If \(X\) and \(Y\) are simply-connected CW complexes such that \(\tilde{H}_i(X)\) and \(\tilde{H}_j(Y)\) are finite and of relatively prime orders for all pairs \((i, j)\), show that the inclusion \(X \vee Y \hookrightarrow X \times Y\) is a homotopy equivalence and \(X \wedge Y\) is contractible.
Solution
By the reduced Künneth theorem for smash products, \[ \widetilde H_k(X\wedge Y;\mathbb Z) \] is built from the groups \[ \widetilde H_i(X)\otimes\widetilde H_j(Y) \quad\text{and}\quad \operatorname{Tor}(\widetilde H_i(X),\widetilde H_j(Y)). \] Each \(\widetilde H_i(X)\) and \(\widetilde H_j(Y)\) is finite, and their orders are relatively prime. Therefore both the tensor product and Tor vanish for every pair \((i,j)\). Hence \[ \widetilde H_*(X\wedge Y)=0. \]
Since \(X\) and \(Y\) are simply connected, the smash product is simply connected as well. Thus the map \[ X\wedge Y\to * \] is a homology equivalence between simply-connected CW complexes. The homology Whitehead theorem gives \[ \boxed{X\wedge Y\simeq *.} \]
The inclusion \[ X\vee Y\hookrightarrow X\times Y \] is a cofibration with quotient \(X\wedge Y\). Since the quotient is acyclic, the relative homology groups \[ H_*(X\times Y,X\vee Y) \] vanish. Both \(X\vee Y\) and \(X\times Y\) are simply connected, so the inclusion is a homology equivalence between simply-connected CW complexes. Again by homology Whitehead, \[ \boxed{X\vee Y\simeq X\times Y.} \]