Show that \(S^1 \times S^1\) and \(S^1 \vee S^1 \vee S^2\) are not homotopy-equivalent.
Solution
1. The torus has degree-one classes \(\alpha,\beta\in H^1(T^2;\mathbb Z)\) with\[ \alpha\smile\beta \]a generator of \(H^2(T^2;\mathbb Z)\cong\mathbb Z\).
Take \(\alpha,\beta\) Poincaré dual to the two standard circle factors. Their representing cycles intersect transversely once, so their cup product evaluates to \(1\) on the fundamental class.
2. In\[ Y=S^1\vee S^1\vee S^2, \]the cup product of any two degree-one classes is zero.
Every degree-one class is supported on the two \(S^1\) wedge summands. Products between classes from different wedge summands vanish, and each individual \(S^1\) has no degree-two cohomology. Hence all products in \(H^1(Y)\otimes H^1(Y)\to H^2(Y)\) are zero.
3. Therefore \(T^2\) and \(S^1\vee S^1\vee S^2\) are not homotopy-equivalent.
A homotopy equivalence induces an isomorphism of graded cohomology rings. The nonzero product in 1 cannot correspond under any ring isomorphism to the identically zero degree-one product in 2.