Is any CW complex with only finitely many nonzero homology groups homotopic to a finite dimensional CW complex?

Question

My question is

Is any CW complex with only finitely many nonzero homology groups homotopic to a finite dimensional CW complex?

(My thoughts on this which might not be useful at all.) Since an infinite dimensional CW complex could have only finitely many nontrivial homology groups($S^\infty$ for example), it seems to me that the relation between the dimension and the zeroness of the CW complexes is not very strong. On the other hand, we know that Moore spaces are unique up to homotopy equivalence and any CW complex with prescribed homology groups can be construct by taking the wedge sum of Moore spaces. If this statement above is true, then it means every CW complex with only finitely many nonzero homology groups is essentially built up in this way...

Please note that I meant finite dimensional CW complex instead of finite CW complex. Otherwise the infinite dimensional discrete space will serve the purpose. Also, I really appreciate it if someone can point it out whether the statement can become true buy adding some small conditions(One of my professors said we need $X$ to be simply connected).

Answer 1

In the simply connected case, the answer is yes.

In the general case, the theory was worked out in complete detail by Wall in the paper:

Wall, C. T. C. Finiteness conditions for CW-complexes. Ann. of Math. (2) 81 1965 56–69.

See "Theorem E."

Here is a description of the result:

In the non-simply connected case, to get that your complex $X$ is equivalent to an $n$-dimensional one, you need to know that Wall's condition $D_n$ is satisfied:

$D_n$: $H_i(\widetilde X) =0$ for $i > n$ and $H^{n+1}(X;\mathscr B) = 0$.

In the above, we $\widetilde{X}$ is the universal covering and $\mathscr B$ denotes any local coefficient system on $X$.

Answer 2

As requested I am writing this as an answer.

No there are spaces with vanished homology which are not homotopy equivalent to finite CW-complexes.

For example if $G$ is an acyclic group, then the classifying space will be a space with vanishing homology. It is not contractible as $\pi_1 BG = G$.

Now a finite CW-complex will always have $\pi_1$ a finitely generated group (in fact finitely presentable). So if $G$ is not finitely generated, then $BG$ is not homotopy equivalent to a finite CW-complex.

There are many examples of large infinitely generated acyclic groups. For example can take the group of bijections of a countably infinite set. This is an uncountably infinite group and so it cannot be finitely generated by size considerations.

[edit: It was pointed out in the comments that I misread the question. I thought the question was asking if the space is homotopy equivalent to a finite CW-comples, when actually it asks for a finite dimensional CW-complex. Nevertheless BG where G is the group of bijections of an infinite set is still a counter example to the asked question, see Mark Grant's answer]

Answer 3

Supplementing Chris's answer, you could take $G$ to be any acyclic group of infinite cohomological dimension. Such groups exist, e.g. binate groups. See

Berrick, A.J., The acyclic group dichotomy., J. Algebra 326, No. 1, 47-58 (2011). ZBL1253.20055.

Then any model of $BG$ will have trivial homology but must be infinite-dimensional. Taking wedge sums or cartesian products of $BG$ with a finite complex will give plenty more counter-examples.

For the (positive) simply-connected statement mentioned by John Klein, search for "minimal CW structures".