I am looking to prove that $\mathbb{RP}^2$ is not homeomorphic to $\Sigma X$ for any space $X$, where $\Sigma$ denotes suspension of a space. I am stumpted as the context of the question is homology, and while homology is homeomorphism invariant, it's certainly not the case that non-homeo. spaces must have different homology groups. If I suppose that there is a space $X$ satisfying this requirement, what other properties must $\Sigma X$ that $\mathbb{RP}^2$ does not to achieve a contradiction?
My only other thought so far is that $H_2(\mathbb{RP^2})=0$, which might be interesting… but as I can make no assumptions about $X$, I can't see why $\Sigma X$ must have a non-trivial second homology group.
Best Answer
The reduced homology of a space $X$ and its suspension are related by $\tilde{H}_n(X)\simeq\tilde{H}_{n+1}(\Sigma X)$. Now $\tilde{H}_0(X)$ is always torsion-free but $\tilde{H}_1(\mathbb{RP}^2)\simeq\mathbb{Z}/2\mathbb{Z}$, so $\mathbb{RP}^2$ cannot be homeomorphic to $\Sigma X$.