# Tensor Products and Maps Factoring Through

Gold Member
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?

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fresh_42
Mentor
Let's go back to the modules. Here is the formal definition:

The tensor functor ##T\, : \,(\operatorname{Mod}_R \times \operatorname{Mod}_R)\times \operatorname{Mod}_R \longrightarrow \operatorname{Set} ## defines a couniversal transformation problem which is solvable. A representing pair ##(M \otimes_R N\, , \,u) ## of ##T(M\times N \times -)\, : \,\operatorname{Mod}_R \longrightarrow \operatorname{Set}## is called tensor product of ##M## and ##N##.

Let ##T## be a contravariant in the first and covariant in the second variable functor.
The question whether there for any object ##M \in \operatorname{Mod}_R## the covariant functor ##T(M,-,-)\, : \,\operatorname{Mod}_R \longrightarrow \operatorname{Set}## is representable is called the couniversal transformation problem.

So you see there are some commutative diagrams involved, although hidden in the contra- and covariance, or all quantors. The connection is the following theorem:

The couniversal transformation problem ##T## is solvable, if and only if for any module ##M\in \operatorname{Mod}_R## there is a module ##M \times N\in \operatorname{Mod}_R\times \operatorname{Mod}_R## and a module homomorphism ##\varphi\, : \,M \longrightarrow M\otimes N## such that: Every module homomorphism ##\psi\, : \,M \longrightarrow P## determines exactly one module homomorphism ##\overline{\psi}\, : \,M\otimes N \longrightarrow P## with ##\psi = \overline{\psi} \circ \varphi ## (from right to left).

This is why it's called "(co)universal". It connects the tensor product with an arbitrary module homomorphism.

i don't really understand ur question. what do u mean what theorem is need? this is an elementary result that u use for non trivial theorems, u dont prove elementary properties with other theorems. the tensor product V(x)W is the "freest" way of having formal sums (v1,w1)+(v2,w2)+...(vn,wn) such that always (v,w)+(y,w)=(v+y,w). this is the general method for constructing "free objects" that satisfy certain conditions, by taking formal strings/sums then modding out by whatever relation u need. the existence of L is because B induces a linear map T: F(VxW)->Z, where F(VxW) is the free module on the set VxW, and clearly T maps everything u mod out by in the tensor product V(x)W=F(VxW)/~ to 0, so T induces a map from V(x)W to Z, the map u want.

also this is only for commutative rings. its been ages so i cant remember, but im pretty for non commutative rings it's a different story.

EDIT: just checked wikipedia u also need the relation r(v,w)=(rv,w)=(v,rw) because bilinear maps preserve scalar multiplication in either variable. https://en.wikipedia.org/wiki/Tensor_product#The_definition_of_the_abstract_tensor_product been ages so i cant remember anything.

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Gold Member
Yes,this was 5 years ago when I first read on the topic.

fresh_42
fresh_42
Mentor
Yes,this was 5 years ago when I first read on the topic.
I could have deleted it, but I couldn't restrain from giving a lesson (not to you, as I just checked my list and not even recognized it was you, but to other readers) on universality of functors (hoping I didn't mess with the translation and adaptation on ##T##).

##\operatorname{Hom} \ldots ##

WWGD
mathwonk
Homework Helper
2020 Award
To just try to answer the original question as directly as possible, in the spirit of blackhole's answer, the desired theorem says that every bilinear map from VxW factors through a linear map on VtensW. This is done in two stages: first the theorem that every function on a basis of a free module induces a linear map on the module gives you a linear map on F(VxW). Then next, the theorem WWGD quotes in his 4th paragraph, about descending a homomorphism, is used. I.e. note that if the original map on VxW was bilinear, then the submodule of relations he gave is contained in the kernel of the induced linear map on F(VxW), hence by the theorem in his 4th paragraph, there is a descended linear map defined on VtensW. That's all, 1) extending a map on a basis, plus 2) descending a map to a quotient.

WWGD
mathwonk
Homework Helper
2020 Award
Somewhat in the spirit of fresh's answer, i.e. of representable functors, given modules V and W, we start from the functor that assigns to a module *, the module Bil(VxW, *) of bilinear maps VxW-->*, and we ask to "represent" it by a "Hom" functor. I.e. we ask for (the existence of) a module X such that there is a natural isomorphism Hom(X,*) ≈ Bil(VxW, *) for all modules *, where Hom(X,*) = the module of linear maps X-->*. I.e. changing Bil(VxW, *) into Hom(X,*), is replacing bilinear maps on VxW, by linear maps on X, as stated in the OP's first sentence.

The key to this is to observe that a bilinear map on VxW defines a linear map from V to linear maps on W, i.e. there is a natural isomorphism Bil(VxW, *) ≈ Hom(V, Hom(W, *)). This is just the definition of "bilinear"; i.e. for each element v of V, the restriction of the bilinear map to {v}xW is linear on W, and this assignment of a linear map on W to each element of V is linear on V. Thus our functor of bilinear maps is actually a composition of Hom functors. Then there is a general theorem that any composition of Hom functors is itself a Hom functor. I.e. there does exist a module X such that Bil(VxW, *) ≈ Hom(V, Hom(W, *)) ≈ Hom(X, *). That X is then the tensor product of V and W.

I.e. the general theorem is that a Hom functor is characterized by the fact that it preserves injections and "inverse limits" (a generalization of direct products). I.e. any operation on modules and maps that preserves injections and inverse limits must be isomorphic to the operation of taking linear maps on some new module, i.e. is a Hom functor. Now since each Hom functor preserves these things, so does their composition. Hence the composition of two (or more) Hom functors is also a Hom functor.

(If Y-->Z is an injection, so is the induced map Hom(X,Y)-->Hom(X,Z), and if Z = ∏Uj, then Hom(X,Z) ≈ ∏Hom(x,Uj), and similarly for inverse limits.)

Notice this more sophisticated argument gives you less, in this specific case, than the concrete one above, since it does not tell you how to actually find the representing module VtensW . I.e. this answers the question of whether bilinear maps on VxW can be represented by linear maps from some module X, with "yes", but it does not discuss how to find that module X. Presumably the proof of the general representability theorem would do so.

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