Assune A,B,C,D are sets such that $|A |= |B |$ and $|C |= |D |$. Show that $|C^A |= |D^B |$

Question

Assume $A$,$B$,$C$,$D$ are sets such that $|A|= |B |$ and $|C |= |D |$. Show that $|C^A |$= $|D^B |$,where $C^A$ is the set of all functions from $A$ to $C$, and $D^B$ is the set of all functions from $B$ to $D$.

Answer 1

Let be $f:A\rightarrow B$ and $g:C\rightarrow D$ bijections. Define $F:C^A\rightarrow D^B$ as $$ F(h):=g\circ h\circ f^{-1} $$ Edit: As $A$ and $B$ has the same cardinality, then exists a bijection $f$ between them. In the same form you conclude that $g$ exists.

Now, we need to prove that $F$ is a bijection:

$F$ is one to one: Let $h,h'\in C^A$ such that $F(h)=F(h')$ so, $$ \begin{array}{rcl} g\circ h\circ f^{-1} & = & g\circ h'\circ f^{-1}\\ g^{-1}\circ g\circ h\circ f^{-1}\circ f & = &g^{-1}\circ g\circ h'\circ f^{-1}\circ f\\ h & = & h' \end{array} $$ F is onto: Let $h\in D^B$. Is easy to see that $g^{-1}\circ h\circ f:A\rightarrow C$ and $F(g^{-1}\circ h\circ f)=h$.

With this, we have that $|C^A|=|D^B|$.