$S(RP^2),S(CP^2)$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?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ $\pi_1($ any suspension) is trivial, isn't it ? $\endgroup$– Nikita KalininCommented Aug 26, 2012 at 15:05
-
4$\begingroup$ @Nikita Kalinin: Not quite any suspension; the space you suspend had better be connected. $\endgroup$– Andreas BlassCommented Aug 26, 2012 at 15:07
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Aug 26, 2012 at 15:46
-
4$\begingroup$ 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? $\endgroup$ Commented Aug 27, 2012 at 21:11
-
$\begingroup$ @GregFriedman This is that argument, just phrased differently. (In particular $\Omega \Sigma \mathbb{R}P^2 \times \mathbb{R}P^2$ being connected is equivalent to $\pi_0(\mathbb{R}P^2)$ and $\pi_1(S(RP^2))$ being trivial.) $\endgroup$ Commented Nov 27, 2017 at 20:08