Preliminaries: Let $\Sigma$ be a closed manifold, $X$ be a CW complex and $f:\Sigma \to X$ be a map. We say that the pair $(\Sigma,f)$ is null-homologous (over $\mathbb{Z}_2$) if $f_*[\Sigma] = 0 \in H_*(X;\mathbb{Z}_2)$, and we say that $(\Sigma,f)$ is null-bordant if there exists a manifold with boundary $Y$ and a map $g:Y \to X$ such that $\partial Y = \Sigma$ and $g|_\Sigma = f$.

The (non-oriented) bordism group $\Omega_*(X)$ of $X$ is the graded group generated by bordism classes $[\Sigma,f]$ of maps $f:\Sigma \to X$, with addition given by disjoint union. Note that there is a natural map $\Omega_*(X) \to H_*(X;\mathbb{Z}_2)$ given by $[\Sigma,f] \mapsto f_*[\Sigma]$ where $[\Sigma]$ is the $\mathbb{Z}_2$-fundamental class of $\Sigma$.

The Stieffel-Whitney numbers $sw_{\alpha,I}[\Sigma,f] \in \mathbb{Z}_2$ of a bordism class $[\Sigma,f] \in \Omega_n(X)$ are numerical invariants associated to $[\Sigma,f]$. Given a choice of a cohomology class $\alpha \in H^k(X;\mathbb{Z}_2)$ and a finite sequence of numbers $I = (i_j)_1^N$ with $i_j \in \{0,\dots,n\}$ such that $|\alpha| + i_1 + \dots + i_N = n$, the associated Stieffel-Whitney number is given by the following formula: $$ sw_{\alpha,I}[\Sigma,f] = \langle w_{i_1}(\Sigma)\dots w_{i_k}(\Sigma) f^*\alpha,[\Sigma]\rangle $$ Here $w_i(\Sigma) \in H^i(\Sigma;\mathbb{Z}_2)$ denotes the $i$-th Stieffel-Whitney class of $\Sigma$.

Main Part: I am interested in the following result, which is Proposition 17.2 in [1]. It generalizes Thom's characterization of (non-oriented) null-bordant manifolds.

Theorem 1: Suppose that $X$ is a finite CW complex. Then a class $[\Sigma,f]$ is 0 if and only the Stieffel-Whitney numbers $sw_{\alpha,I}[\Sigma,f]$ are zero for all $\alpha$ and $I$.

This theorem yields lots of useful corollaries relating the property of being null-bordant to the property of being null-homologous. For instance, we have the following result.

Corollary 2: Let $X$ be a finite CW complex and let $[\Sigma,f] \in \Omega_n(X)$ be a bordism class such that $w(\Sigma) = 1 \in H^*(\Sigma;\mathbb{Z}_2)$. Here $w(X)$ is the total Stieffel-Whitney class of $\Sigma$. Then $[\Sigma,f] = 0$ if and only if $f_*[\Sigma] = 0$.

My question is about extending these results to infinite CW complexes.

Question: Can the finiteness hypothesis on $X$ in Theorem 1 and/or Corollary 2 be weakened to admit (some) non-finite CW complexes?

Remark On Proving Corollary 2: One possible way to try to prove Corollary 2 for infinite CW complexes is the following.

Suppose that $X$ and $[\Sigma,f]$ were as in Corollary 2, but $X$ can now be infinite. Clearly $[\Sigma,f] = 0$ implies $f_*[\Sigma] = 0$ under no hypotheses at all, so we need to show the other way. By cellular approximation, we can choose a representative $(\Sigma,f)$ of $[\Sigma,f]$ such that $f(\Sigma) \subset A$ with $A \subset X$ a finite sub-complex. If we could choose A so that the map $\subset_*:H_*(A;\mathbb{Z}_2) \to H_*(X;\mathbb{Z}_2)$ were injective, then we would get a map into a finite complex with $f_*[\Sigma] = 0$ and Corollary 2 applies.

I got a bit stuck while pursuing some inductive proof that such an $A$ can be chosen, and now I'm not so convinced either way that you can find such an $A$.

Motivation: This discussion is pertinent to the homology theory of $\infty$-dimensional spaces, e.g. spaces of diffeomorphisms, symplectomorphisms, embeddings etc from a closed manifold $M$ to a closed manifold $N$. In particular, every metrizable Banach manifold is homotopy equivalent to a (possibly infinite) CW complex by a result of Palais, see [2].

When one can show that a sub-manifold $\Sigma$ of such a space $\mathcal{X}$ is non-null-bordant, it would be nice to have criteria to check that $\Sigma$ is actually non-null-homologous.

