Let \(q : S^3 \to \mathbb{RP}^3\) be the usual quotient map which identifies antipodal points. It can be used to attach a \(4\)-ball to \(\mathbb{RP}^3\), forming a space \(X = \mathbb{RP}^3 \cup_q B^4\). Compute \(\pi_4(X)\).
Solution
1. The standard CW construction of \(\mathbb{RP}^4\) is obtained from \(\mathbb{RP}^3\) by attaching a \(4\)-cell via the antipodal quotient\[ q:S^3\to\mathbb{RP}^3. \]
In the usual CW structure on real projective space, the characteristic map of the top cell descends from a hemisphere of \(S^4\), and its boundary attaching map is exactly the double covering \(S^3\to\mathbb{RP}^3\).
2. Hence the given space is\[ X\cong\mathbb{RP}^4. \]
It has precisely the standard cell attachment described in 1.
3. The universal covering map \(S^4\to\mathbb{RP}^4\) induces an isomorphism on \(\pi_4\).
Covering maps induce isomorphisms on all homotopy groups in degrees at least \(2\).
4. Therefore\[ \boxed{\pi_4(X)\cong\pi_4(S^4)\cong\mathbb Z.} \]
The identity class generates \(\pi_4(S^4)\cong\mathbb Z\).