One to One Correspondence between the Set of All functions Denote by $F(X , Y)$ the set of all functions from $X$ to $Y$. For sets $A$, $B$, and $C$ prove that:
B) $F(C , F(B , A))$ is in one-to-one correspondence with $F(B \times C , A)$.
Let's give this bijective function between $F(C,F(A\times B))$ and $F(B \times C , A)$ a name, for example $M$. If you start out with a function $g$ which maps out every $c$ to the set of functions from $A$ to $B$, then $M(g)=$the set of functions from ordered pair $(b,c)$ to $A$. 
I cannot seem to prove that $M$ is a bijection. Could someone please lay out how I can start with this since we are dealing with a function that maps a function to an ordered pair of functions. 
What mapping could I define for this function to prove it is a bijection?
 A: First let's count elements to see if there should be a bijection.  We have $|F(X,Y)|=Y^X$, so $|F(C,F(B,A))|=(|A|^{|B|})^{|C|}$ and $|F(B \times C,A)|=|A|^{|B||C|}$ so there should be one.  
An element of $F(C,F(B,A))$ takes an element of $C$ and assigns a function $B \to A$ to it.  So for each $g \in  F(C,F(B,A))$ and $c \in C$ you would have $g(c)$ is a function $B \to A$ and for $b \in B, c \in C$ you have $g(c)(b)$ is some element in $A$  
An element of $F(B \times C,A)$ is a function $h$ that takes an ordered pair $(b,c) \in B \times C$ and returns an element of $A$ so $h((b,c))$ is some element in $A$  
$g$ and $h$ naturally match up seen this way.
A: $M(g)$ is not equal but it is an element of 'the set of functions $B\times C\to A$'.
So, $M(g)$ needs to be a function $B\times C\to A$.
Well, we have $g\in F(C,\,F(B,A))$, that is, $g:C\to F(B,A)$, so, for any element $c\in C$, $\ g(c)$ is again a function, $g(c):B\to A$.
So, which function should $M(g):B\times C\to A\ $ be? If $(b,c)\in B\times C$ is given, where should it be mapped, based on $g$?
To prove $M$ is a bijection, perhaps the easiest is to find its inverse.
