Trouble proving that if $A \approx B$ then $A^C \approx B^C$ 
Show that if $A \approx B$ then $A^C \approx B^C$

According to my textbook, $A^C$ is the set of all functions from $C$ to $A$, ie. $A^C=\{F\in \mathscr P(Y \times X): F$ is a function$\}$.
So I need to prove that there is a function $g: A^C \to B^C$ that is one-one and onto. ie. For one-one, assume $g(h)=g(h')$ where $h$ is a function in $A^C$, and show that $h=h'$. Also assume that there is a $f: A\to B$ that is one-one and onto. 
I am stuck because I have no idea how to define $g$; I think this is partly because I am slightly confused by what the definition of $g: A^C \to B^C$ is. I know it is a set of functions, but what exactly does that mean/ what does a member look like? Is it a set of ordered pairs consisting of functions? eg. $<f,g>$?
Also it seems that some info about $C$'s relation to other sets is needed, but is not provided.
Could anyone help or point me at the right direction please? 
(Schröder–Bernstein theorem is available, but I am not sure if it will be useful here)
 A: Given an element of $A^C,$ it is some map $u:C \to A.$ Then define $g(u):C \to B$ by composition. That is, $g(u)(c)=f(u(c)).$
A: Technically functions can be modeled by certain sets of pairs, but that is not a very productive intuition about what is going on for your purpose here.
I think it is more fruitful to think of a function $X\to Y$ as a machine with the property that when you put an element of $X$ into it, somehow an element of $Y$ comes out the other end -- and this happens in a repeatable fashion such that if you put the same element of $X$ into the machine multiple times, the same element of $Y$ will be produced several times.
When you're looking for a function $g:A^C \to B^C$, then you want to constructs a machine that makes machines into other machines. The input to your $g$ will be a machine $C\to A$. You want use that to produce a machine $C\to B$.
The natural way to do that will be to bolt the $C\to A$ machine you get as input together with a copy of the the $f:A\to B$ you have assumed exists. The combined machine will first use the input to convert the incoming $c$ into an $a$, and then use $f$ to convert the $a$ into a $b$. The net effect is that something from $C$ comes in and something from $B$ comes out -- in other words we do get a function $C\to B$.
Once you have used the machine analogy to work out what your $g$ should do, you can go back to symbolic reasoning to try to prove that this $g$ is actually a bijection.
A: Since A$\approx$B, there exists a one-to-one onto function f:A$\rightarrow$B. That is the key to the exercise.
Let $\phi\in A^C$. Then f$\circ \phi$:C$\rightarrow$B. So we define a function $\psi_f$:$A^C \rightarrow B^C$ such that  $\phi \mapsto f\circ\phi$. 
1:$\psi_f$ is onto. Let g$\in B^C$. Then $f^{-1}\circ g \in A^C$. So 
$$\psi_f(f^{-1} \circ g) = f \circ f^{-1} \circ g = g. \checkmark $$
2.$\psi_f$ is one-to-one. Suppose $\phi_1,\phi_2 \in A^C$ such that $f \circ \psi_1 = f \circ \psi_2$. Thus for every c $\in$ C, then 
$$f(\psi_1(c)) = f(\psi_2(c))$$
But since f is invertible, that forces 
$$\psi_1(c) = \psi_2(c)\text{  }  \forall c \checkmark$$