  • 1
    $\begingroup$ The point of the weak topology is that any map from a compact space factors through a finite subcomplex. So if $f_* \Sigma = 0$ in $X$, pick a map from a pseudomanifold bounding it. It factors through some $A$, and therefore $f_*\Sigma \in H_*(A) = 0$. This is the general argument that homotopy/homology of an infinite complex is the colimit of those of its finite subcomplexes, and works for anything defined by mappings out of compact things. $\endgroup$
    – mme
    Oct 21, 2018 at 5:28
  • $\begingroup$ @MikeMiller Hi Mike. I see, it's a good observation that you don't need a sub-complex $A \subset X$ such that $H_*(A) \to H_*(X)$ is injective to get $f_*[\Sigma] = 0 \in H_*(A)$. That seems to prove Corollary 2, thanks! It would still be nice to know something about the theorem. $\endgroup$ Oct 21, 2018 at 5:39
  • 1
    $\begingroup$ I think the following formal argument works. Your Stiefel-Whitney numbers define a map $\Omega_n(X) \to \left(H^*(X;\Bbb Z/2)^\vee\right)^{p(n)} = \left(H_*(X;\Bbb Z/2)\right)^{p(n)}$. Theorem 1 is that if $X$ is a finite complex, this is injective. If $X$ is an infinite complex, $\Omega_n(X) = \colim_A \Omega_n(A)$, the colimit running over every finite subcomplex $A$, and you still have the map to $\left(H^*(X;\Bbb Z/2)^\vee\right)^{p(n)} = \left(\lim H^*(A;\Bbb Z/2)^\vee\right)^{p(n)} = \left(\colim H_*(A;\Bbb Z/2)\right)^{p(n)}$. Finally, a colimit of injective maps of v.s. is injective. $\endgroup$
    – mme
    Oct 21, 2018 at 9:53
  • 1
    $\begingroup$ there are some unclear parentheses in the last line: $(\lim H^*(A;\Bbb Z/2))^\vee = \text{colim }(H^*(A;\Bbb Z/2))^\vee = \text{colim } H_*(A;\Bbb Z/2)$ $\endgroup$
    – mme
    Oct 21, 2018 at 9:58
  • $\begingroup$ @MikeMiller Maybe that could work. My qualms are: (1) $H^*(X;\mathbb{Z}_2)$ is not finitely dimensional as a $\mathbb{Z}_2$-vectorspace, so $H^*(X;\mathbb{Z}_2)^{\vee} \neq H_*(X;\mathbb{Z}_2)$and (2) it isn't clear to me that the Stieffel-Whitney numbers define a linear map $\Omega_n(X) \to (H^*(X;\mathbb{Z}_2)^\vee)^{p(n)}$ even for finite $X$ because of the dependence of the Stieffel-Whitney numbers non-linearly on the Stieffel-Whitney classes, which depend on your choice of $[\Sigma,f] \in \Omega_n(X)$. I would need to check that both of these things are actually problems though. $\endgroup$ Oct 21, 2018 at 10:11

1 Answer 1


I would like to share a careful proof of the generalized Corollary 2 from my question.

The essential idea is to factor the map $f:Z \to X$ through a map to a finite sub-complex of $X$, like I originally had in mind. However, I use Mike Miller's observation that you only need to know that $f_*[Z]$ is 0 in the homology of the finite sub-complex. Also I use a stratifold as my compact bounding object instead of pseudo-manifolds (Mike suggested the latter in his comment).

Lemma 1: Let $X$ homotopy equivalent to a CW complex, and let $f:Z \to X$ be a continuous map from a closed manifold $Z$ with Stieffel-Whitney class $w(Z) = 1 \in H^*(Z;\mathbb{Z}_2)$. Then $f_*[Z] = 0 \in H_*(X;\mathbb{Z}_2)$ if and only $[Z,f] = 0 \in \Omega_*(X;\mathbb{Z}_2)$.

Proof: ($\Rightarrow$) Suppose that $f_*[Z] = 0 \in H_2(Z;\mathbb{Z}_2)$. Pick a homotopy equivalence $\varphi:X \simeq X'$ with a CW complex $X'$. Such an equivalence induces an isomorphism of unoriented bordism groups $\Omega_*(X;\mathbb{Z}_2) \simeq \Omega_*(X';\mathbb{Z}_2)$, so it suffices to show that the pair $(Z,\varphi \circ f)$ is null-bordant, or equivalently to assume that $X$ is a CW-complex to begin with.

