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?

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