# cohomology of suspension

Let $X$ be a topological space. Let $\Sigma$ be suspension.

Does $H^n(X;\mathbb{Z})\cong H^{n+1}(\Sigma X;\mathbb{Z})$ isomorphic or not?

Does $H^n(X;\mathbb{Z}_2)\cong H^{n+1}(\Sigma X;\mathbb{Z}_2)$ isomorphic or not?

Does $H^n(X;\mathbb{Z}_p)\cong H^{n+1}(\Sigma X;\mathbb{Z}_p)$ isomorphic for odd prime $p$ or not?

I tried to use long exact sequence of cohomology induced from $S^1\vee X\to S^1\times X\to \Sigma X$ and apply Kunneth formula on $H^*(S^1\times X;\mathbb{Z}_2)$, but I cannot work out.

• Consider the long exact sequence associated to the pair $(CX,X)$, where $CX$ is the cone on $X$.
– user98602
Commented Nov 5, 2014 at 5:33
• Commented Nov 5, 2014 at 7:38
• Just use Mayer-Vietoris... Besides you switched $n$ and $n+1$. PS: please wait some time before posting to MO if you need to post to MO at all. This kind of problem isn't research-level mathematics (which is what MO is about). Commented Nov 5, 2014 at 10:10
• Thanks. I already understand by Prof. Mike's comment :) Commented Nov 5, 2014 at 11:18
• Is the same also true for reduced cohomology? Commented Sep 19, 2023 at 18:27

## 1 Answer

Here's the argument given in the comment as an answer. Coefficients should be understood to be $$\Bbb Z$$.

Consider the pair $$(CX,X)$$, where $$CX = X \times [0,1]/\sim$$ , where $$(x,1) \sim (x',1)$$ for all $$x,' \in X$$, and $$X$$ is considered as the subspace $$X \times \{0\}$$ of $$CX$$. Then $$CX/X \cong \Sigma X$$.

Now we have a long exact sequence in reduced homology $$\dots \to H_{n+1}(CX,X) \to H_n(X) \to H_n(CX) \to H_n(CX,X) \to \dots$$

Now $$CX$$ is contractible so has trivial homology, so by exactness $$H_{n+1}(CX,X) \cong H_n(X)$$. Because $$X$$ is a closed subset of $$CX$$, we have that $$H_{n+1}(CX,X) \cong H_{n+1}(CX/X) \cong H_{n+1}(\Sigma X)$$, hence $$H_{n+1}(\Sigma X) \cong H_n(X)$$. Finally, thanks to the universal coefficient theorem for cohomology (Theorem 3.2 in https://pi.math.cornell.edu/~hatcher/AT/AT.pdf) we get the desired results.