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.

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    $\begingroup$ It is sufficient to know that every topological manifold has the homotopy type of a CW complex. (This is a theorem, I believe, but I don't know the proof.) This implies that a weak equivalence $X\to Y$ induces bijections of sets of homotopy classes $[M,X]\to [M,Y]$. Isn't there also an issue about excision? For smooth bordism excision uses a transversality argument. $\endgroup$ Mar 5, 2012 at 17:49
  • $\begingroup$ I believe there's a proof that every manifold has the homotopy type of a CW complex here: Milnor, John On spaces having the homotopy type of a CW-complex. Trans. Amer. Math. Soc. 90 1959 272–280. $\endgroup$ Mar 5, 2012 at 19:51
  • $\begingroup$ Adding slightly to Tom's comment, the invariance under weak equivalence of topological bordism (as you define it) should follow from the following two facts: (1) any topological manifold has the homotopy type of a CW-complex, and (2) the inclusion of the boundary of a topological manifold is a cofibration (since it is collared). $\endgroup$ Mar 6, 2012 at 7:13
  • $\begingroup$ @Tom: there is no issue with excision. In the book I mentioned Conner and Floyd prove it without transversality, they use "straightening of angles" which roughly amounts to the observation that if you take the product of two smooth manifolds with boundary, then you can "smooth the corners" so that it becomes a smooth manifold with boundary itself. As to your other remark, it is indeed true that a weak equivalence $X \to Y$ induces a bijection $[M, X] \to [M, Y]$ for any manifold $M$. However the bordism relation identifies more than the homotopy relation... $\endgroup$ Mar 6, 2012 at 11:13
  • $\begingroup$ so I don't see how the axiom follows. Lennart Meier suggested to me that it could be verified by looking at a handle decomposition of $M$, but I don't see the details yet (and it would still pose issues for the topological case). $\endgroup$ Mar 6, 2012 at 11:15

2 Answers 2


This answer is simply to write the details for my comment above. It amounts to doing a little more work with homotopy equivalences, so as to carry out essentially the argument you gave in your comment regarding the smooth case.

Assume that $f:X\to Y$ is a weak equivalence of topological spaces. We want to show that the map induced on topological bordism (as you define it) is a bijection.

Regarding the surjectivity, I will simply repeat the content of Tom Goodwillie's comment. First, we observe that any topological manifold has the homotopy type of a CW-complex (you can find references to proofs by Hanner in Milnor's "On spaces having the homotopy type of a CW-complex", as remarked in the comment of Greg Friedman). It follows that $f$ induces a bijection on homotopy classes of maps $f_\ast :[M,X]\to [M,Y]$. Therefore, any map $g:M\to Y$ is homotopic and thus cobordant to a map which lifts to $X$. This proves surjectivity.

To prove injectivity, we need to know that the inclusion of the boundary of a topological manifold is a closed Hurewicz cofibration, that is a NDR-pair. This holds since the boundary of a topological manifold is collared (see chapter 2 of Ferry's notes). With this fact at hand, we can show that for any topological manifold $M$, the pair $(M,\partial M)$ is homotopy equivalent rel $\partial M$ to a pair $(A,\partial M)$ such that the inclusion $\partial M\to A$ is a Serre cofibration.

Proof of claim: Simply factor the inclusion $\partial M\to M$ as a Serre cofibration followed by a weak equivalence $\partial M\to A\to M$. Since $\partial M$ is a topological manifold, it has the homotopy type of a CW-complex, and it follows that $A$ also has the homotopy type of a CW-complex. Then the map $A\to M$ is a weak equivalence of spaces with the homotopy type of CW-complexes, and thus a homotopy equivalence. The map $A\to M$ is therefore a homotopy equivalence rel $\partial M$, since both maps $\partial M\to A$ and $\partial M \to M$ are Hurewicz cofibrations.

