Proof that $M\otimes N$ and $M^*\times N^* \longrightarrow \mathbb{R}$ are isomorphic

Question

I am familiarizing myself with the study of tensors between modules (for a further treatment on vector spaces). I understand how to think about tensor in terms of universal property. In physics I see the tensors being defined as a multilinear maps between vectors spaces and I understand that it is enough. I would like to prove that : \begin{align} M\otimes N \simeq M^*\times N^* \longrightarrow \mathbb{R} \end{align}

I found a reference that proceed the following way :

1) Proof that $M^* \otimes N \simeq M \longrightarrow N$ by finding a map between those two spaces that is a bijection.

2) Using the dual of M we have that $M \otimes N \simeq M^* \longrightarrow N$ ( is this even legal ? )

3) using the double dual of N we have $M^* \longrightarrow N \implies M^* \longrightarrow (N^*)^*$

4) Proof that $M^* \longrightarrow (N^*)^* \simeq M^*\times N^* \longrightarrow \mathbb{R}$

The proof are pretty involve and I got the grasp of them. But this is not good for my intuition. Basically I am working on a math book which define tensor using the universal property and treating object directly coming from the tensor product. On the physic side everything is presented in terms of multilinear map. I am just trying to find out why the isomorphism in question is true. Now that I found the proof I think there might be some easier way to prove it that would build more intuition.

I don't see myself putting on paper a multitude of proof just to remember that the isomorphism is valid.

So my question : Is there an simpler proof ? in one or 2 steps ? Thanks.

Note : The number 1) proof is the one that obscur me the most because it involve an other proof passing from using a trilinear map, applying the universal property on it to find a bilinear map.

Answer 1

You want to show there is an isomorphism

$$M\otimes N\simeq (M^\ast\otimes N^\ast)^\ast .$$

Because everything is finite dimensional, you may as well show that

$$(M\otimes N)^\ast \simeq M^\ast\otimes N^\ast.$$

If you have two functionals $\varphi\in M^\ast,\psi\in N^\ast$, you can define a bilinear map $F : M\times N\to \mathbb R$ so that $F(m,n) = \varphi(m)\psi(n)$, and by the universal property of tensor products this descends to a unique linear map such that $$F(m\otimes n)=\varphi(m)\psi(n).$$

This defines an arrow $M^\ast \times N^\ast \longrightarrow (M\otimes N)^\ast$ that you can check is also bilinear, so in fact we have obtained a map $M^\ast \otimes N^\ast \longrightarrow (M\otimes N)^\ast$ such that

$$\varphi\otimes \psi \mapsto (m\otimes n\mapsto \varphi(m) \psi(n))$$

This works for any pair of modules over a ring $A$, so you now have to use you're working over a field (i.e. with vector spaces) to construct an inverse, and you'll have to use bases.

Answer 2

The proof you mentioned seems quite natural already. The only major step is the first one. For the second step, set $M$ to be its dual, so that $M^*$ becomes $M^{**}$, which you can identify naturally with $M$. That is, one has an isomorphism $M\rightarrow M^{**}$ defined by mapping a vector $v$ to the function $M^*\rightarrow\mathbb{R}$ that takes $f\in M^*$ to $f(v)$. This is the usual way to identify a finite-dimensional vector space with its double dual.