Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
$\pi_1($ any suspension) is trivial, isn't it ? –  Nikita Kalinin Aug 26 '12 at 15:05
3  
@Nikita Kalinin: Not quite any suspension; the space you suspend had better be connected. –  Andreas Blass Aug 26 '12 at 15:07

1 Answer 1

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.

share|improve this answer
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 '12 at 21:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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