Why does a space with finite homotopy groups [for every n] have finite homology groups? How can I proof this [not only for connected spaces with trivial fundamental group]? The converse is false. $\mathbb{R}P^2$ is a counterexample.

Do finitely generated homotopy groups imply finitely generated homology groups? I can proof this only for connected spaces with trivial fundamental group. The converse is false. $S^1\vee S^2$ is a counterexample.


3 Answers 3


(This answer has been edited to give more details.)

Finitely generated homotopy groups do not imply finitely generated homology groups. Stallings gave an example of a finitely presented group $G$ such that $H_3(G;Z)$ is not finitely generated. A $K(G,1)$ space then has finitely generated homotopy groups but not finitely generated homology groups. Stallings' paper appeared in Amer. J. Math 83 (1963), 541-543. Footnote in small print: Stallings' example can also be found in my algebraic topology book, pp.423-426, as part of a more general family of examples due to Bestvina and Brady.

As Stallings noted, it follows that any finite complex $K$ with $\pi_1(K)=G$ has $\pi_2(K)$ nonfinitely generated, even as a module over $\pi_1(K)$. This is in contrast to the example of $S^1 \vee S^2$.

Finite homotopy groups do imply finite homology groups, however. In the simply-connected case this is a consequence of Serre's mod C theory, but for the nonsimply-connected case I don't know a reference in the literature. I asked about this on Don Davis' algebraic topology listserv in 2001 and got answers from Bill Browder and Tom Goodwillie. Here's the link to their answers:


The argument goes as follows. First consider the special case that the given space $X$ is $BG$ for a finite group $G$. The standard model for $BG$ has finite skeleta when $G$ is finite so the homology is finitely generated. A standard transfer argument using the contractible universal cover shows that the homology is annihilated by $|G|$, so it must then be finite.

For a general $X$ with finite homotopy groups one uses the fibration $E \to X \to BG$ where $G=\pi_1(X)$ and $E$ is the universal cover of $X$. The Serre spectral sequence for this fibration has $E^2_{pq}=H_p(BG;H_q(E))$ where the coefficients may be twisted, so a little care is needed. From the simply-connected case we know that $H_q(E)$ is finite for $q>0$. Since $BG$ has finite skeleta this implies $E_{pq}^2$ is finite for $q>0$, even with twisted coefficients. To see this one could for example go back to the $E^1$ page where $E_{pq}^1=C_p(BG;H_q(E))$, the cellular chain group, a finite abelian group when $q>0$, which implies finiteness of $E_{pq}^2$ for $q>0$. When $q=0$ we have $E_{p0}^2=H_p(BG;Z)$ with untwisted coefficients, so this is finite for $p>0$ by the earlier special case. Now we have $E_{pq}^2$ finite for $p+q>0$, so the same must be true for $E^\infty$ and hence $H_n(X)$ is finite for $n>0$.

Sorry for the length of this answer and for the multiple edits, but it seemed worthwhile to get this argument on record.


It sounds like maybe you can prove your first statement for simply connected spaces. In that case, you can use the homotopy orbit spectral sequence (the Serre spectral sequence associated to the fibration $\tilde X \to X \to BG$ where $\tilde X$ is the universal cover of $X$ and $G = \pi_1(X)$). The only terms in the $E^2$ page of the spectral sequence which have a chance of being non-finite are the $H_i(BG, \mathbb{Z})$. But these are finitely generated, and for $i > 0$ rationally are zero by a transfer argument. So $H_i X$ is finite for $i > 0$.


Using transfer map the question "why finite homotopy groups implies finite homology groups" can be reduced to the simply-connected case, where Serre's mod finite groups theory applies.

Recall that for a finitely-sheeted covering map $p:\tilde X\to X$ the homology transfer $t: H_*(X)\to H_*(\tilde X)$ is the homomorphism induced by associating to an oriented cell in $X$ the sum of its (finitely many) lifts in $\tilde X$. So the composition $p_* t$ is the self-map of $H_*(X)$ that simply multiplies by the order of the cover; this map is an isomorphism when we use rational coefficients. Thus the tranfer is injective on rational homology.

In particular, if $X$ is a space with finite homotopy groups and some rational homology group nonzero, then the same is true for its universal (finite) cover.

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    $\begingroup$ To go from trivial rational homology to finite integral homology one needs to know the integral homology groups are finitely generated. $\endgroup$ Dec 21, 2009 at 4:07
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    $\begingroup$ Yes, I was implicitly assuming that $X$ is a finite cell complex. $\endgroup$ Dec 21, 2009 at 5:18

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