How can the existence of the tensor product be proven in Federer's construction?

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SUMMARY

The existence of the tensor product in Herbert Federer's construction is established through the vector space F, which consists of real-valued functions on the Cartesian product of vector spaces V1, V2, V3, ..., Vn. The linear map φ is defined to map tuples from these vector spaces to F, and the subspace G is generated by specific linear combinations of these mappings. The quotient space F/G is identified as the tensor product, with the n-linear map μ linking it to any vector space W. The existence of the linear map g from F/G to W is proven by demonstrating that g maps elements of G to zero, allowing for a unique factorization.

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turiya
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Hi all,

I was reading the book by Herbert Federer on Geometric Measure Theory and
it seems he proves the existence of the Tensor Product quite differently
from the rest. However it is not clear to me how to prove the existence of the linear map "g" in his construction.

He defines F as the vectorspace consisting of all real valued functions on
V1 x V2 x V3 x V4 ... x Vn which vanish outside some (varying) finite set.
(Here, V1, V2, V3, V4, ... Vn are all vector spaces). Now consider the map
\phi: V1 x V2 x V3 x V4 ... Vn -> F where \phi(v1,v2,v3,...,vn) is the function
with value 1 at (v1,v2,...,vn) and zero elsewhere.

If we let G to be the vector space generated by all elements of two types:
1) \phi(v1,..,vi-1,x,vi+1,..,vn) + \phi(v1,..,vi-1,y,vi+1,..,vn) - \phi(v1,..,vi-1,x+y,vi+1,..,vn)
2) \phi(v1,..,vi-1,cvi,vi+1,..,vn) - c\phi(v1,..,vi-1,vi,vi+1,..,vn) (c is a real number)
then G a sub-space of F.

Now F/G is considered to be the the Tensor product of V1, V2, .. Vn and \mu = r \dot \phi is the n-linear map associated with the Tensor product, where "r" is the quotient map from F to F\G. Ofcourse, the theorem says that for every n-linear map f : V1xV2x..xVn -> W there exists a unique linear map g : F/G -> W such that f = g \dot \mu for any vector space W.

I have tried and searched a lot to prove the existence of such a "g" but to no avail. Most other texts use a different definition of Tensor product and so any help in this is greatly appreciated.

Thanks in advance
Phanindra
 
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The point is that \{\varphi(v_1,...,v_n)~\vert~v_i\in V_i\} is a basis for F.
So, given a map f:V_1\times...\times V_n\rightarrow W, we can make a map g:F\rightarrow W. It is enough to put g(\varphi(v_1,...,v_n))=f(v_1,...,v_n) and extend this linearly. So, this gives a map g:F\rightarrow W.

Now, it suffices to show that g(G)=\{0\}. Since in that case, g will factor as h:F/G\rightarrow W by h([v])=g(v)...
 
Thanks a lot micromass. Your steps indeed prove that "h" is the unique linear map I am looking for.
 

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