TITLE: Uniqueness of tensor product
QUESTION [5 upvotes]: The uniqueness property of tensor product $M ⊗ N$ of two $A-$modules $M$ and $N$ specifies the following:
for sake of simplicity we will write $M⊗N$ as $T$.
A tensor product of $M$ and $N$ is pair $(T,g)$ where $T$ is an $A-$module and $g:M × N → T$ is a bilinear map such that any bilinear map $f:M × N → P$ factors through an $A-$module homomorphism $h$ such that $f = h \circ g$.
uniqueness says that if you have another pair $( T',g' )$  then $∃  \  j: T \to T' $ an isomorphism s.t $g' = j \circ g$. 
While proving this, 
we use the factoring property of tensor product for both $(T,g)$ and $(T' , g')$ to get maps $j$ and $j'$ s.t   $g' = j \circ g$ and $g = j' \circ g'$.
Now to show $j$ is an $A-$module isomorphism,  since $g' = j \circ g ⇒ g' = j \circ (j' \circ g') ⇒ g' = (j \circ j') \circ g'$ ,hence  $j \circ j' =Id_{T'}$ and similarly,  $j' \circ j = Id_{T}$, hence $j$ is an isomorphism.
my problem is with this statement  $g' =  j \circ (j' \circ g') ⇒ g' = (j \circ j') \circ g'$ 
how can we change the order of brackets when $g'$ is a bilinear map , $j  ∈ End_A(T,T')$ , $j' ∈ End_A(T' , T)$ and we do not know if associativity is valid for whatever algebraic object(if any) these together live in.

REPLY [4 votes]: The composition of three functions is always an associative operation. It doesn't have anything to do with additional algebraic structure.
You can prove it independently of any algebraic structure:
Given $W\xrightarrow{h}X \xrightarrow{g}Y\xrightarrow{f}Z$, then $(f\circ g)\circ h=f\circ( g\circ h)$
In our case it's just
$M\times N\xrightarrow{g'}T' \xrightarrow{j'}T\xrightarrow{j}T'$