I want to show that the spaces $ S^1 \vee S^1 \vee S^2$ and $S^1 \times S^1 $ do not have the same homotopy type.
I calculated their homologies and cohomologies and they turn out to be equal.
So I want to employ the cup product to get the desired result, but I need a hint as to how that can be done.
Thank you!
By the Künneth formula, we have $$ H^*(S^1 \times S^1; \Bbb Z) \cong \Lambda_\Bbb Z[\alpha, \beta], $$ the exterior algebra on two variables, with $|\alpha| = |\beta| = 1$. This can also be shown by direct computation using simplicial homology.
By the formula for the cohomology ring of a wedge sum, we have $$ \tilde H^*(S^1 \vee S^1 \vee S^2; \Bbb Z) \cong \tilde H^*(S^1; \Bbb Z) \oplus \tilde H^*(S^1; \Bbb Z) \oplus \tilde H^*(S^2; \Bbb Z). $$
Now we can see that the cup product of the two generators of $H^1(S^1 \times S^1; \Bbb Z)$ is nontrivial, whereas the cup product of the two generators of $H^1(S^1 \vee S^1 \vee S^2; \Bbb Z)$ is trivial. It follows that the two spaces are not homotopy equivalent.