$\tilde{H}^i(\sum X) \cong \tilde{H}^{i-1}(X)$

Question

I need help with this problem

Let $\sum X =C^+X \cup_X C^-X$ be the union of two cones on $X$ with a common base. Show that $\tilde{H}^i(\sum X) \cong \tilde{H}^{i-1}(X)$. Dose this give an isomorphism of cohomology rings of $X$ and $\sum X$ with a shift in dimension. Where $\sum X$ is the suspension of $X$.

I have no idea where to start. any help is greatly appreciated.

Answer 1

1. Use the Mayer-Vietoris sequence to prove the first part. The two cones are contractible, and their intersection is homotopy equivalent to $X$.

2. For the second part, consider any space whose cohomology ring has nontrivial products and check what the dimension of the products would be. For example $H^*(\mathbb{CP}^2) = \mathbb{Z}[x]/(x^3)$ with $|x| = 2$. This means that if $X = \Sigma \mathbb{CP}^2$, then: $$H^i(X) = \begin{cases} \mathbb{Z} & i = 0, 3, 5 \\ 0 & \text{otherwise} \end{cases}$$ Let $u$ be a generator of $H^3(X)$, then $|u^2| = 2 |u| = 6$ so necessarily $u^2 = 0$. But $x^2 \neq 0$ in $H^*(\mathbb{CP}^2)$, so it's not an isomorphism of rings (the cup product is not preserved).