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Can you explain to me why relative homotopy groups $\pi_{n}(X, A; x_0)$ are commutative for $n \geq 3$? It would be great if you will show me explicit homotopy.

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No. The reason is that this is a research forum. You rather ask this question in math.stackexchange. –  Fernando Muro Aug 9 '13 at 9:22
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Can this not be seen as a consequence of the long exact sequence of a pair together with the fact that the non-relative homotopy groups are abelian for $n\geq 2$ ? –  Daniel Rust Aug 9 '13 at 14:39
    
I think that is right idea. Unfortunately, I can't prove it. Can you explain this to me? –  Gleb Aug 10 '13 at 13:55

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up vote 2 down vote accepted

It is easiest to see this using a model of relative homotopy groups $\pi_n(X,A,a)$ in terms of maps $f$ of an $n$-cube $I^n \to X$ which maps the face $\partial^-_1$ to $A$ and all other faces $\partial^\pm _i$ to the base point $a$. (Here $I^n$ consists of points $x =(x_1, \ldots, x_n) \in \mathbb R^n$ such that $0 \leqslant x_i \leqslant 1$ and the face $\partial^-_i$ has $x_i=0$, while $\partial ^+_i$ has $x_i=1$.) Given $i >1$ and two such maps $f,g$ such that $f(x)=g(y) $ if $x_i=1, y_i=0$ and $x_j=y_j $ for $j \ne i$ (this condition can be written $ \partial^+_i f= \partial^-_i g$) we can define $h=f+_ig: I^n \to X$ by

$$h(x) = f(x_1, \ldots, x_{i-1},2x_i ,x_{i+1}, \ldots, x_n) \; \text{if} \; x_i \leqslant 1/2,$$

$$h(x) = g(x_1, \ldots, x_{i-1},2x_i-1 ,x_{i+1}, \ldots, x_n) \; \text{if}\; x_i \geqslant 1/2. $$

(Compare the addition of paths.) If $n \geqslant 3$ and $i\ne j$ are $>1$ then we find the interchange law

$$(f+_i g)+_j (h+_ik) = (f+_jh)+_i (g+_j k) $$ whenever both sides are defined, and so both sides can be represented by the matrix

$$\begin{bmatrix} f& h \\ g & k \end{bmatrix} \; \;\; \begin{matrix} & \rightarrow &j\\\downarrow && \\ i & & \end{matrix} $$

where down is direction $i$ and across is direction $j$. This rule passes to homotopy classes, and we can now apply the standard interchange law argument to get the result asked for. The usual addition on relative homotopy groups is given by $+_2$ but for $n \geqslant 3$ we can also use $+_3$ and the argument says they agree, and are commutative.

It is not quite so easy to extract an explicit homotopy from this argument!

The above can be put in the explicit context of the cubical singular complex of a filtered space, and so obtain further useful properties of relative homotopy groups. See the book, with pdf available, Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids.

Later: on second thoughts, the homotopy required is the composite of the homotopies, in which $0$ stands for the constant map with value $a$:

$$\begin{bmatrix} f&g \end{bmatrix} \simeq \begin{bmatrix} f&0\\0&g \end{bmatrix} \simeq \begin{bmatrix} 0&f\\g&0 \end{bmatrix}\simeq \begin{bmatrix} g&f \end{bmatrix}. $$

Discussion of the use of matrix notation and of the use of the interchange law is on p. 148-149 of the above book.

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I get it now. Thanks. –  Gleb Aug 10 '13 at 11:27

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