Hi all. This question is related to my previous one on tensor products: Is there a way of "well-defining" a function on a tensor product M(x)N (where M,N are both R-modules) ? This is the motivating example for my question : Say we want to define a map f: M(x)M-->M by f(m(x)m')=m+m' . But then we have that (m(x)m')=(-m(x)-m') , so that f(m(x)m')= f(-m(x)-m')=-m-m' . But in most cases we do not have m+m' =-m-m' , so this function is not well-defined, i.e., its value depends on the choice of representative . Clearly we must find a way of defining a function that is constant in the classes (m(x)n) , but, how do we do that? One way would be to find a bilinear map defined on MxN and determining its image on M(x)N under the map (m,n)--> (m(x)n) , since the commutative triangle guarantees that the map is well-defined. Is there some other way of doing this, i.e., of "well-defining" maps on M(x)N ? I suspect a necessary and sufficient condition is that a candidate function satisfies the tensor relations: i ) ((a+a')(x)b) = (a(x)b)+(a'(x)b) ii) (a (x) (b+b'))=(a(x)b)+ a(x)b' iii) r(a(x)b)=(ra(x)b)= (a(x)rb) but I can't see a good way of proving this. Anyone? Thanks.