TITLE: Trouble proving that if $A \approx B$ then $A^C \approx B^C$
QUESTION [1 upvotes]: 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)

REPLY [2 votes]: 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 
$$\psi_1(c) = \psi_2(c)\text{  }  \forall c \checkmark$$
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