Prove $\psi: L(E,F;G)\to L(E; L(F;G))$ is an isomorphism 
Let $E, F, G$ be vector spaces and
$\psi: L(E,F;G)\to L(E; L(F;G))$
the transformation that associates to each bilinear application $B\in
 L(E,F;G)$ the linear application $T = \psi(B)\in L(E; L(F;G))$ defined
  like this: for each $u\in E,T\cdot u\in L(F;G)$ is such that $[T\cdot
 u]\cdot v = B(u,v)$ for all $v\in F$. Show that $\psi$ is an
  isomorphism

As I understood, $\psi$ takes a bilinear application from $E\times F$ to $G$ and associates with it the linear application $T$ that goes from $E$ to $L(F;G)$, which is itself a space of transformations that goes from $F$ to $G$. Am I right?
In order to show that this is an isomorphism, of course we need to show that there's a bijection. How am I supposed to do even this part? I'm very confused by all of this. 
I think that to show that there's an isomorphism it's easier because it's a linear transformation so I could use that linear properties but I'm very confused
 A: $\psi$ is well defined (easy to check).
To see is linear take $A,B \in L(E,F;G)$ and $c \in K$ where $K$ is the underlying field, we want to prove that $\psi(cA+B)=c\psi(A)+\psi(B)$ (these are functions).
Let $u \in E$, arbitrary, we need to show that $$\psi(cA+B)(u)=(c\psi(A)+\psi(B))(u),$$ or equivalently, $$\psi(cA+B)(u)=c\psi(A)(u)+\psi(B)(u)$$ (but these are all functions).
Let $v \in F$, arbitrary, we need to show that $$\psi(cA+B)(u)(v)=(c\psi(A)(u)+\psi(B)(u))(v),$$ or equivalently, $$\psi(cA+B)(u)(v)=c\psi(A)(u)(v)+\psi(B)(u)(v)$$ (now these are vectors).
Now, by definition $\psi(cA+B)(u)(v)=(cA+B)(u,v)=cA(u,v)+B(u,v)$. Also, $\psi(A)(u)(v)=A(u,v)$ and $\psi(B)(u)(v)=B(u,v)$, then $c\psi(A)(u)(v)+\psi(B)(u)(v)=cA(u,v)+B(u,v)$. Thus $$\psi(cA+B)(u)(v) = c\psi(A)(u)(v)+\psi(B)(u)(v),$$ since $v$ was arbitrary, this holds for every element in $F$
$$\Rightarrow \psi(cA+B)(u)=c\psi(A)(u)+\psi(B)(u),$$ since $u$ was arbitrary this holds for every element in $E$
$$\Rightarrow \psi(cA+B)=c\psi(A)+\psi(B)$$
Therefore $\psi$ is linear.
To see is one-to-one take $A,B \in L(E,F;G)$ such that $\psi(A)=\psi(B)$. Let $u \in E, v \in F$ then $\psi(A)(u) = \psi(B)(u)$ since $\psi(A)=\psi(B)$ is a function, again, $\psi(A)(u)$ is a function, then $\psi(A)(u)(v)=\psi(B)(u)(v)$, but by definition $\psi(A)(u)(v)=A(u,v)$ and $\psi(B)(u)(v)=B(u,v)$, then $A(u,v)=B(u,v)$ for all $u\in E,v\in F$, that is, $A=B$ and $\psi$ is one-to-one.
To see is onto, take $A \in L(E;L(F;G))$, then define $B:E\times F \rightarrow G$ by $B(u,v)=A(u)(v)$ for all $u \in E$ and $v \in F$. Is easy to see that $B \in L(E,F;G)$, since $A(u)$ is linear, $B$ is linear in its second component, and because $A$ is linear $B$ is linear in its first component and clearly $\psi(B)=A$. Then $\psi$ is onto.
Therefore $\psi$ is bijective and is an isomorphism.
