Why does Z have a universal mapping property in tensor product existence?

[Math Amateur]

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:

In the above text we read the following:" ... ... Recall that $V_1 \times \ ... \ \times V_m = X$ and that $Z$ is a vector space based on $X$. Since $W$ is a vector space and $f$ is a map from $X$ to $W$, by the universal property of $Z$ there exists a unique linear transformation $S \ : \ Z \longrightarrow W$ such that $S$ restricted to $X$ is $f$. ... ..."Now I have summarised the mappings involved in Theorem 10.1 in Figure 1 below ... ...

My question is as follows:

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

Peter

*** NOTE ***

... ... oh no! ... ... I think I have just realized the answer to my question ... hmm ... embarrassingly simple ... ... I think that $Z$ has a UMP because $( Z, \iota )$ is assumed to be the vector space based on the set $X$... 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:

 

[WWGD]

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.

 

[Math Amateur]

Thanks WWGD ... most helpful ...

Peter

 

[WWGD]

Thanks WWGD ... most helpful ...

Peter

Glad it helped, Peter, I went through my own pain trying to understand it -- I feel your pain :).

 

[Math Amateur]

Thanks again WWGD ... good to have your support ...

Peter