Detail in proving existence of tensor products I am reading the proof that Tensor products exist (Prop 2.0.2) on page 393 of Paul Garrett's Abstract Algebra (page 5 of this pdf). A few lines into the proof it says

Given a bilinear map $\varphi:M \times N \to X$, by properties of free modules there is a unique....

Unfortunately, I don't know which properties of free modules are being used for that statement. I have studied some basic module theory before so perhaps I just need to be reminded of the right thing, or maybe I need to learn more. Can someone please tell me what I am missing?
 A: What's one of your favorite properties about vector spaces? Surely one must be that to define a linear transformation from a vector space $V$ to another space $W$, all you have to do is choose a basis $B$ for $V$ and then specify where the basis vectors go. Once you have done this, you have uniquely defined a linear transformation $V \to W$ (this is usually called extending by linearity).
There is a technical term for this property: We say $V$ is "free" on $B$.
A free module shares this property with vector spaces. When the author says that $F$ is free on $M\times N$, he means that if you are given another module $X$ and you would like to define an $R$-linear map $F \to X$, all you have to do is choose a destination (in $X$) for the elements of $M\times N$. This amounts to choosing a function $\phi:M \times N \to X$ (in this case it's not just a set function, but bilinear too). Once you have chosen a target for the elements of $M\times N$, you extend by linearity to a unique $R$-linear map (he calls it $\Phi$).
A: Note that $F$ is defined as the free module on the set $M\times N$, i.e. $F=\{r_1x_1+\cdots+r_nx_n : x_1,\ldots,x_n\in M\times N\}$. Define $\Psi:F\to X$ by $\Psi(r_1 x_1+\cdots+r_1x_n)=r_1\varphi(x_1)+\cdots+r_n\varphi(x_n)$. Note that since $i(x)=x$, the diagram is commutative, and that for any $\Psi'$ for which the diagram commutes we must have $\Psi'(x)=\varphi(x)$ so $$\Psi'(r_1x_1+\cdots+r_nx_n)=r_1\Psi'(x_1)+\cdots+r_n\Psi'(x_n)=r_1\varphi(x_1)+\cdots+r_n\varphi(x_n)$$
thus $\Psi'=\Psi$.