We can now prove injectivity by carrying out an argument similar to the one you wrote in your comment. Given bordism classes $g_0:M_0\to X$, $g_1:M_1\to X$ in $X$ and a cobordism $g:M\to Y$ ($\partial M=M_0\coprod M_1$) between $f\circ g_0$ and $f\circ g_1$, we need to find a cobordism $g':M\to X$ between $g_0$ and $g_1$. For that purpose, we replace $(M,\partial M)$ with $(A,\partial M)$ as described above. Since $\partial M\to A$ is a Serre cofibration, we can find a lift up to homotopy $A\to X$ of the composite $A\to M\to Y$, which furthermore extends the map $(g_0,g_1):\partial M\to X$. Composing with the homotopy equivalence $M\to A$ rel $\partial M$, we get the desired map $g':M\to X$.

  • $\begingroup$ This answer makes everything clear, thank you. If I'm not mistaken the claim you prove in your second to last paragraph could be restated by saying that $\partial M \to M$ is a mixed cofibration in the sense of Cole. I wonder if your last paragraph has some nice interpretation in these terms. $\endgroup$ Mar 8, 2012 at 6:56
  • $\begingroup$ You are correct. I don't know much about mixed model structures, other than what is at the nlab page (ncatlab.org/nlab/show/mixed+model+structure). Nevertheless, a simple relative homotopy lifting argument for the Hurewicz/Strom model structure shows that any Strom cofibration which is homotopy equivalent to a Serre cofibration is also a mixed cofibration (i.e. has the left lifting property with respect to Hurewicz fibrations which are weak equivalences). The argument in the penultimate paragraph of the answer shows any Strom cofibration between spaces of the homotopy type... $\endgroup$ Mar 8, 2012 at 9:02
  • $\begingroup$ ... of CW-complexes is homotopy equivalent to a Serre cofibration, and thus must be a mixed cofibration. In particular, for any topological manifold $M$, the inclusion $\partial M\to M$ is a mixed cofibration. Using this, the last paragraph of the answer can be replaced by a single application of the lifting axioms in the mixed model structure. That is, after factoring $f:X\to Y$ as a trivial Strom cofibration (a strong deformation retract) followed by a Hurewicz fibration --- this factorization actually coincides with the one in the mixed model structure. $\endgroup$ Mar 8, 2012 at 9:27
  • $\begingroup$ I was looking at the section 17.3 on mixed model structures in the book "More concise algebraic topology", and it seems that the mixed cofibrations are precisely the Strom cofibrations $f:A\to B$ which are homotopy equivalent rel $A$ to Serre cofibrations $g:A\to C$ (theorem 17.3.5). In fact, proposition 17.3.4 proves what I stated above: a Strom cofibration between mixed cofibrant objects (i.e. with the homotopy type of a CW-complex) is a mixed cofibration, which was sufficient for our purposes. $\endgroup$ Mar 8, 2012 at 23:31

Thom directly constructed a natural isomorphism $MO_n(X) \rightarrow \pi_n (MO \wedge X_+)$ with no assumptions on $X$. The latter sends weak equivalences to isomorphisms since it is computed as a colimit over homotopy groups.

  • $\begingroup$ This answer seems to be about bordism of smooth manifolds, and not of topological manifolds. Am I misunderstanding something? $\endgroup$ Aug 7, 2020 at 15:31
  • $\begingroup$ @DenisNardin Sorry, interpreted "bordism homology theory" as the standard one, but indeed he specified it later. I believe Thom's proof works similarly replacing $MO$ with $MTop$ because we still have notions of topological transverality, but there is probably difficulty since $MTop$ isn't built out of Thom spaces over manifolds. I imagine this is probably known whether or not this works. $\endgroup$ Aug 7, 2020 at 15:59
  • $\begingroup$ To maybe be more explicit, I think the argument is as follows (where for simplicity we take $X$ to be a point). If we embed our topological manifold into $R^n$ nice enough we have a normal disk bundle. Applying the Pontryagin-Thom construction we get a map from the sphere to $MTop(m)$ for some $m$ depending on the codimension of the embedding. This gives us a map in one direction. The inverse map depends on if we can make a map from the sphere transverse to the infinite dimensional manifold $Top(m)$. This I would have to think about but suspect is possible. $\endgroup$ Aug 7, 2020 at 16:13
  • $\begingroup$ See people.math.harvard.edu/~kupers/notes/toplectures.pdf for the topological statement which gives the topological analog of my answer. $\endgroup$ Aug 18, 2020 at 23:07

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