The "characteristic property" of the tensor space

Question

This relates to page 309 of John Lee's textbook on Smooth Manifolds. Specifically Proposition 12.7, the characteristic property of the tensor product space.

I fundamentally fail to see how this would uniquely characterise the tensor product space up to isomorphism, even if the proposition is true:

Proposition 12.7 (Characteristic Property of the Tensor Product Space)

Let $V_1,...,V_k$ be finite-dimensional real vector spaces. If $A:V_1\times ...\times V_k\rightarrow X$ is any multilinear map into a vector space $X$, then there is a unique linear map $\tilde A:V_1\otimes ...\otimes V_k\rightarrow X$ such that the following diagram commutes:

where $\pi$ is the map $\pi(v_1,...,v_k)=v_1\otimes...\otimes v_k$.

Could someone explain in principle how this would uniquely characterise the tensor product space? The book provides a proof of the proposition, I'm just confused on the underlying logic being employed.

I think an emphasis on why the existence of "another candidate for the tensor product space" must be isomorphic to the first given the truth of the above.

Answer 1

Denote the tensor product by $T$. Assume there is another space $T'$ and a multilinear map $\pi': V_1\times...\times V_k\to T'$ with the same universal property: if $f:V_1\times...\times V_k\to X$ is multilinear then then there is a unique linear map $\tilde{f}:T'\to X$ such that $\tilde{f}\circ\pi'=f$.

Since $\pi: V_1\times...\times V_k\to T$ is multilinear, there is a unique linear map $\varphi:T'\to T$ such that $\varphi\circ\pi'=\pi$. Similarly, since $\pi':V_1\times...\times V_k\to T'$ is multilinear, there is a unique map $\psi:T\to T'$ such that $\psi\circ\pi=\pi'$.

So we get $(\varphi\circ\psi)\circ\pi=\pi$. So $\varphi\circ\psi:T\to T$ is a linear map which "extends" the multilinear map $\pi:V_1\times...\times V_k\to T$ to $T$. On the other hand, $id_T$ is another linear map which satisfies $id_T\circ\pi=\pi$, and so it also "extends" the same multilinear map. By the uniqueness of the extension, it follows that $\varphi\circ\psi=id_T$. Similarly, it can be shown that $\psi\circ\varphi=id_{T'}$. So $\psi:T\to T'$ is an isomorphism.

Answer 2

Yes! What you have remarked at the end is how we show that.

Let $W$ satisfies the characteristic property of tensor product. This means :

1. There is a map $\theta: V_1 \times \cdots \times V_k \to W$
2. Given any map $T : V_1 \times \cdots \times V_k \to X$, there exist a unique map $t : W \to X$ such that $$T = t \circ \theta$$

Now there should be a unique map $t_\pi : W \to V_1 \otimes \cdots \otimes V_k$ such that $\pi = t_\pi \circ \theta$ and a unique map $A_\theta : V_1 \otimes \cdots \otimes V_k \to W$ such that $\theta = A_\theta \circ \pi$.

Hence we get $$\pi = (t_\pi \circ A_\theta ) \circ \pi \quad \text{and} \quad \theta = (A_\theta \circ t_\pi) \circ \theta$$

Also we have $$\pi = \text{Id}_{V_1 \otimes \cdots \otimes V_k} \circ \pi \quad \text{and} \quad \theta = \text{Id}_W \circ \theta$$

But due to uniqueness, $$\text{Id}_{V_1 \otimes \cdots \otimes V_k} = t_\pi \circ A_\theta \quad \text{and} \quad \text{Id}_W = A_\theta \circ t_\pi$$

So, $V_1 \otimes \cdots \otimes V_k$ is isomorphic to $W$.