Construct a bijection $F(C, A \times B)\to F(C ,A) \times F(C, B)$ 
Let $A, B, C$ be sets. Construct a bijective function $F(C, A \times B)\to F(C ,A) \times F(C, B)$.

Here, $F(C, A \times B)$ is the set of functions $f: C \to A \times B$.
Could someone show me the way how to do this? What questions should I ask myself?
I understand that $F(C, A \times B)$ is also denoted as $(A\times B)^C$ and that $F(C ,A) \times F(C, B)$ is also denoted as $A^C \times B^C$, so that indeed I need to construct a bijective function $(A\times B)^C \to A^C \times B^C$. 
But how to do this?
 A: Hint: What relation is there between a function $f: C \to A \times B$ and a pair of functions $f_A: C \to A$ and $f_B: C \to B$?

In arduous detail, as the hint seems insufficient:
We want to construct a bijection $\Psi: (A\times B)^C \to A^C \times B^C$. That is to say, to each function $f:C \to A \times B$, we want to associate a pair $(g,h)$ such that:
$$g: C \to A\quad \text{and}\quad h:C \to B$$
Let us see what happens, for a given $c \in C$. We only have access to $f(c)$, and we know that $f(c) = (a,b)$ for some $a\in A,b\in B$.
From $f$, and in particular, from $f(c)$ (do you see why it suffices to consider one $c$ at a time), we need to cook up new functions $g: C \to A$ and $h: C \to B$, respectively $g(c)$ and $h(c)$.
The obvious choice is $g(c) = a, h(c) = b$, where $f(c) = (a,b)$. We see that the pair $(g,h)$ is uniquely determined by $f$ (for it is at each $c \in C$), and conversely, that we can recover $f$ as $f(c) = (g(c),h(c))$. So this process determines $\Psi: (A\times B)^C \to A^C \times B^C$, and shows it to be a bijection.
