Is the splitting $\pi_{k}(X,A)\simeq\pi_{k}(X)\times \pi_{k-1}(A)$ a $\pi_1(A)$-modules isomorphism? Let $(X,A)$ be a pair of topological spaces with $A\subset X$. Fix a basepoint $x_0$ of $X$ which lies in $A$. Assume that the inclusion $(A,x_0)\to (X,x_0)$ is homotopic to a constant map relatively to the basepoint, that is there is a homotopy $H:A\times [0,1]\to X$  with
$H(x,0)=x$ and $H(x_0,t)=H(x,1)=x_0$. Then, from a map
$f:(D^k,S^{k-1},s_0)\to(X,A,x_0)$
with $s_0 \in S^{k-1}$ one can construct an extension
$$\tilde{f}:(D^k,S^{k-1})\to(A,x_0)$$
by "gluing" a cylinder $S^{k-1}\times [0,1]$ to the disk $D^k$ and by following the homotopy radially from the boundary. After verification, this gives a well defined morphism
$$r:\pi_k(X,A,x_0)\to\pi_k(A,x_0)$$
given by $r([f])=[\tilde{f}]$. Since the inclusion $A\to X$ is homotopically trivial, the long exact sequence of the pair $(X,A)$ becomes, for all $k\geq 2$,
$$0\to\pi_{k}(X,x_0)\overset{j_*}{\longrightarrow}\pi_{k}(X,A,x_0)\overset{\partial_*}{\longrightarrow} \pi_{k-1}(A,x_0)\to0.$$
By construction we have $r\circ j_*=id$, so $r$ is a retraction of the above short exact sequence and it gives a splitting
$$\boxed {\pi_{k}(X,A,x_0)\simeq\pi_{k}(X,x_0)\times \pi_{k-1}(A,x_0)}.$$
Also the three groups above are $\pi_1(A,x_0)$-modules and the maps $j_*$ and $\partial_*$ are both $\pi_1(A,x_0)$-equivariant.

Is the above splinting an isomorphism of $\pi_1(A,x_0)$-modules ?

This amounts to proving that the retraction $r$ is $\pi_1(A,x_0)$-equivariant. Basically I tried to do drawings and interchange the action of $\pi_1(A,x_0)$ and the operation of gluing $H$ but it failed miserably. Maybe someone has a more straightforward approach or a counterexample ? Any help will be greatly appreciated.
 A: Recall that then $F\to X\to Y$ is a fibration sequence, you have an induced long fibration sequence $$\cdots\to\Omega F\to\Omega X\to\Omega Y\to F\to X\to Y$$
meaning that each series of three consecutive spaces forms a fibration sequence, called "exact Puppe sequence".
For a fibration $f:E\to B$ with fiber $F$ you have $\pi_k(E,F,*)\cong\pi_k(B,*)$ so that if $i:A\to X$ is an inclusion with homotopy fiber $F$, there is an isomorphism
$$\pi_k(\Omega X,\Omega A)\cong\pi_k(F)$$
induced from the fibration sequence $\Omega A\to\Omega X\to F$. Of course you also have
$$\pi_k(\Omega X,\Omega A)=\pi_{k+1}(X,A)$$
because the adjunction $(\Sigma,\Omega)$ induces an adjunction on the homotopy category of pairs.
In short, for any pair $(X,A)$, there is an isomorphism $\pi_{k+1}(X,A)\cong \pi_k(F)$ where $F$ is the homotopy fiber of the inclusion.
Unraveling the different isomorphisms, one can make the latter explicit : take a representative $f\colon (D^{k+1},S^k,*)\to (X,A,*)$ of an element of $\pi_{k+1}(X,A,*)$, then the associated element of $\pi_k(F)$ is represented by the map $f^\prime\colon (S^k,*)\to (F,*)$
$$f^\prime(z)=(a(z),\gamma(z))$$
with $a(z)=f(z)$ and $\gamma(z)(t)=f(z,t)$ under the identification $D^{k+1}=CS^k$.
We should make sure that is is $\pi_1(A)$-equivariant, I will just accept it, Tyrone claimed it in his comment. The action of $\pi_1(A)$ on $\pi_k(F)$ is the action of the fundamental group of the total space on the homotopy groups of the fiber: there is a monodrony action $\Omega X\times F\to F$ which is not basepoint preserving, but precomposing with $A\to X$ gives a well defined pointed map
$$\Omega A\times F\to F$$
inducing an action on homotopy groups.
Now if the inclusion $A\subseteq X$ is nulhomotopic, i.e. if it is homotopic to the trivial map $*\colon A\to X$, then the homotopy fibers of the inclusion and of the trivial map are weakly equivalent (by naturality of the long exact sequence of homotopy groups of a fibration, and the five lemma). But the homotopy fiber of the trivial map is $$A\times \Omega X$$ Also, the long exact sequences are $\pi_1(A,*)$-equivariant as well as the morphism between them. For a trivial map $*\colon A\to X$, the additional piece of data we have is that the fiber sequence
$$\Omega X\to F\to A$$
is such that $\Omega X\to F$ has a retract. So, the splitting is equivariant in this case, and hence is always equivariant.
Little check: we used that a commutative square of fibrations induces a map of long exact sequences, which is true in the strict setting. It is also true in the case of a diagram which only commutes up to homotopy: if $f\colon A\to X$ and $g\colon B\to C$ are fibrations and we have maps $k\colon A\to C$, $l\colon B\to D$ making the square commute up to homotopy, take a homotopy
$$H:A\times[0,1]\to D$$
and its dual
$$K:A\to D^{[0,1]}$$
You can form the following two strictly commutative squares:
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
%
\begin{array}{ccc}
A & \ra{f} & B \\
\da{a\mapsto (a,0)} & & \da{l}\\
A\times[0,1] & \ra{H} & D
\end{array}
$$
and
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
%
\begin{array}{ccc}
A & \ra{K} & D^{[0,1]} \\
\da{k} & & \da{\mathrm{ev}_1}\\
C & \ra{g} & D
\end{array}
$$
They induce long exact sequence in homotopy, and the lower long sequence of the first diagram is exactly the same as the upper long sequence of the second one (aka the long exact sequences of the maps $l\circ f\simeq k\circ g$), so they can be glued forming the exact sequence we were looking for.
