$\pi_q(S^n) \cong \pi_q(\Bbb RP^n)$

Question

I'd like to prove that $\pi_q(S^n) \cong \pi_q(\Bbb RP^n)$ when $q>1$ and $n>1$. Do I have to prove this by cases? I mean, the cases $q<n$, $q=n$ and $q>n$? Or can I prove it directly? Which theorem or result could I use?

Answer 1

In general, if $p : X \to Y$ is a covering map, then $p_* : \pi_q(X) \to \pi_q(Y)$ is an isomorphism for $q > 1$.

To see this, first note that any map $f : S^q \to Y$ lifts to a map $f' : S^q \to X$ because $S^q$ is simply connected. Therefore $f = p\circ f'$ and hence $[f] = [p\circ f'] = p_*[f']$; that is, $p_*$ is surjective. For injectivity, suppose $[g] \in \pi_q(X)$ is in the kernel of $p_*$ so that $p_*[g] = [p\circ g] = 0$. Then there is a homotopy $H : S^q\times[0, 1] \to Y$ between $p\circ g$ and a constant map $c$. As $g$ is a lift of $p\circ g$ there is a homotopy $H' : S^q\times[0, 1]$ between $g$ and $c'$, a lift of $c$ - here we have used the homotopy lifting property, see Proposition $1.30$ of Hatcher's Algebraic Topology for example. As $S^q$ is connected, $c'$ must also be constant, and hence $g$ is nullhomotopic, i.e. $[g] = 0$. Therefore $p_*$ is injective and hence an isomorphism.

As $p : S^n \to \mathbb{RP}^n$ is a covering map for any $n$, so we see by the above that $p_* : \pi_q(S^n) \to \pi_q(\mathbb{RP}^n)$ is an isomorphism for $q > 1$.