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. ..

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.

• briansdumb

Ok, on re-reading the lecture notes I have changed my mind. I’ll revise my hw. Thanks for the post!

• Sebastián O.

Moreover, we have to check that is a $\mathbb{F}$-algebra homomorphism, more than a simple homomorphism, am I right?

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).