Fix a space $X$, which I want to assume is a manifold. Under the assumption of simple-connectivity, Hurewicz's theorem tells us that $$ \pi_3(X)\to H_3(X,\mathbb{Z})\qquad (*) $$ is surjective, hence that every homology class is represented by a map $S^3\to X$. There is lots of hard work gone into worrying when homology classes are represented by embedded submanifolds, but that is not what I'm interested in here. What I want to know is:

Are there are nontrivial examples of manifolds $X$ with $(*)$ surjective, but with infinite $\pi_1(X)$?

I don't need the case when $\pi_1$ is finite, since I have a different proof that doesn't need this more subtle property.

An equivalent question is asking whether $H_3(\tilde{X},\mathbb{Z})\to H_3(X,\mathbb{Z})$ is surjective, for $\tilde{X}$ the universal covering space. Since $\pi_1(X)$ is nontrivial, then questions about the Eilenberg–Moore spectral sequence become more subtle, but I'm only looking at such a low-dimensional group that maybe things are not so bad (I don't really understand the EM spectral sequence, even relative to my background knowledge of spectral sequences, so I don't quite know how to start extracting information from that).

Added: I was interested to know if there are general hypothesis that allow me to conclude $(\ast)$ is onto, but given a comment below by user51223, I think all I care about is the weaker statement that $$ \pi_3(X)\to H_3(X,\mathbb{Z}) \to H_3(X,\mathbb{Z})/\text{torsion} \qquad (**) $$ is surjective. So:

Are there general conditions that guarantee this weaker statement is true?

  • $\begingroup$ Does "nontrivial" mean $H_3(X)$ is nontrivial? If so, sure. Take $S^1 \vee S^3$, you can thicken that up to a manifold with boundary or "thicken" it to a closed manifold, provided it's suitably high-dimensional. $\endgroup$ Apr 9, 2021 at 4:38
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    $\begingroup$ Could you give a reference for your version of Hurewicz's theorem? For me it says that $\pi _2(X)\cong H_2(X,\mathbb{Z})$, but nothing about the next degree. $\endgroup$
    – abx
    Apr 9, 2021 at 4:45
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    $\begingroup$ @theHigherGeometer: regarding your added question, there are at least lots of examples. In fact, Kervaire showed that any finitely-presented group can be realized as the fundamental group of a closed, oriented smooth 4-manifold. So, for any finitely-presented, infinite group $G$ with finite abelianization there is a closed, oriented $4$-manifold $X$ with $\pi_1(X)=G$. For such a manifold, $H_1(X, \, \mathbb{Z})$ is a torsion group, and so the same is true for $H^3(X, \, \mathbb{Z})$. $\endgroup$ Apr 9, 2021 at 6:49
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    $\begingroup$ @FrancescoPolizzi oh, good point! That's a lovely class of examples. $\endgroup$
    – David Roberts
    Apr 9, 2021 at 6:53
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    $\begingroup$ Also, by the argument of my answer. If $X$ is any space with $H: \pi_{3}(X) \rightarrow H_{3}(X,\mathbb{Z})$ surjective then $X \times S^1$ also satisfies this condition. So combining with Francesco Polizzi's answer you can get $\pi_{1}(X)$ to be almost anything whilst making $b_{3}(X)$ arbitrarily large. So there will be almost no restriction on the algebraic invariants of $X$ $\endgroup$
    – Nick L
    Apr 9, 2021 at 6:57

3 Answers 3


Take $M = S^1 \times S^3$ . Then $\tilde{M} = \mathbb{R} \times S^3$ and clearly $\pi_{*} : H_{3}(\tilde{M},\mathbb{Z}) \rightarrow H_{3}(M,\mathbb{Z})$ is surjective, since it maps the $3$-cycle $\{x\} \times S^3$ to the generator ${[x]} \times S^3$.


For closed 3-manifolds, this holds iff it is the 3-sphere. If the fundamental group of a closed 3-manifold $X$ is infinite, then $H_3({\tilde X})$ is trivial. If $\pi_1(X)$ is finite, then $\tilde{X}\cong S^3$, and $H_3(\tilde{X})\cong \pi_3(\tilde{X})$. So the map $\pi_3(X) \to H_3(X)$ will have range a subgroup of index $|\pi_1(X)|$.

For a closed 4-manifold $X$, $H_3({\tilde X})$ will be torsion-free. If it’s non-trivial, then $\pi_1 X$ will have more than one end, and hence by Stallings’ theorem $\pi_1 X$ will split over a finite group. Take a maximal splitting of $\pi_1 X$ as a graph of groups with finite edge groups. Then if the vertex groups have $H^1( ;\mathbb{Q})=0$, then your condition (*) should hold with rational coefficients. In fact I think this is also necessary, but one might have to analyze Stallings’ proof to show this. Equivalently, the map from the manifold to the graph defining the graph of groups should induce a surjection on $H^1( ; \mathbb{Z})$ from the graph to the manifold. As in the 3-manifold case, I’m not sure that this can hold with integral coefficients when the edge groups of the graph of groups are non-trivial finite groups (in this case one might only hit a finite-index subgroup of $H_3(X)$). So I’m guessing that a necessary and sufficient condition is that the manifold is a connect sum of a manifold $X$ with $H_3(X)=0$ and $\#^k (S^1\times S^3)$ (or the twisted version $S^1\tilde{\times} S^3$ if non-orientable manifolds are included).


Let $X$ be a fake projective plane. It is a quotient of the unit ball in $\mathbb{C}^2$ by a co-compact subgroup of automorphisms, hence $\pi_1(X)$ is infinite and non-abelian.

By Poincaré duality, we have $H^4(X, \mathbb{Z})=H_0(X, \, \mathbb{Z}) \simeq \mathbb{Z}$ and so, by the Universal Coefficient Theorem, the torsion part of $H_3(X, \, \mathbb{Z})$ vanishes.

Since a fake projective plane has the same Betti numbers of $\mathbb{P}^2(\mathbb{C})$, it follows $H_3(X, \, \mathbb{Z})=0$ and so your Hurewicz map $(\ast)$ is automatically surjective.

Comment on the added question. There are at least lots of examples. In fact, Kervaire showed that any finitely-presented group can be realized as the fundamental group of a closed, oriented smooth $4$-manifold. So, for any finitely-presented, infinite group $G$ with finite abelianization, there is a closed, oriented $4$-manifold $𝑋$ with $\pi_1(X)=G$. For such a manifold, $H_1(𝑋, \, \mathbb{Z})$ is a torsion group, so the same is true for $H_3(X, \, \mathbb{Z})$ and subsequently $(**)$ is automatically surjective.

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    $\begingroup$ Heh, ironically, that also rules out the interesting behavior I was hoping to see! But thanks for the example. $\endgroup$
    – David Roberts
    Apr 9, 2021 at 6:12

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