$k$ is a field. Let $X$ be a connected pointed $CW$-complex such that the homology $H_{n}(X;k)$ is a finite dimensional $k$-vector space for any $n\in \mathbb{N}$. Suppose that we have continuous mapping $$r: X\rightarrow K(\pi_{1}(X),1) $$ from the space $X$ to the Eilenberg-MacLane space $K(\pi_{1}(X),1)$ inducing an isomorphism on fundamental group $\pi_{1}$. Suppose also that there is a mapping $$ i: K(\pi_{1}(X),1) \rightarrow X $$ such that:

  1. $r\circ i= id$
  2. the homotopy cofiber of $i$ is homotopy equivalent to a finite $CW$-complex.
  3. and $H_{n}(r;k): H_{n}(X;k)\rightarrow H_{n}(K(\pi_{1}(X),1);k)$ is an isomorphism for any $n\in \mathbb{N}$.

My question is the following: Is the homology of $\tilde{X}$ (the universal covering of $X$) finite dimensional? i.e. $H_{n}(\tilde{X};k)$ is a finite dimensional $k$-vector space for any $n\in \mathbb{N}$?

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    $\begingroup$ What about $S^2 \vee S^1 \to S^1$? Am I missing something? $\endgroup$ Jun 11, 2019 at 20:26
  • $\begingroup$ @NajibIdrissi Sorry, I forgot the most important condition (3). $\endgroup$
    – GSM
    Jun 11, 2019 at 20:30

2 Answers 2


If we replace the field $k$ with the ring of integers $\Bbb Z$, then no.

There are non-trivial high dimensional knots $K: S^n \to S^{n+2}$, whose complements $X = S^{n+2}-K(S^n)$ have $\pi_1(X) \cong \Bbb Z$ ($n > 2$). The map $$ X \to S^1 $$ defining the generator of $H^1(X) \cong \Bbb Z$ has a section, and the mapping cone of this map has the homotopy type of a finite complex.

The homotopy fiber of this map is $\tilde X$, the universal abelian cover. It has the homotopy type of a finite complex iff $K$ is a fibered knot (meaning that there is a representative of the map which is a fiber bundle having compact manifold fibers which are Seifert surfaces of $K$; this is a result of Browder and Levine).

Since $\tilde X$ is $1$-connected, it has the homotopy type of a finite complex iff its homology is finitely generated over $\Bbb Z$ in each degree (this is an easy case of a result due to Wall). Since there are non-fibered knots with $\pi_1(X) \cong \Bbb Z$, any such knot will have the property that the homology of $\tilde X$ in some degree will fail to be finitely generated over $\Bbb Z$.


(1) It could very well be that these examples work over any field $k$, but I do cannot seem to deduce that statement.

(2) Observe what we really constructed is an example satisfying your criteria such that $\tilde X$ fails to have the homotopy type of a finite complex.

Addendum (June 12, 2019): Danny Ruberman points out that Milnor settled Remark (1) in the negative. In other words, there are no examples satisfying the original question having $\pi_1(X) \cong \Bbb Z$.

  • $\begingroup$ In this case $X\rightarrow S^1$ is a homology isomorphism ? $\endgroup$
    – GSM
    Jun 11, 2019 at 22:36
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    $\begingroup$ For any knot the map $X\to S^1$ is a homology isomorphism by Alexander duality. $\endgroup$
    – John Klein
    Jun 11, 2019 at 23:51
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    $\begingroup$ Milnor (Infinite Cyclic Coverings, Assertion 5) shows that with field coefficients, the homology of the infinite cyclic covering of any knot complement is finitely generated. The proof is a nice application of the Milnor exact sequence in homology derived from the SES $0 \to C_{*}(\tilde{X}) \overset{(t_*-1)}{\to} C_{*}(\tilde{X}) \overset{p_*}{\to} C_{*}(X) \to 0$. $\endgroup$ Jun 12, 2019 at 0:59
  • $\begingroup$ @DannyRuberman Yes, I had forgotten that. So these examples are not finitely generated over $\Bbb Z$ in some degree but are so over any field, implying that there is torsion at in infinite number of primes in that degree. $\endgroup$
    – John Klein
    Jun 12, 2019 at 2:06
  • $\begingroup$ @DannyRuberman I misphrased my previous comment. The statement about primes was rubbish. I meant to write: 1) The universal cover, which is 1-connected, is finite dimensional. 2) The total homology with Z coefficients is not finitely generated. 3) The total homology with any field coefficients is finitely generated. so we could get an example wit, say, a copy of $\Bbb Q$ in its homology and 2 and 3 will hold. $\endgroup$
    – John Klein
    Jun 14, 2019 at 17:49

Take $Y=S^1\times S^2$. Take a map from a sphere $S^2\to Y$ given by one of the fibers $*\times S^2$ and glue on a disk: $X=S^1\times S^2\cup_{S^2} D^3$. Then $H_•(X)=H_•(S^1)$. There is a section from $S^1$ inducing a homology isomorphism, so the cofiber is contractible. For an alternate description, think of gluing as $X=S^1\times S^2\cup_{S^1\vee S^2}S^1\vee D^3$. From this point of view, the universal cover is seen to be a pushout of universal covers: $$\tilde X =\widetilde{S^1\times S^2}\bigcup_{\widetilde{S^1\vee S^2}}\widetilde{S^1\times D^3} =S^2\times\mathbb R^1\bigcup_{\widetilde{S^1\vee S^2}}\mathbb R^1\times D^3$$ The three constituents are homotopy equivalent to $S^2$, $\bigvee_\infty S^2$, and a point, so we can compute the homology with Mayer-Vietoris and it is infinite rank, all in degree $3$.

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    $\begingroup$ The cofiber of the map $\ast \times S^2 \to S^1 \times S^2$ is $S^1 \wedge (S^2_+) \simeq S^1 \vee S^3$ (which is still a finite complex). So $H_\bullet(X) \neq H_\bullet(S^1)$. Moreover it's clear from this that $\tilde X$ is homotopy equivalent to countable wedge of copies of $S^3$. But your example fails to satisfy condition (3). $\endgroup$
    – John Klein
    Jun 13, 2019 at 2:22
  • $\begingroup$ Oh, yeah. But it would work if the fundamental group were acyclic... $\endgroup$ Jun 13, 2019 at 22:11
  • $\begingroup$ @BenWieland How? could you say more, please ? $\endgroup$
    – GSM
    Jun 14, 2019 at 10:15

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