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I'm going over some notes I took from the blackboard, and reached a slight hitch. I thought that maybe someone could help.

Let $E,F$ be vector spaces over a field $\mathbb K$. A tensor product of $E$ and $F$ is a pair $(E\otimes F, r)$, where $E\otimes F$ is a vector space over $\mathbb K$ and $r:E\times F\to E\otimes F$ is bilinear, and the following property holds: if $G$ is a $\mathbb K$-vector space, and $\varphi: E\times F\to G$ is bilinear, then $\exists ! f_{\varphi}:E\otimes F\to G $ such that $f_{\varphi}\circ r = \varphi$.

I've understood the proof that, if a tensor product exists, it is unique save vector space isomorphism, so from now on I'll say the tensor product. The next thing we prove is that $\mathcal L_2(E\times F,G)$ and $\mathcal L(E\otimes F,G)$ are isomorphic (for any $\mathbb K$-vector space $G$). To do so we simply construct the following isomorphism: $${\Phi:\mathcal L_2(E\times F,G)\to\mathcal L(E\otimes F,G)} \atop {\varphi\mapsto f_{\varphi}}$$

using the third property in the definition.

  • Injectivity: $f_{\varphi}=f_{\psi} \Rightarrow r\circ f_{\varphi}=r\circ f_{\psi} \Rightarrow \varphi = \psi$
  • Exhaustivity: for $f\in \mathcal L(E\otimes F,G)$, it is clear that we can take $g = f\circ r \in \mathcal L_2(E\times F,G)$, and $\Phi(g) = f$
  • Linear: All I have written is $\Phi(\lambda f + \mu g)(r(u,v)) = \dotsb = (\lambda \Phi(f) + \mu\Phi(g))(r(u,v))$ (how convenient).

So I have some holes to fill in here. The linearity I think is pretty straightforward, so I don't specifically require that anyone clear it up, but there's a note that says: "we actually still have to prove that $r$ is exhaustive!". I sort of understand why but not completely, and certainly don't know how to go about proving it. I'd take this up with my professor but we're on break, so if someone could help I'd appreciate it.

Edit: Exhaustive = Surjective

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I found this [math.stackexchange.com/questions/348940/… –  janmarqz Jan 11 '14 at 20:12
Aahh that helps yes –  GPerez Jan 11 '14 at 20:19

1 Answer 1

up vote 1 down vote accepted

You can prove that $\Phi$ is surjective: For if $T\in{\cal{L}}(E\otimes F,G)$ then you take $\overline{T}\in{\cal{L}}_2(E\times F,G)$ defined as $$\overline{T}(f,g)=T(f\otimes g).$$ That $\overline{T}$ is bilinear is easy to check, then $\Phi(\overline{T})=T$.

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Maybe I'm just not seeing this, but how have you proven that $r$ is exhaustive? Also aren't you only proving this for the model $(E^*\times F^*,\cdot\otimes\cdot)$ where $\cdot\otimes\cdot$ is the tensor product of maps? –  GPerez Jan 11 '14 at 19:07
for each $f\otimes g$ of base vector in $E$ and $F$ respectively you have $(f,g)$, right?? –  janmarqz Jan 11 '14 at 19:10
Yes, but you'd be defining $r$ as $\cdot\otimes\cdot$, so it's not proven in general? –  GPerez Jan 11 '14 at 19:12
And, my answer proves surjectivity of $\Phi$... my fault!! –  janmarqz Jan 11 '14 at 19:12
No problem :) Do you see the difficulty though? We only have the three properties given... how is one to prove that any vector in $E\otimes F$ is $r(u,v)$ for $(u,v)\in E\times F$?? –  GPerez Jan 11 '14 at 19:17

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