How to calculate relative homotopy groups of real projective spaces

Solution 1:

The sequence continues one to the left as $\pi_n(\mathbb{R}P^{n-1}) \rightarrow \mathbb{Z} \rightarrow \pi_n(\mathbb{R}P^n,\mathbb{R}P^{n-1}) \rightarrow \mathbb{Z} \rightarrow 0$ for $n>2$. We want to understand the image of this map $\pi_n(\mathbb{R}P^{n-1}) \rightarrow \pi_n(\mathbb{R}P^n)=\mathbb{Z}$. To do this, we may lift this to a map of universal covers, in this case the map lifts to the inclusion of $S^{n-1}$ into $S^n$. This map is nullhomotopic, hence on $\pi_n$, $n>1$, it induces the zero map.

From this we deduce that we have a short exact sequence $0\rightarrow \mathbb{Z} \rightarrow \pi_n(\mathbb{R}P^n,\mathbb{R}P^{n-1}) \rightarrow \mathbb{Z} \rightarrow 0$ since the image of the zero map must be the kernel of $\mathbb{Z} \rightarrow \pi_n(\mathbb{R}P^n,\mathbb{R}P^{n-1})$. Since $\mathbb{Z}$ is free, this short exact sequence splits, hence $\pi_n(\mathbb{R}P^n,\mathbb{R}P^{n-1})=\mathbb{Z}\oplus \mathbb{Z}$, at least if $n>2$. This gives a nice example of the failure of excision for homotopy, since otherwise this group would be $\mathbb{Z}$ since the quotient is homeomorphic to $S^n$.