TITLE: Spaces that are both homotopically and cohomologically finite
QUESTION [18 upvotes]: Is it true that every connected space with
1) just finitely many nontrivial homotopy groups, all finite, 
and 
2) just finitely many nontrivial rational cohomology groups, all finite rank, 
is weakly homotopy equivalent to a point?  
In 1953 Serre proved that any noncontractible simply-connected finite CW-complex has infinitely many nontrivial homotopy groups.  That kills off a lot of possible counterexamples.  
In 1998, Carles Casacuberta wrote:


However, we do not know any example of a finite CW-complex with finitely many nonzero homotopy groups which is not a $K(G, 1)$, and the results of this paper suggest that it is unlikely that there exist any.


I'm interested in my question because the spaces it asks about are the connected spaces whose homotopy cardinality and Euler characteristic are both well-defined.  These concepts are morally 'the same', but it seems the spaces on which they're both defined are in very short supply, unless we stretch the rules of the game and use tricks for calculating divergent alternating products or sums.
For some further discussion of these issues see the comments starting here:
http://golem.ph.utexas.edu/category/2011/05/mbius_inversion_for_categories.html#c038299
and also these slides and references:
http://math.ucr.edu/home/baez/counting/
Edit: Condition 1) was supposed to say our space is "cohomologically finite", while 2) was supposed to say it's "homotopically finite".  It's been pointed out that condition 1) is too weak: spaces like $\mathbb{R}P^\infty = K(\mathbb{Z}/2,1)$ exploit this weakness and serve as easy counterexamples to my question.  They are cohomologically infinite in some sense, but not in a way detected by rational cohomology.
So let me try again.  I can think of two ways:
Fix #1: Is it true that every connected space with
1) just finitely many nontrivial homotopy groups, all finite, 
and 
2) just finitely many nontrivial integral cohomology groups, all finitely generated,
is weakly homotopy equivalent to a point?  
Fix #2:  Is it true that every connected finite CW complex with just finitely many nontrivial homotopy groups, all finite, is homotopy equivalent to a point?

REPLY [7 votes]: Yes to Fix #2.
A space with finitely many homotopy groups, all finite, is in the Bousfield class generated by the Eilenberg-MacLane spaces for those groups in those dimensions. The Sullivan Conjecture---Miller's Theorem---implies that any such space has no nontrivial maps into any finite-dimensional space. Thus if the space is finite (-dimensional) its identity function is homotopic to the constant function. 
There's a lot more to say about this point of view.
The fundamental relation is that the space of pointed maps $\mathrm{map}_*(X,Y)$ is weakly contractible.  Thus $X$ can't 'see' $Y$ at all from a homotopy-theoretical perspective.    
Sullivan Conjecture/Miller's theorem:  If $G$ is a (locally) finite group, then $BG$ cannot see any finite-dimensional space.
Then we have 
Zabrodsky Lemma:  If $F \to E \to B$ is a fibration sequence (with $B$ path-connected) and if $F$ cannot see $Y$, then $B$ can see $Y$ if and only if $E$ can see $Y$.
Now a simple induction shows that for any $n\geq 1$,  $K(G,n)$ cannot see any finite-dimensional space; and then that any space $X$ with finitely many nonzero homotopy groups, all finite, also cannot see finite-dimensional spaces.