Converse of Uniqueness of Universal Property of Tensor Products? In Atiyah Macdonald, for a commutative and unital ring $A$ the tensor product of $A$-modules, $M,N$ is defined by the usual universal property:

A pair $(T,g)$ with $T$ an $A$-module and $g$ an $A$-bilinear map
  $g\colon M \times N \to T$ such that for any $A$-module $P$ with an
  $A$-bilinear map $f\colon M \times N \to P$ there exists a unique
  $A$-module homomorphism $h\colon T \to P$ such that $h \circ g = f$.

And then, as usually goes with universal properties:

Moreover, if $(T,g), (T',g')$ are any two pairs satisfying this
  condition, then there is a unique isomorphism $j \colon T \to T'$ such
  that $j \circ g = g'$

My question is, in the second highlighted selection, is the converse true? If I have already established the existence of the usual tensor product, $M \otimes_A N$, via quotient of free module, and I encounter some other pair $(T', g')$ and a unique isomorphism $j: T \to T'$ such that $j \circ g = g'$, can I conclude that $T'$ also satisfies the universal property defining the tensor product, thus I can regard both $T$ and $T'$ as the tensor product of this $M$ and $N$?
I believe I proved that I can indeed do this, but I need reassurance because I have overlooked small details with tensor products before, and don't want to make that mistake again. My proof was as follows:
Suppose $(T,g)$ satisfies the universal property and $(T',g')$ is a pair where $T'$ is an $A$-mod and $g'$ an $A$-bilinear map $g' \colon M \times N \to T'$. Suppose $j$ is the unique isomorphism $j\colon T \to T'$ such that $j \circ g = g'$, or $g = j^{-1} \circ g'$. Suppose $P$ is any $A$-mod with an $A$-bilinear map $f\colon M \times N \to P$. We wish to show there exists a unique $A$-module homomorphism $\phi\colon T' \to P$ such that $\phi \circ g' = f$. 
Since $(T,g)$ already satisfies the property by hypothesis, we know there exists a unique $A$-module homomorphism $\varphi \colon T \to P$ such that $\varphi \circ g = f$. So define the map $\phi \colon T' \to P$ by $\varphi \circ j^{-1}$. Then $\phi \circ g' = \varphi \circ j^{-1} \circ g' = \varphi \circ g = f$. So the diagram commutes as desired and the map $\phi$ is defined as the composition of two unique maps so is unique as well, thus $(T',g')$ satisfies the universal property.
Sorry that got a little (unnecessarily) complicated, I don't know how to draw diagrams on here, which really makes everything easier. As an aside, I feel like this probably is true and Atiyah & Macdonald didn't mention it in the text because they deemed it trivial.
 A: This is correct, although your argument for uniqueness of $\phi$ is not very clear: you chose to define it as a composition with certain maps that are unique, but how do you know you couldn't get a $\phi'$ that works and is not given by those compositions?  What you want to say is that if $\phi'$ makes the diagram commute for $(T',g')$, then the map $\varphi'=\phi'\circ j$ would make the diagram commute for $(T,g)$ by just swapping the roles of $(T,g)$ and $(T',g')$ in your argument.  Therefore by uniqueness of $\varphi$, $\varphi=\varphi'$, and so $\phi'=\varphi'\circ j^{-1}=\varphi\circ j^{-1}=\phi$.
Intuitively, what's going on here is that the isomorphism $j$ says that $(T,g)$ and $(T',g')$ have exactly the same structure (just the elements of $T$ have to be renamed via $j$ to get $T'$).  So any property of $(T,g)$ (which is defined using only the module structure on $T$ together with the map $g$) will also be true of $(T',g')$.  In particular, this applies to the universal property of the tensor product.
