeness, the first comment here says that the only cohomology theory that can be obtained as homology of chain complexes is the ordinary one. – user17786 Dec 22 '11 at 9:44
@user17786 That's true (btw, 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. – Grigory M Dec 22 '11 at 10:15

Also, any complex oriented cohomology theory will have the Hurewicz map factor through $MU_*X$, the "complex bordism" of X.

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