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Tensor Products and Maps Factoring Through

  1. May 23, 2014 #1

    WWGD

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    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?
     
    Last edited: May 23, 2014
  2. jcsd
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