Compute the cohomology groups and cohomology ring \(H^*(S^2 \vee S^2 \vee S^4; \mathbb{Z})\) of the wedge sum \(X = S^2 \vee S^2 \vee S^4\).
Solution
Let \[ X=S^2_a\vee S^2_b\vee S^4. \]
1. The reduced cohomology of a finite wedge is the direct sum of the reduced cohomologies of the summands. Hence\[ H^k(X;\mathbb Z)\cong \begin{cases} \mathbb Z,&k=0,4,\\ \mathbb Z^2,&k=2,\\ 0,&\text{otherwise}. \end{cases} \]
2. Let \(\alpha,\beta\in H^2(X;\mathbb Z)\) be the classes coming from the two \(2\)-sphere summands, and let \(\gamma\in H^4(X;\mathbb Z)\) be the class coming from the \(4\)-sphere summand.
3. Every product of two positive-degree classes is zero.
For each wedge summand, the square of its positive-degree generator vanishes for dimensional reasons. Products of classes supported on distinct wedge summands vanish because the reduced diagonal of a wedge has no mixed component between distinct summands. Thus \[ \alpha^2=\beta^2=\alpha\beta=0, \] and every product involving \(\gamma\) also vanishes by degree.
4. Therefore, as a graded ring,\[ H^*(X;\mathbb Z) \cong \mathbb Z[\alpha,\beta,\gamma]/ (\alpha^2,\beta^2,\alpha\beta,\alpha\gamma,\beta\gamma,\gamma^2), \]with \(|\alpha|=|\beta|=2\) and \(|\gamma|=4\).