Complete Derivation: Universal Property of the Tensor Product

Previous tensor product video: • A Concrete Introductio...
The universal property of the tensor product is one of the most important tools for handling tensor products. It gives us a way to define functions on the tensor product using bilinear maps. However, the statement of the universal property can be confusing if it is presented without background. This video is an explanation of the universal property that proves it for a concrete instantiation of the tensor product of vector spaces (or modules over a commutative ring).
Tensor Products playlist: • Tensor Products
0:00 Introduction
3:04 Constructing the Tensor Product
7:54 Bilinear Maps
10:39 Maps on the Tensor Product
16:17 Defining g
26:24 Linearity and Uniqueness
29:44 Universal Property
30:50 Example
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Music: C418 - Pr Department

[@paoloemankin4486]

I've always said that things thought to be difficult are very often simply poorly explained. This video is a confirmation: it makes things considered difficult easy. Thank you.

[@Happy_Abe]

This has got to be the best video on tensor products on KZbin. I hope you make more videos on this stuff!

[@tomlopfer]

Very well done! I really liked how you discussed all details 100% concisely.

[@SanuIITM15]

Fantastic explanation ❤️

[@kdr1895]

Always as clear as crystal🎉🎉🎉

[@zhuolovesmath7483]

Great skills of explaining!

[@StratosFair]

Thank you for this enlightening lecture

[@keerthanajaishankar1883]

I thank you so much for this video, helped me a lot!

[@luisaim27]

Awesome video!!! Thank you so much!!!

[@davidescobar7726]

Great video! ❤

[@tmjz7327]

amazing video.

[@madmath1971]

Consider the vector space V=RxR and a bilinear map f from V to R. Fix the standard basis e_i so that f is represented by the matrix A with element a_ij =f(e_i,e_j). This way f(x,y)= in the standard way being the standard scalar product. So basically your correspondence takes A to f and is a bijection. The tensor product is then the space the linear map represented by A acts uniquely. This work for finite dimensional vector spaces or even finitely generated R modules over a commutative ring R. Do you agree?

[@andresxj1]

But does the converse hold? This is, every linear well-defined map on the tensor product comes from a bilinear map on the cartesian product?

[@MuPrimeMath]

Yes, the converse holds. In other words, for every linear map g on V⊗W there exists a unique bilinear map f on V×W such that g ∘ τ = f. Uniqueness is obvious from the equation g ∘ τ = f because this specifies that, as functions, f is equal to g ∘ τ. We can write the map as f(v,w) = g(τ(v,w)) = g(v⊗w). Checking that this map is bilinear follows straightforwardly from the construction of the tensor product as a quotient space shown in this video and the fact that g is assumed to be linear.

[@andresxj1]

@@MuPrimeMath I see. Thanks a lot. Keep up the good work!

[@marcuschiu8615]

4:40 Is v*w not a multiple of 2v*2w?

[@MuPrimeMath]

Remember that * is not denoting ordinary multiplication in this case. v*w just means "the basis vector associated with the pair (v,w)". Since each distinct pair has its own basis vector, v*w and 2v*2w are distinct basis vectors, hence not multiples of each other.

[@sonamanimaity5543]

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