I have a question about this passage of Hatcher book "Algebraic Topology":
A sum operation is defined in $\pi_n(X,A,x_0)$ by the same formulas as for $\pi_n(X,x_0)$, except that the coordinate $s_n$ now plays a special role and is no longer available for the sum operation.
I'm trying to understando what is this "special role" that Hatcher is talking about. First, in the definition of sum in $\pi_n(X,x_0)$, Hatcher define only in the first coordinate, furthermore, I'm having troubles to visualize this definition geometricaly.
Let's consider the case $n=1$, so $s_1 \in I^1 = [0,1], I^0 = {0}$ and $J^0 ={1}$, so let $\alpha,\beta: ( I^1, I^0, J^0) \to (X,A,x_0)$, be two homotopic paths $\text{rel.}A$, so is such that $\alpha(I^1)$ and $\beta(I^1)$ are paths in $X$, $\alpha(I^0) = \alpha(0)$ and $\beta(I^0) = \beta(0)$ are subsets of $A$, and $\alpha(J^0) = \beta(J^0) = x_0$. I'm understand this as a homotopy between $\alpha,\beta$ that's fix only the end points of this paths. In this case, the coordinate $s_n$ is $s_1$, then this paths need's not to have the same initial points. The "product" of paths is able, but at least that $A$ is $\{x_0\}$, $\alpha$ and $\beta$ are not loops, then I understand why $\pi_1(X,A,x_0)$ is not a group.
Now, I trying to understand the case that $n>1$. I'm study the case when $n =2$, for example, consider the image below (taken from the book non abelian algebraic topology our good friend R. Brown)
My interpretation of this is as follows;$(s_1,s_2)\in I^2$, $J^1 = I\times \{0,1\} \cup {1}\times I$, so $\alpha(J^1) = x_0$, $\alpha(I)$ is a loop in $A$ whith base point $x_0$, and $\alpha(I^2)$ is a shape homeomorphic to a sphere in $X$. This interpretation leads to belive why $\pi_2(X,A,x_0)$ is not a abelian group, since we are working with homotopy of paths in $A$ which is generally noncommutative. The conclusion I draw from this is that the fundamental group $\pi_1(A,x_0)$ influence in some way the relative homotopy group $\pi_2(X,A,x_0)$. But, my question is: whats is the role of $s_2$ in this case??
As far as I have understood, your questions are:
The group operation on $\pi_n(X,A,x_0)$ (for $n\geq 2$) is, as Hatcher says, induced by the concatenation of representatives, i.e. induced by the operation $$(f+g)(s_1,\ldots,s_n):=\begin{cases}f(2s_1,s_2,\ldots,s_n)&\text{for }s_1\in[0,1/2]\\ g(2s_1-1,s_2,\ldots,s_n)&\text{for }s_1\in[1/2,1].\end{cases}$$ An important property is the continuity of $f+g$, which is given since $(f+g)(1/2,s_2,\ldots,s_n) = f(1,s_2,\ldots,s_n) = g(0,s_2,\ldots,s_n) = x_0$. Every coordinate, except the last coordinate, is available for the sum operation. This is not the case for the last coordinate because the concatenation would lead to $f(s_1,\ldots,s_{n-1},1) = x_0$ and $g(s_1,\ldots,s_{n-1},0)\in A$. Now, in the case of $n = 1$, we only know that $f(1) = x_0$ and $g(0)\in A$ and thus continuity is not readily present. Hence a natural group operation cannot be defined on $\pi_1(X,A,x_0)$ except for the case of for example $A = \{x_0\}$ because $\pi_1(X,x_0,x_0) = \pi_1(X,x_0)$.
The visualization is exactly the same as for absolute homotopy groups, i.e. as Hatcher shows in his first square on p. 340 of his book on Algebraic Topology. In this figure he shows commutativity for which you need, also algebraically, at least two coordinates to define appropriate concatenations and obtain the corresponding homotopies. Since the last coordinate for relative homotopy groups is not available for concatenations, relative homotopy groups are abelian for $n\geq3$.
The picture just shows which parts of the boundary of $I^2$ are sent to the base point $x$ and which part is arbitrarily mapped into the subspace $A$ by any representative of any relative homotopy class.