# Proof that $\mathbb{RP}^2$ is not the suspension of a space $X$.

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.

• If you know co-homology, then you can use that the cup product in any suspension is trivial. Commented May 8, 2022 at 16:37
• I do know cohomology, but not this fact - is there an 'obvious' reason why it's true? Commented May 8, 2022 at 16:43
• math.stackexchange.com/questions/1298938/… Commented May 8, 2022 at 16:46

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 $$\Sigma X$$.