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WWGD

Science Advisor

Gold Member

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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?

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?

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