# Checking the universal property of tensor products of modules

I'm wondering about a general procedure for checking that an object satisfies the universal property of tensor products. (Similar questions already asked seem to consider only special cases and use, to various extent, ad hoc solutions. I don't feel confident about my understanding of the general principles.)

Let $M$ and $N$ be modules over a ring $R$. Assume that we want to show that an abelian group, $T$, together with a bihomomorphism (aka bilinear map or balanced map) $\tau:M\times N\rightarrow T$, satisfies the universal property of tensor products. I would proceed in the following way:

1. Consider an arbitrary bihomomorphism $\sigma:M\times N\rightarrow G$, where $G$ is an abelian group. We need to show that there is a unique group morphism $\sigma':T\rightarrow G$ fulfilling $\sigma'\circ\tau=\sigma$. This of course forces us to define, for $x\in M\times N$, $\sigma'(\tau(x))=\sigma(x)$.
2. Show that $\sigma'$ as given above is a well-defined map. This proves existence of the factorization.
3. Show that $\sigma'$ is a group morphism.
4. Show that the image of $\tau$ generates $T$ as a group. By the previous point, this shows uniqueness of the factorization.

My first question: Is the above a good general strategy (sufficient and irredundant)?

My second question: In the solution of problem 1 found in http://www2.math.uu.se/~khf/LoesM+H6.pdf a seemingly different approach is used, and it feels like several important things are omitted (for instance my step 3 above, and step 2 is not very clear either). Am I missing something here?

 To check the universal property of an object, usually we have to first define or construct it explictly, and in this process, we implicitly use the universal property, and the proof of that the object constructed above satisfies the universal property is usually easy. So, in your example, we often first define $T$ as the quotient module of $M\times N$ module the submodule generated by the bilinear relations. – ougao Dec 13 '12 at 18:54