Finding a Pinch Map from the Given Multiplications This is an exercise problem and I could'n know how to tackle with it and need someone's help.
Notation: Let $(A, a_{0})$ and $(X , x_{0})$ be pointed spaces, then $[A,X]_{*}$ denotes the set of all pointed homotopy classes of maps $A \rightarrow X$.
Question: Suppose $(A , a_{0})$ is a space such that for any $(X , x_{0})$


*

*$[A , X]_{*}$ has a group structure

*for each pointed map $f: (X , x_{0}) \rightarrow (Y , y_{0})$, the induced map $f_{*} : [A , X]_{*} \rightarrow [A , Y]_{*}$ is a group homomorphim.


Then prove that there is a pinch map $p : A \rightarrow A \vee A$ so that the multiplication in the group $[A , X]_{*}$ is given by
\begin{equation*}
[f][g] = [(f \vee g) \circ p].
\end{equation*}
where $[f] , [g] \in [A ,X]_{*}$ and the the multiplication between them is given by the condition 1. above.
Can someone give me a hint to start with? Thank you!
 A: Let $i_1:A\hookrightarrow A\vee A$, $a\mapsto (a,\ast)$, $i_2:A\hookrightarrow A\vee A$, $a\mapsto (\ast,a)$ be the inclusion maps (we are viewing $A\vee A\subseteq A\times A$). Take $X=A\vee A$. Then by assumption the set $[A,A\vee A]$ has group structure, so define
$c=i_1+i_2\in[A,A\vee A]$
(we'll use additive notation for this product although it is not necessarily abelian). Choose a representative for this homotopy class (which we'll continue to write as $c$), and then for arbitrary maps $f:A\rightarrow X$, $g:A\rightarrow X$ consider the composition
$\nabla\circ(f\vee g)\circ c:A\rightarrow A\vee A\rightarrow X\vee X\rightarrow X$
where $\nabla:X\vee X\rightarrow X$ is the folding map. Now observe that
$\nabla\circ(f\vee g)\circ c=\nabla_*(f\vee g)_*c=\nabla_*(f\vee g)_*(i_1+i_2)=\nabla_*(f\vee g)_*i_1+\nabla_*(f\vee g)_*i_2=\nabla_*(f\vee\ast)+\nabla_*(\ast\vee g)=f+g$.
There are some steps missing, which I'll leave you to fill in (you need to show that $\ast:A\rightarrow \ast\rightarrow X$ is the unit element in $[A,X]$, for instance). If you need some more help please just let me know in a comment.
