Suspension of $\mathbb{R}P^2$ Contractible?

Question

Is the suspension of $\mathbb{R}P^2$ Contractible? And if it is, How would you prove it. Thank you!

Answer 1

The homology of the suspension is easy to calculate, since we know the two pieces are contractible individually and their intersection is $\mathbb{R}\mathbb{P}^{2}$. So use Mayer-Vietoris you should have $$0\rightarrow H_{3}(X)\rightarrow H_{2}(\mathbb{R}{P}^{2})\rightarrow 0\rightarrow H_{2}(X)\rightarrow H_{1}((\mathbb{R}{P}^{2})\rightarrow 0...$$

So the space should be non-orientable and has non-trivial homology group. Potato's hint is not difficult and I suspect the fundamental group may be $\mathbb{Z}_{2}$ as well. This showed the space is not contractible since contractible space has trivial homology groups.

A way to visualize it may be thinking of the CW structure. If you can imagine how $\mathbb{D}^{2}$ is glued to $\mathbb{RP}^{1}$ with the 2 fold gluing map, then you can certainly imagine how the suspension works. Obviously the gluing would not be undone and so the space remains "twisted" in some sense. So it cannot be contractible. But I admit such a picture is vague and not dependable.