Hi, I understand the tensor product of modules as a new module in which every bilinear map becomes a linear map. But now I am trying to see the Tensor product of modules from the perspective of maps factoring through, i.e., from properties that allow a commutative triangle of maps. As a concrete example of what I mean: For homomorphisms f: A-->B and g: A-->C , a condition of the kernels allow the existence of a commutative triangle , i.e., the conditions allow the existence of a map h with the necessary properties so that hog=f. More specifically: If f: A -> B is a homomorphism of groups, and g: A -> C is a surjective homomorphism of groups, we have that f "descends to a homomorphism" of groups h: B -> C iff the kernel of g is contained in the kernel of f. In other words, there is an h with hog=f, which allows f to "factor through" and the associated triangle is commutative . Sorry, I don't know how to draw triangles here in ASCII. ** Now ** I'm curious as to how moding out the free module on a product VxW of R-modules by the standard necessary relations on the tensor product: i ) (a+a',b)=(a,b)+(a',b) and: ii) (a,b+b')=(a,b)+(a,b') allows for the existence of the linear map L that completes the commutative triangle, so, my question is: what specific algebraic result/theorem are we using to guarantee that imposing the above relations allow the existence of a linear map L : V(x)W --> Z , for an R-module Z, to factor thru the maps: (x): V x W --> V(x)W , and B: V x W --> Z i.e., L is the linear map defined on the tensor product V(x) W that represents the bilinear map B defined on the product V xW into Z ? Thanks, WWGD: What Would Gauss Do?