Editing Tensor product (section)

=== Universal property ===
[[File:Another universal tensor prod.svg|class=skin-invert-image|right|thumb|200px|Universal property of tensor product: if {{math|''h''}} is bilinear, there is a unique linear map {{math|{{overset|lh=0.7|~|''h''}}}} that makes the diagram [[commutative diagram|commutative]] (that is, {{math|1=''h'' = {{overset|lh=0.7|~|''h''}} ∘ ''φ''}}).]]

In this section, the [[universal property]] satisfied by the tensor product is described. As for every universal property, two objects that satisfy the property are related by a unique [[isomorphism]]. It follows that this is a (non-constructive) way to define the tensor product of two vector spaces. In this context, the preceding constructions of tensor products may be viewed as proofs of existence of the tensor product so defined.

A consequence of this approach is that every property of the tensor product can be deduced from the universal property, and that, in practice, one may forget the method that has been used to prove its existence. 

The "universal-property definition" of the tensor product of two vector spaces is the following (recall that a [[bilinear map]] is a function that is ''separately'' [[linear map|linear]] in each of its arguments):
:The ''tensor product'' of two vector spaces {{mvar|V}} and {{mvar|W}} is a vector space denoted as {{tmath|1= V\otimes W }}, together with a bilinear map <math>{\otimes} : (v,w)\mapsto v\otimes w</math> from <math>V\times W</math> to {{tmath|1= V\otimes W }}, such that, for every bilinear map {{tmath|1= h : V\times W\to Z }}, there is a ''unique'' linear map {{tmath|1= \tilde h : V\otimes W\to Z }}, such that <math>h=\tilde h \circ {\otimes}</math> (that is, <math>h(v, w)= \tilde h(v\otimes w)</math> for every <math>v\in V</math> and {{tmath|1= w\in W }}).<!-- Commenting out considerations that are misplaced here; kept for a possible use elsewhere.

This characterization can simplify proofs about the tensor product. For example, the tensor product is symmetric, meaning there is a [[canonical isomorphism]]:
<math display="block">V \otimes W \cong W \otimes V.</math>

To construct, say, a map from <math>V \otimes W</math> to {{tmath|1= W \otimes V }}, it suffices to give a bilinear map <math>h: V \times W \to W \otimes V</math> that maps <math>(v,w)</math> to {{tmath|1= w \otimes v }}. Then the universal property of <math>V \otimes W</math> means <math>h</math> factors into a map {{tmath|1= \tilde{h}:V \otimes W \to W \otimes V }}.
A map <math>\tilde{g}:W \otimes V \to V \otimes W</math> in the opposite direction is similarly defined, and one checks that the two linear maps <math>\tilde{h}</math> and <math>\tilde{g}</math> are [[Inverse function|inverse]] to one another by again using their universal properties.

The universal property is extremely useful in showing that a map to a tensor product is injective. For example, suppose we want to show that <math>\R \otimes \R</math> is isomorphic to {{tmath|1= \R }}. Since all simple tensors are of the form {{tmath|1= a \otimes b = (ab) \otimes 1 }}, and hence all elements of the tensor product are of the form <math>x \otimes 1</math> by additivity in the first coordinate, we have a natural candidate for an isomorphism <math>\R \rightarrow \R \otimes \R</math> given by mapping <math>x</math> to {{tmath|1= x \otimes 1 }}, and this map is trivially surjective.

Showing injectivity directly would involve somehow showing that there are no non-trivial relationships between <math>x \otimes 1</math> and <math>y \otimes 1</math> for {{tmath|1= x \neq y }}, which seems daunting. However, we know that there is a bilinear map <math>\R \times \R \rightarrow \R</math> given by multiplying the coordinates together, and the universal property of the tensor product then furnishes a map of vector spaces <math>\R \otimes \R \rightarrow \R</math> which maps <math>x \otimes 1</math> to {{tmath|1= x }}, and hence is an inverse of the previously constructed homomorphism, immediately implying the desired result. A priori, it is not even clear that this inverse map is well-defined, but the universal property and associated bilinear map together imply this is the case.

Similar reasoning can be used to show that the tensor product is associative, that is, there are natural isomorphisms
<math display="block">V_1 \otimes \left(V_2\otimes V_3\right) \cong \left(V_1\otimes V_2\right) \otimes V_3.</math>
Therefore, it is customary to omit the parentheses and write {{tmath|1= V_1 \otimes V_2 \otimes V_3 }}, so the ''ijk-th'' component of <math> \mathbf{u} \otimes \mathbf{v} \otimes \mathbf{w} </math> is 
<math display="block">(\mathbf{u} \otimes \mathbf{v}\otimes \mathbf{w})_{ijk} = u_i v_j w_k,</math>
similar to the first [[#Tensors in finite dimensions, and the outer product|example]] on this page.

The category of vector spaces with tensor product is an example of a [[symmetric monoidal category]].
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