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

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