1

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

Susan
  • 1,205
  • Do you know how $\pi_1$ relates to contractibility? Can you compute $pi_1$ of this space by decomposing it into two contractible spaces (the "top" and "bottom" of the suspension) and using the Van Kampen theorem? – Potato Apr 05 '13 at 05:03
  • Not sure how to compute this. All I know is that it is the pushout of $\mathbb{Z}$ going to $\mathbb{Z}/2\mathbb{Z}$ both ways – Susan Apr 05 '13 at 05:10

1 Answers1

2

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.

Bombyx mori
  • 19,638
  • 6
  • 52
  • 112
  • Actually, $\pi$ of the cone is nontrivial. See one of Susan's other questions. – Jason DeVito - on hiatus Apr 05 '13 at 12:32
  • @JasonDeVito: I suspect you mean the fundamental group of the suspension instead. The cone is contractible. – Bombyx mori Apr 05 '13 at 12:46
  • Yes, you're correct. Sorry about that! In fact, rereading my original comment, it was suppose to read "Actually, $\pi_1$ of the suspension is nontrivial. See one of Susan's other questions." (I left off the subscript on $\pi_1$!). That'll teach me to type math without caffeine! – Jason DeVito - on hiatus Apr 05 '13 at 13:08