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.

  • $\begingroup$ I don't understand why associativity may not be valid. If you are talking about usual module homomorphisms, then they are functions, and function composition is associative. $\endgroup$
    – Tunococ
    May 14, 2014 at 11:26
  • 1
    $\begingroup$ You seem to have (slightly) misquoted the definition of tensor product; the homomorphism $h$ should be required to be unique. $\endgroup$ May 14, 2014 at 11:31
  • $\begingroup$ yes you are right. $\endgroup$ May 14, 2014 at 13:12
  • $\begingroup$ Is the $A-$bilinear mapping $g:M\times N \to T$ onto? $\endgroup$
    – Saikat
    Feb 26, 2021 at 10:23

1 Answer 1


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

$M\times N\xrightarrow{g'}T' \xrightarrow{j'}T\xrightarrow{j}T'$


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