No. If X × Y has the homotopy type of a finite-dimensional CW complex, then X and Y each do as well.

Here is a clean argument.

• Let Z ≃ X × Y be homotopy equivalent to an m-dimensional CW complex. Then for every local coefficient system L on Z we have H^i(Z; L) = 0 for i > m, since cohomology with local coefficients is computed by a length-m cochain complex.

• Fix any local coefficient system A on X. Consider the pullback local system p_X^*A on X × Y (where p_X: X × Y → X is the projection). There is a natural cohomological cross product with the unit in H^0(Y; Z): × 1: H^k(X; A) → H^k(X × Y; p_X^A), characterized by the property that for the inclusion i: X ≅ X × {y0} → X × Y we have i^(u × 1) = u. Hence ×1 is injective.

• Since H^i(X × Y; p_X^*A) = 0 for i > m, the injectivity above forces H^i(X; A) = 0 for all i > m and all local systems A on X. Similarly, H^i(Y; B) = 0 for all i > m and all local systems B on Y.

• A standard obstruction-theoretic characterization (see, e.g., Hatcher, Algebraic Topology, §4F) says: a connected space has the homotopy type of an n-dimensional CW complex if and only if H^i(–; M) = 0 for all i > n and all local coefficient systems M. Applying this with n = m shows that X and Y each have the homotopy type of an m-dimensional CW complex.

Consequences and comments:

• This subsumes the simple Künneth obstruction you noted: if X has infinitely many nontrivial ordinary homology groups, then certainly X × Y cannot be finite-dimensional. The argument above shows that even without such a homology hypothesis, there is no way for a product to be finite-dimensional unless each factor already is.

• In particular, there is no example with X = K(G,1) for a group G of torsion. Indeed, BG = K(G,1) cannot have finite-dimensional CW type (groups with torsion have infinite cohomological dimension), and by the argument above, no Y can make X × Y finite-dimensional up to homotopy.

So the answer to the original question is negative: if X × Y has the homotopy type of a finite-dimensional CW complex, then X and Y each do as well.