Suppose a sum \(f +' g\) of maps \(f, g: (I^n, \partial I^n) \to (X, x_0)\) is defined using a coordinate of \(I^n\) other than the first coordinate as in the usual sum \(f + g\). Verify the formula \((f + g) +' (h + k) = (f +' h) + (g +' k)\), and deduce that \(f +' k \simeq f + k\) so the two sums agree on \(\pi_n(X, x_0)\), and also that \(g +' h \simeq h + g\) so the addition is abelian.
Solution
Theorem.
For \(n\ge1\), any coordinate choice
for the concatenation sum on \(\pi_n(X,x_0)\) is equivalent and gives an
abelian group law.
Proof.
Let \(q_1: (I^n,\partial I^n)\to (I^n\vee I^n,\*)\) be the standard pinch map collapsing \(\{(x_1,\dots,x_n):x_1=1/2\}\) to the wedge point, so \(f+g=(f\vee g)\circ q_1\).
Let \(q_j\) be the analogous pinch map using coordinate \(x_j\); then \(f +' g=(f\vee g)\circ q_j\).
The coordinate permutation \(\sigma:(I^n,\partial I^n)\to(I^n,\partial I^n)\) that sends \(x_1\leftrightarrow x_j\) and is identity on the boundary is homotopic rel \(\partial I^n\) to the identity. Therefore \(q_j\simeq q_1\circ \sigma\) rel boundary.
Hence \[ f +' g \simeq (f\vee g)\circ q_1\circ \sigma. \] Since \(\sigma\) is homotopic to \(\mathrm{id}\), this gives \(f +' g \simeq f+g\).
Expanding the definitions gives \[ (f + g) +' (h + k)=((f\vee g)+(h\vee k))\circ q_j \simeq (f +' h)+(g +' k). \]
Taking \(h=k\) and swapping coordinates gives \(f +' k \simeq f+k\), so the operations are the same in \(\pi_n\).
Swapping the order of the two summands in \(q_j\) gives a homotopy to the sum with reversed order, so \(g +' h \simeq h + g\).
Thus the addition is independent of coordinate choice and is abelian. ∎