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I Proof of Existence of Tensor Product ... Further Question ..

  1. Mar 17, 2016 #1
    I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

    I am focused on Section 10.1 Introduction to Tensor Products ... ...

    I need help with another aspect of the proof of Theorem 10.1 regarding the existence of a tensor product ... ...


    The relevant part of Theorem 10.1 reads as follows:


    ?temp_hash=b2111ef8ae3decfb3487a4288419e3ab.png
    ?temp_hash=b2111ef8ae3decfb3487a4288419e3ab.png
    ?temp_hash=b2111ef8ae3decfb3487a4288419e3ab.png



    In the above text we read the following:


    " ... ... Recall that [itex]V_1 \times \ ... \ \times V_m = X[/itex] and that [itex]Z[/itex] is a vector space based on [itex]X[/itex]. Since [itex]W[/itex] is a vector space and [itex]f[/itex] is a map from [itex]X[/itex] to [itex]W[/itex], by the universal property of [itex]Z[/itex] there exists a unique linear transformation [itex]S \ : \ Z \longrightarrow W[/itex] such that [itex]S[/itex] restricted to [itex]X[/itex] is [itex]f[/itex]. ... ..."


    Now I have summarised the mappings involved in Theorem 10.1 in Figure 1 below ... ...




    ?temp_hash=103d7bc653fad9a1cc4300e9c88a4a21.png





    My question is as follows:

    Why does [itex]Z[/itex] have a universal mapping property ...? ... ... and indeed if [itex]Z[/itex] has one, why doesn't [itex]V[/itex] ... ... giving us the relationship [itex]T \gamma = f[/itex] that we want ... what is special about [itex] Z [/itex]?



    Hope someone can help ...

    Peter

    *** NOTE ***

    ... ... oh no! ... ... I think I have just realised the answer to my question ... hmm ... embarrassingly simple ... ... I think that [itex] Z [/itex] has a UMP because [itex]( Z, \iota )[/itex] is assumed to be the vector space based on the set [itex] X [/itex]... and vector spaces based on a set have a UMP ... is that right? ... see Cooperstein Definition 10.1 on the first page of Section 10.1 provided below ...

    Can someone confirm that this is the reason Z has a Universal Mapping Property ...

    Peter



    ==========================================================


    *** NOTE ***


    It may help readers of the above post to be able to read Cooperstein's introduction to Section 10.1 where he covers, among other things, the notion of a vector space being based on a set and the idea of the universal mapping problem ... ... so I am providing this text as follows:


    ?temp_hash=c9f6a4cb768e80949ddfbbb6334eb0e0.png
    ?temp_hash=c9f6a4cb768e80949ddfbbb6334eb0e0.png
    ?temp_hash=c9f6a4cb768e80949ddfbbb6334eb0e0.png
    ?temp_hash=c9f6a4cb768e80949ddfbbb6334eb0e0.png
     

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    Last edited: Mar 17, 2016
  2. jcsd
  3. Mar 18, 2016 #2

    WWGD

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    I think a helpful thing to know/understand is the concept of maps factoring through (the quotient, here): https://en.wikipedia.org/wiki/List_of_mathematical_jargon in algebra, there are conditions on the kernel of a map. Factoring through is in analogy with the factoring of numbers as products.

    You are given functions f:A-->C , g: A-->B . Then f factors through g if there is an h: with f=hg ; h is a map from B-->C. This is I think the clearest way of understanding the tensor product. For vector spaces V,W , the tensor product , ## V \ Oline W ## is a vector space in which every bilinear map defined on VxW into a third vector space Z factors through a linear map from the tensor product into Z. The conditions on the kernel guarantee that maps factor through. I will look up the conditions on the kernel and image of the respective groups and get back with it.
     
  4. Mar 18, 2016 #3
    Thanks WWGD ... most helpful ...

    Peter
     
  5. Mar 18, 2016 #4

    WWGD

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    Glad it helped, Peter, I went through my own pain trying to understand it -- I feel your pain :).
     
  6. Mar 18, 2016 #5
    Thanks again WWGD ... good to have your support ...

    Peter
     
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