For basepoint-preserving maps \(f: S^1 \to X\) and \(g: S^n \to X\) with \(n > 1\), show that the Whitehead product \([f, g]\) is \(\pm(g - fg)\), where \(fg\) denotes the action of \(f\) on \(g\).
Solution
Let \[ f:S^1\to X,\qquad g:S^n\to X, \quad n>1. \] The product \(S^1\times S^n\) is obtained from the wedge \[ S^1\vee S^n \] by attaching one \((n+1)\)-cell. Its attaching map is, up to the standard sign convention, the Whitehead product \[ [\iota_1,\iota_n]. \]
Lift the wedge to the universal cover. The lift is a bi-infinite string of \(n\)-spheres indexed by the deck group generated by the loop \(\iota_1\). A lift of the top cell joins two consecutive \(n\)-spheres, so its boundary represents the difference \[ \iota_n-(\iota_1\cdot\iota_n) \] in \(\pi_n\) of the lifted wedge, up to orientation sign. Applying the map induced by \(f\vee g\) gives \[ [f,g]=\pm\bigl(g-fg\bigr), \] where \(fg\) denotes the action of \([f]\in\pi_1(X)\) on \([g]\in\pi_n(X)\). Hence \[ \boxed{[f,g]=\pm(g-fg).} \]