So assume that $X$ is a CW-complex. By Lemma 2, we can find a finite sub-complex $A \subset X$ such that $f(Z) \subset A$ and $f_*[Z] = 0 \in H_*(A;\mathbb{Z}_2)$. By Theorem 17.2 of \cite{conner1964}, $[Z,f] = 0 \in \Omega_*(A;\mathbb{Z}_2)$ if and only if the Stieffel-Whitney numbers $\text{sw}_{\alpha,I}[Z,f]$ are identically $0$. Recall that the Stieffel-Whitney number $\text{sw}_{\alpha,I}[Z,f]$ associated to $[Z,f]$, a cohomology class $\alpha \in H_k(A;\mathbb{Z}_2)$ and a partition $I = (i_1,\dots,i_k)$ of $\text{dim}(Z) - k$ is defined to be: $$ \text{sw}_{\alpha,I}[Z,f] = \langle w_{i_1}(Z)w_{i_2}(Z) \dots w_{i_k}(Z) f^*\alpha,[Z]\rangle \in \mathbb{Z}_2 $$ Here $w_j(Z) \in H^j(Z;\mathbb{Z}_2)$ denotes the $j$-th Stieffel-Whitney class of $Z$. By assumption, $w(Z) = 1$ and so $w_j(Z) = 0$ for all $j \neq 0$. In particular, the only possible non-zero Stieffel-Whitney numbers have $I = (0)$. But we see that: $$ \text{sw}_{\alpha,(0)}[Z,f] = \langle f^*\alpha,[Z]\rangle = \langle \alpha,f_*[Z]\rangle = 0 $$ Therefore, $\text{sw}_{\alpha,I}[Z,f] \equiv 0$ and $[Z,f]$ must be null-bordant.

($\Leftarrow$) This direction is completely obvious, since the map $\Omega_*(X) \to H_*(X;\mathbb{Z}_2)$ given by $[Z,f] \mapsto f_*[Z]$ is well-defined.

Lemma 2: Let $X$ be a CW complex, and let $f:Z \to X$ be a map from a closed manifold $Z$ with $f_*[Z] = 0 \in H_*(X;\mathbb{Z}_2)$. Then there exists a finite sub-complex $A \subset X$ with $f(Z) \subset A$ and $f_*[Z] = 0 \in H_*(A;\mathbb{Z}_2)$.

Proof: A very convenient tool for this is the stratifold homology theory of [1], which we now review briefly.

Given a space $M$, the $n$-th stratifold group $sH_n(M;\mathbb{Z}_2)$ with $\mathbb{Z}_2$-coefficients (see Proposition 4.4 in [1]) is generated by equivalence classes of pairs $(S,g)$ of a compact, regular stratifold $S$ and a continuous map $g:S \to M$. Two pairs $(S_i,g_i)$ for $i \in \{0,1\}$ are equivalent if they are bordant by a $c$-stratifold, i.e. if there is a pair $(T,h)$ of a compact, regular $c$-stratifold and a continuous map $g:T \to M$ such that $(\partial T,h|_{\partial T}) = (S_0 \sqcup S_1,g_0 \sqcup g_1)$ (see Chapter 3 and Section 4.4 of [1]). Given a map $\varphi:M \to N$ of spaces, the pushforward map $\varphi_*:sH(M;\mathbb{Z}_2) \to sH(M;\mathbb{Z}_2) $ on stratifold homology is given (on generators) by $[S,g] \mapsto [S,\varphi \circ g] = \varphi_*[\Sigma,g]$.

Stratifold homology satisfies the Eilenberg-Steenrod axioms (see Chapter 20 of [1]), and thus if $M$ is a CW complex then there is a natural isomorphism $sH_*(M;\mathbb{Z}_2) \simeq H_*(M;\mathbb{Z}_2)$. If $M$ is a closed manifold of dimension $n$, the fundamental class $[M] \in sH_n(M;\mathbb{Z}_2)$ is given by the tautological equivalence class $[M] = [M,\text{Id}]$.

The proof of the lemma is simple with the above machinery in place. Since $f_*[Z] = 0$, the pair $(Z,f)$ must be null-bordant via some compact $c$-stratifold $(Y,g)$. Since $Y$ and its image $g(Y)$ are both compact, we can choose a sub-complex $A \subset X$ such that $g(T) \subset A \subset X$. Then the pair $(Z,f)$ are null-bordant by $(Y,g)$ in $A$ as well, so that $[Z,f] = 0 \in sH_*(A;\mathbb{Z}_2)$ and thus $f_*[Z] = 0 \in H_*(A;\mathbb{Z}_2)$ via the isomorphism $sH_*(A;\mathbb{Z}_2) \simeq H_*(A;\mathbb{Z}_2)$.

  • $\begingroup$ Stratifolds are very nice, and deserve to be better known. I'm glad you wrote this up! I still think that Theorem (1) should hold more generally, but I got stuck in my initial attempts to write down a cleaner proof attempt than the comments above, and got distracted with other things. $\endgroup$
    – mme
    Oct 24, 2018 at 20:09
  • $\begingroup$ No problem, thanks for giving this some thought! I'll accept this answer, but if you come up with something for the general case and post it, I'll accept that one instead :) $\endgroup$ Oct 25, 2018 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.