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?).

arefinitely 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$7more comments