About tensor algebras homomorphisms

Hi, just want to clear up something,

In class we saw an example of proving an isomorphism involving tensor algebras, something of the form M \otimes N \cong L. We used the universal property of tensor product, finding a bilinear map which give us map f:M \otimes N\rightarrow L, then we tried to find an inverse map or just to prove that f was surjective and one to one.

My question is, ¿Aren’t we forgetting to check that f satisfies f(ab)=f(a)f(b)? I mean, the universal property doesn’t yield an homomorphism of algebras at all. I understand that it’s just a linear map. ..

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5 comments

  1. katrina888

    Servando and Cecilia pointed that fact to me yesterday, so yes, we have to check that.

  2. briansdumb

    I disagree that you must check that f is a homomorphism.
    I’m open to being corrected, but I understood that the bilinear map “descends” to a homomorphism. If this is correct, then we do not have to validate that f is a homomorphism because the existence of homomorphism f is implied by the bilinearity of the map from the direct product.
    I think that if you want to show that f is an isomorphism, then yes you must check that it is bijective.

  3. yimp34, this is a great post, thank you! You are absolutely right.
    Moral of the story: Don’t be sloppy! (I was. Sorry. 🙂 I am not very lucid in the morning.)

    Let me explain this better:

    When you say that something is a “homomorphism”, it is crucial to clarify what kind of homomorphism it is: a vector space homomorphism (which is the same as a linear map), a K-algebra homomorphism, etc.
    Also, when you write something like V or V \otimes W, it is crucial to know whether you are thinking of these as K-vector spaces, as K-algebras, etc.

    When V,W, and L are K-vectors spaces and you have a bilinear map f: V \times W \rightarrow L, the universality theorem tells you that it descends to a *linear map* \tilde{f}: V \otimes W \rightarrow L of K-vector spaces. We proved this. We used this in examples 1, 2, and 3 to produce linear maps:

    K \otimes A \rightarrow A (for A a K-algebra),

    \mathbb{C} \otimes \mathbb{R}[x] \rightarrow \mathbb{C}[x], and

    V \otimes W \rightarrow V for any K-vector spaces V and W.

    We also checked that the first two are in fact isomorphisms *as K-vector spaces*. (In example 2, K=R.)

    Now, in examples 1 and 2, I also claimed that these are isomorphisms as K-algebras. To prove that, I should have checked that these are also *K-algebra homomorphisms*. For that I need them to also respect the ring structure, i.e., to satisfy f(ab)=f(a)f(b). This is straightforward, but must be checked.
    Once we do this, since we already checked that they are bijective, we know they are in fact *isomorphisms* of K-algebras.

    In example 3, I only claimed that I had a homomorphism of K-vector spaces f_g: V \otimes W \rightarrow V for each linear map g: W \rightarrow K. This is all I needed to prove the next lemma. In fact, if V and W are also K-algebras, f_g is not necessarily a K-algebra homomorphism. Here is a nice exercise for you:
    Prove that f_g is a homomorphism of K-algebras if and only if g is a homomorphism of K-algebras (i.e. it is also multiplicative).

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