- What are the integer cohomology rings of \(S^2 \times S^2\) and \(\mathbb{CP}^2\)?
- Show that there is no continuous map \(f: S^2 \times S^2 \to \mathbb{CP}^2\) having odd degree.
Solution
1. Let \(u,v\in H^2(S^2\times S^2;\mathbb Z)\) be the pullbacks of the positive generator of \(H^2(S^2;\mathbb Z)\) under the two projections. Then\[ H^*(S^2\times S^2;\mathbb Z) \cong \mathbb Z[u,v]/(u^2,v^2), \qquad |u|=|v|=2, \]and \(uv\) generates \(H^4(S^2\times S^2;\mathbb Z)\) after choosing the product orientation.
2. If \(x\in H^2(\mathbb{CP}^2;\mathbb Z)\) is the standard generator, then\[ H^*(\mathbb{CP}^2;\mathbb Z) \cong \mathbb Z[x]/(x^3), \qquad |x|=2, \]with \(x^2\) generating \(H^4(\mathbb{CP}^2;\mathbb Z)\) after choosing the usual orientation.
3. Let \(f:S^2\times S^2\to\mathbb{CP}^2\) be continuous.
3.1. Since \(H^2(S^2\times S^2;\mathbb Z)=\mathbb Zu\oplus\mathbb Zv\), write\[ f^*(x)=au+bv \]for integers \(a,b\).
3.2. Naturality of cup products gives\[ f^*(x^2)=(au+bv)^2 =a^2u^2+2ab\,uv+b^2v^2 =2ab\,uv. \]
3.3. By the definition of degree,\[ f^*(x^2)=\deg(f)\,uv. \]Thus \(\deg(f)=2ab\), which is even.
4. Therefore no map \(S^2\times S^2\to\mathbb{CP}^2\) has odd degree.