Explicit homotopy on showing that $\pi_n(X,x_0)$ is abelian for $n \geq 2$ I am trying to show that $\pi_n(X,x_0)$ is abelian for $n \geq 2$. Given two maps $f,g:(I^n,\partial I^n)\to(X,x_0)$, I want to construct an explicit homotopy rel $\partial I^n$ from $f + g$ to $g + f$. I saw the following picture below, in Hatcher, but how can I write down this homotopy in formulas?

 A: It is not difficult. For the sake of "symmetric" formulas let us replace $I^n$ by $J^n$, where $J = [-1,1]$. Then
$$(f + g)(s_1,\ldots,s_n) = \begin{cases} f(2s_1+1,s_2,\ldots,s_n) & s_1 \le 0 \\ g(2s_1-1,s_2,\ldots,s_n) & s_1 \ge 0 \end{cases} $$
Write $J^n = J^2 \times J^{n-2} $. Define a homeomorphism $h : J^2 \to D^2$ by linearly compressing the line segment between $z \in \partial J^2$ and $0$ to the line segment between $\frac{z}{\lvert z \rvert}$ and $0$. Let $G : D^2 \times I \to D^2$ denote the homotopy $G(z,t) = e^{t\pi i} z$, which rotates the disk $D^2$ by an angle of $\pi$. We have $G_1(z) = -z$. Define $H = h^{-1} \circ G \circ (h \times id_I)$ and $R = H_1 \times id_{J^{n-2}}$. This is a homeomorphism homotopic to $id_{J^n}$ via a homotopy of pairs on $(J^n,\partial J^n)$. We have $R(s_1,s_2,s_3,\ldots,s_n)  = (-s_1,-s_2,s_3,\ldots,s_n)$. 
Then $\phi : (J^n,\partial J^n) \to (X,x_0)$ and $\phi \circ R$ represent the same element in $\pi_n(X,x_0)$. But now we get
$$(f + g) \circ R = g \circ R + f \circ R .$$
To see this, let $s_1 \le 0$. Then
$$(g \circ R + f \circ R)(s_1,\ldots,s_n)) = (g \circ R)(2s_1+1,\ldots,s_n) = g(-2s_1-1,-s_2,s_3,\ldots,s_n)$$
and because $-s_1  \ge 0$
$$(f + g)(R(s_1,\ldots,s_n)) = (f+g)(-s_1,-s_2,s_3,\ldots,s_n) = g(2(-s_1)-1,-s_2,s_3,\ldots,s_n) = (g \circ R + f \circ R)(s_1,\ldots,s_n) .$$
The case $s_1 \ge 0$ is treated similarly.
We conclude
$$[f  + g] = [(f + g)\circ R] = [g \circ R + f \circ R] = [ g + f].$$
Remark: Hatcher's homotopy is more complicated because he first "compresses" the $f,g$-parts into the interior of $I^n$ and then makes a deformation of $I^n$ not rotating the smaller $f,g$-parts. Explicitly describing this deformation is probably not easy. 
