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I'll say $K$ has "finitely generated" homotopy groups if there is a finite wedge of spheres $W = \bigvee S^{n_i}$ and a map $f: W\to K$ which induces a surjection on $\pi_*$.

It seems likely that this is equivalent to saying that $\pi_*(K)$ is finitely generated as a $\Pi$-algebra, but I'm not sure of this.

Write $\mathcal{F}$ for the collection of all finite complexes with finitely generated homotopy groups. Clearly every finite wedge of spheres is in $\mathcal{F}$, as are the finite products of spheres and the projective spaces. $\mathcal{F}$ is closed under products.

QUESTION 1: I wonder if it is conceivable that every simply-connected finite complex $K$ has finitely generated homotopy groups in this sense.

EDIT 1: I said $\mathcal{F}$ is closed under wedges earlier, but I don't see why now.

EDIT 2: If the answer to original question is "yes", then it is also true rationally. And since the rational question may be easier (I wouldn't be shocked if the experts know the answer to be "no"), I'm explicitly adding it here

QUESTION 2: If $X$ is the rationalization of a simply-connected finite complex, is there a wedge of rational spheres $W$ and a map $W\to X$ which is surjective on $\pi_*$?

FROM A COMMENT BY BEN WIELAND: Question 2 has the following algebraic reformulation, using the Lie model. If we have a differential graded Lie algebra that is finitely generated as a graded Lie algebra, is its homology finitely generated as a graded Lie algebra? (See this question: Is homology finitely generated as an algebra?).

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    $\begingroup$ This is an obvious comment, but stably a finite complex with finitely generated homotopy groups that was not a wedge of spheres would violate the generating hypothesis, so is very unlikely to exist. So my guess is that there is no finite complex with finitely generated homotopy groups that is not itself a wedge of spheres. $\endgroup$
    – Mark Hovey
    Commented May 29, 2014 at 12:31
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    $\begingroup$ @MarkHovey: Well, projective spaces are finitely generated, as are their products and (I believe) their wedges. I was thinking that Moore spaces would be a good test case; I'll think about your plan. (BTW a proof that $M(p)$ is a counterexample would certainly answer the question.) $\endgroup$
    – Jeff Strom
    Commented May 29, 2014 at 12:56
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    $\begingroup$ Mark, why do you believe the generating hypothesis is true $\endgroup$
    – Peter May
    Commented May 31, 2014 at 0:44
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    $\begingroup$ @PeterMay: well, I don't know about the generating hypothesis but I do believe that there are no finite spectra whose homotopy groups are finitely generated over the stable homotopy ring except for finite wedges of spheres. This would just fit with the general yoga that $\pi_* S$ is maximally bad as a ring. JeffStrom: Of course, you're right. I have a suggestion: look at the spaces for which the unstable Adams-Novikov spectral sequence is well-behaved. I can't remember now what these spaces are called--spherically resolved or something like that. Lie groups, for example (I think). Anyway they $\endgroup$
    – Mark Hovey
    Commented Jun 1, 2014 at 12:20
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    $\begingroup$ When Mark's answer was converted to a comment, the end was chopped off. It was "might have a chance, though I still think the generic answer has to be no." $\endgroup$
    – Andy Putman
    Commented Jun 3, 2014 at 22:00

2 Answers 2

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In Donald Kahn's 1966 paper "On stable homotopy modules" (Inventiones 1, pp 375–379, doi:10.1007/BF01389739) he constructs examples of CW complexes with 3 cells, whose stable homotopy groups are not finitely generated as modules over the ring of stable homotopy groups of spheres. I believe this answers in the negative a stable analogue of the question.

Perhaps a Freudenthal suspension theorem argument, together with the stable result, can imply the desired unstable result. I am sorry but I do not have a few minutes to spare to explore this possibility; perhaps others are interested enough to try it out, though. (I am sorry if this incomplete answer is inappropriate for this site, which I am a beginner to. Please feel free to delete this answer if so. I would certainly have made this all a comment rather than an answer, but don't have enough points on this site to leave comments.)

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This answers a slightly different question. If you do not insist on `simply connected' there are examples to show that this does not happen in general. Fix $k\geq 2$. For any $k$, there is a group $G_k$ that admits an Eilenberg-Mac Lane space $K(G_k,1)$ with finite $k$-skeleton but does not admit any such space with finite $(k+1)$-skeleton.
(The first example for $k=2$ is due to Stallings, the first for higher $k$ to Bieri. In these examples, $G_k$ is the kernel of a homomorphism from the direct product of $(k+1)$ copies of a free group to the infinite cyclic group.)
If $K$ is the finite $k$-skeleton of such an Eilenberg-Mac Lane space, then no map from a finite wedge $W$ of spheres to $K$ can be surjective on $\pi_k$, because $\pi_k(W)$ is a finitely generated module for $\pi_1(W)$, while $\pi_k(K)$ is not a finitely generated module for $\pi_1(K)$.

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