# Existence of isomorphism between tensor products.

In multilinear algebra many maps are usually proven to exist rather than simply defined. For example, commutativity is one such example. In the book I'm studying the author says: let $V_1,\dots,V_k$ be a collection of vector spaces over $K$, then if $\sigma \in S_k$ there is a linear isomorphism

$$f_\sigma : V_1\otimes\cdots\otimes V_k\to V_{\sigma(1)}\otimes\cdots\otimes V_{\sigma(k)}$$

such that $f_\sigma(v_1\otimes\cdots\otimes v_k)=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)}$. Then to show this, he defines

$$g : V_1\times\cdots\times V_k\to V_{\sigma(1)}\otimes\cdots\otimes V_{\sigma(k)}$$

by $g(v_1,\dots,v_k)=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)}$. He argues then that $g$ is multilinear and by the universal property there is a linear $f_\sigma$ such that

$$f_\sigma(v_1\otimes\cdots\otimes v_k)=v_{\sigma(1)}\otimes \cdots\otimes v_{\sigma(k)}$$

and since this turns basis into basis is isomorphism.

That's pretty fine, I understand this proof, I just don't understand the need for it. Why do we need to show that $f_\sigma$ exists this way? Couldn't we simply say: pick $f_\sigma$ defined on the factorizable tensors by

$$f_\sigma(v_1\otimes\cdots \otimes v_k)=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)}$$

and extend by linearity. Then $f_\sigma$ is linear, turn basis into basis and is isomorphism.

Why do we need to first define a multilinear map, then show that $f_\sigma$ exists using the universal property? This just one example, there are many others where the same thing happens (associativity, existance of tensor product of linear maps and so on).

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Define $\newcommand{\RR}{\mathbb{R}}f : \RR \otimes \RR \to \RR$ by $$f(x \otimes y) = x + y$$ Sounds reasonable, right? Think again... Because this map is not well defined! Indeed, $f((1+1) \otimes 1) = (1+1)+1 = 3$. But $(1+1) \otimes 1 = 1 \otimes 1 + 1 \otimes 1$, and if you wanted to extend by linearity, you would find $f(1 \otimes 1 + 1 \otimes 1) = f(1 \otimes 1) + f(1 \otimes 1) = 2+2 = 4$.
I see the problem now. So, instead of defining $f_\sigma$ and then checking it's linear, we simply use the universal property to check linearity of $f_\sigma$ more directly? Thanks for your answer. –  user1620696 Nov 9 '13 at 14:32
Yes, it's partially that. The other problem is that you cannot necessarily directly define $f_\sigma$ directly. –  Najib Idrissi Nov 9 '13 at 15:18