If $A$ is a noetherian local ring and $M$ an $A$-module, then we define the completion $\hat{M}$ of $M$ with respect to the stable $\mathfrak{m}$-filtration $\{M_n\}$ by $$\hat{M}=\left\{(a_1,a_2,...)\in\prod_{i=1}^\infty M/M_i:a_j\equiv a_i\bmod{M_i}\,\,\forall j>i\right\}.$$ See also my previous question.
Now in the book I use (A SINGULAR Introduction to Commutative Algebra by Greuel/Pfister), there is no universal property of this completion mentioned, but once it uses something that looks like one: We have a map from $K[x_1,...,x_n]_{\langle x_1,...,x_n\rangle}$ to some complete ring, hence we got a map from $K[[x_1,...,x_n]]$ to it. Is that the 'universal property of the completion of a ring / module', and if yes, is it somehow obvious from my definition of the completion, so that we could use it directly?
How to prove this property with the above definition? (If it really works for modules; I don't know, at least for rings I guess it should be something like: If $A\to B$ is a ring homomorphism and $B$ is complete, then there is a unique map $\hat{A}\to B$ extending it). Well, I think I maybe got a clue right now, and you could perhaps tell me if this is the correct way (I'd still like to know if this is 'the' universal property of the completion):
If $A\to B$ is a ring homomorphism ($A$ and $B$ noetherian local rings; does this homomorphism have to be local, too?), and $B$ is complete, then I get and induced map $\hat{A}\to\hat{B}=B$ as wanted.