Let $\Sigma$ be suspension. For any CW-complex, or topological space, does the reduced homology satisfy

$$ \tilde H_*(\Sigma^k X)=s^k\tilde H_*(X)? $$ Here $s^k H$ is a copy of $H$ such that an element $s^k x\in S^k H$ has a degree equal to the degree of $x$ plus $k$.

I obtained this for $\mathbb{C}P^n$, $\mathbb{H}P^n$.


1 Answer 1


Yes this is correct. It obviously suffices to show that $\tilde H_{k+1}(\Sigma X) = \tilde H_{k}(X)$ for all $k$.

To show this consider the Mayer Vietors Sequence for suitable neighborhoods of the two cones $C_1X,C_2X$ which are glued together to obtain the suspension. Using homotopy equivalences you will get the following exact sequence:

$$ \cdots \to \tilde H_k(X) \to \tilde H_k(C_1X) \oplus \tilde H_k(C_2X) \to \tilde H_k(\Sigma X)\to \cdots $$

Since $C_iX$ is contractible the long exact sequence splits into very short ones (also known as isomorphisms):

$$ 0\to \tilde H_{k+1} (\Sigma X) \to \tilde H_k(X) \to 0$$


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