There is an interesting and important homology theory called *bordism*. Briefly speaking, a *singular manifold* in a space $X$ is a pair $(M, f)$ where $M$ is a closed smooth manifold and $f : M \to X$ is a map. Two singular manifolds of the same dimension $(M, f)$ and $(N, g)$ in $X$ are *bordant* if there is a pair $(W, H)$ where $W$ is a bordism from $M$ to $N$ and $H : W \to X$ is a map restricting to $f$ and $g$ on $M$ and $N$. We define the *bordism group* $MO_n X$ to be the set of bordism classes of all $n$-dimensional singular manifolds in $X$ (this is indeed an abelian group under the addition induced by the coproduct of manifolds). It turns out that this gives rise to a homology theory, i.e. the functors $MO_*$ satisfy the Eilenberg--Steenrod axioms. Strictly speaking, one needs to define the relative bordism groups which are defined using bordism classes of singular compact manifolds with boundary, i.e. maps $(M, \partial M) \to (X, A)$ and there are some issues with manifolds with corners to be resolved. The complete construction is explained very nicely in *Differentiable Periodic Maps* by Conner and Floyd. (Beware that there are a book and a paper by the same authors and with the same title, I mean the book: *Differentiable Periodic Maps*, LNM 738, 1979.) There are also (even more interesting) variants of the bordism theory like $MU$ and $M \mathrm{Spin}$, which arise by considering additional structure on singular manifolds. I hope that a good answer to my question will handle the general case.

A homology theory $h_*$ satisfies the *weak equivalence axiom* if for every weak equivalence of pairs of spaces $f : (X, A) \to (Y, B)$ (i.e. a map such that both $f : X \to Y$ and $f | A : A \to B$ are weak equivalences) the induced map $h_*(X, A) \to h_*(Y, B)$ is an isomorphism. My question is exactly as in the title.

Does the bordism homology theory satisfy the weak equivalence axiom?

An example of a homology theory that satisfies the weak equivalence axiom is singular homology. The way to prove this is as follows. Fix a pair of spaces $(X, A)$ and a natural number $n$. Consider the *Eilenberg subcomplex* $\mathrm{Sing}^{(n, A)} X$ of the singular complex $\mathrm{Sing} X$ which consist in degree $k$ of maps of pairs $(\Delta^k, (\Delta^k)^{(n)}) \to (X, A)$ where $(\Delta^k)^{(n)}$ is the $n$-skeleton of $\Delta^k$. One can prove that if $(X, A)$ is $n$-connected, then the induced inclusion of singular chain complexes $S_\bullet^{(n, A)} X \to S_\bullet X$ is a chain homotopy equivalence and thus the relative groups $H_*(X, A)$ are zero up to degree $n$.

This gave me an idea that maybe in case of bordism the weak equivalence axiom could be verified by choosing a triangulation on a manifold $M$ and considering maps $(M, M^{(n)}) \to (X, A)$ where $M^{(n)}$ is the $n$-skeleton of $M$ with respect to this triangulation. I was unable to find a proof along these lines. However, if this approach has any merit at all, then it means that the question has something to do with existence of triangulations of manifolds and thus the following may be a subtler variant of the question.

Does the topological bordism homology theory (i.e. the one constructed using compact topological manifolds, which do not necessarily admit triangulations) satisfy the weak equivalence axiom?

Some people may feel that the issue is somewhat immaterial since even if a homology theory doesn't satisfy the weak equivalence axiom, then we simply prolong it from CW-complexes to all spaces by means of CW-replacement. However, I feel that if some geometrically defined (co)homology theory satisfies the weak equivalence axiom for some class of spaces larger than CW-complexes, then it is good to know how large exactly this class is. For example for singular homology this class consists of all spaces which makes the theory unexpectedly well-behaved. An example where it fails very badly is topological K-theory. There the geometric definition is wrong even for non-finite CW-complexes and one needs to use the representing spectrum to prolong the theory to all spaces.

`$X \to Y$`

induces a bijection`$[M, X] \to [M, Y]$`

for any manifold $M$. However the bordism relation identifies more than the homotopy relation... $\endgroup$ – Karol Szumiło Mar 6 '12 at 11:13