Functions and Set Theory Denote by $F(X,Y)$ the set of all functions from $X$ to $Y$. For sets $A$, $B$, and $C$ prove that
a. $F(C,A\times B)$ is in one-to-one correspondence with $F(C,A)\times F(C,B$).
Let's give this bijective function between $F(C,A\times B)$ and $F(C,A)\times F(C,B)$ a name, for example $M$.
If you start out with a function $g$ which maps out every $c$ to some ordered pair $(a,b)$,
then $M(g) = (x,y)$ where $(x,y)$ is an ordered pair of functions. $x(c)$ = $a$ and $y(c) = b$. 
In other words, $M$ looks at the behavior of the input function $g$ for every $c$, and splits $g$'s behavior across two functions.
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.
Also, how would you do: F(C , F(B , A)) is in one-to-one correspondence with F(B × C , A).
Because, C maps to a function.
 A: Set $ \pi_A : A\times B\to A$, $ \pi_B : A\times B\to B$, the projection maps, i.e.,
$$
\pi_A(a,b)=a \quad\text{and}\quad \pi_B(a,b)=b.
$$ 
Let now $f: C\to A\times B$ be a function. Then $\pi_A\circ f : C\to A$ and  $\pi_B\circ f : C\to B$ are also functions.
This allows us to define the mapping 
$$
\varPhi : {\mathcal F}(C, A\times B)\to
{\mathcal F}(C, A)\times {\mathcal F}(C, B)
$$
as 
$$\varPhi(f)=(\pi_A\circ f,\pi_B\circ f).$$
If $f\ne g$, then $f(c)\ne g(c)$, for some $c\in C$, i.e.,
$$
\big(\pi_A\circ f(c),\pi_B\circ f(c)\big)=f(c)\ne g(c)=\big(\pi_A\circ g(c),\pi_B\circ g(c)\big),
$$
which means that $\pi_A\circ f\ne \pi_A\circ g$ or $\pi_B\circ f\ne \pi_B\circ g$, and hence 
$$
\varPhi(f)\ne \varPhi(g).
$$
Thus $\varPhi$ is one-to-one.
A: You've shown how to start with a function from $C$ to $A\times B$, and get a pair of functions, one from $C$ to $A$, the other from $C$ to $B$.
Now show how to start with such a pair of functions and obtain a function from $C$ to $A\times B$.  Also show that these two processes are inverse to each other.
This works, since one way to show that a function $f$ is bijective is to find a function $g$ such that $f\circ g = \text{id}$ and $g\circ f= \text{id}$.
