0

S(RP^2),S(CP^2)denote suspension of real and complex projective space. Then are the first order relative homotopy group pi_1(S(RP^2),RP^2),pi_1(S(CP^2),CP^2) trivial?Why?

flag
$\pi_1($ any suspension) is trivial, isn't it ? – Nikita Kalinin Aug 26 at 15:05
3 
@Nikita Kalinin: Not quite any suspension; the space you suspend had better be connected. – Andreas Blass Aug 26 at 15:07

1 Answer

5

To define the relative homotopy groups of a pair $(X, A)$, let $i:A\to X$ be the inclusion, and write $F_i$ for its homotopy fiber. Then $$ \pi_n(X, A) = \pi_{n-1}(F_i). $$ In your examples, the inclusion maps are nullhomotopic, so the homotopy fibers are $$ \Omega \Sigma \mathbb{R}P^2 \times \mathbb{R}P^2 \qquad \mathrm{and} \qquad \Omega \Sigma \mathbb{C}P^2 \times \mathbb{C}P^2, $$ respectively. Since these spaces are path-connected, the relative homotopy "groups" in question are trivial.

link|flag
1 
This seems like overkill to me. Why not just use the long exact sequence of the pair and that $\pi_0(RP^2)$ and $pi_1(S(RP^2))$ are trivial? – Greg Friedman Aug 27 at 21:11

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.