most recent 30 from http://mathoverflow.net 2013-05-18T16:41:13Z http://mathoverflow.net/feeds/question/105527 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105527/about-relative-homotopy-group jiangsaiyin 2012-08-26T12:52:40Z 2012-08-26T15:46:17Z

<p>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? </p>

http://mathoverflow.net/questions/105527/about-relative-homotopy-group/105546#105546 Jeff Strom 2012-08-26T15:46:17Z 2012-08-26T15:46:17Z

<p>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.</p>