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
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 $\Phi(\overline{T})=T$.