Given that $|A|=|B|$ (the cardinality of set $A$ is equal to the cardinality of $B$).
How can I prove that $|A^C|=|B^C|$ (the cardinality of the set of all functions from $C \longrightarrow A$ is equal to the cardinality of the set of all function from $C \longrightarrow B$).
I understand that I have to find a bijection $h: A^C \longrightarrow B^C$, means that $g$ receives a function $f: C\longrightarrow A$, and return a function $g: C\longrightarrow B$.
I am not sure how to approach finding such a bijection and therefore asking here for help. Much appreciated!
I did something like this: Let $\psi$:$A\longrightarrow B$ be a bijection since A and B are equinumerous.Then; define a function such that; $\phi$:$A^C\longrightarrow B^C$ as $\phi(f)=\psi$$\circ$$f$ then observe that $\psi$$\circ$$f$:$C\longrightarrow$$B$.Since all functions in $A^C$ is different they generate all different functions(hence it's an injection)from C to B(so the output is definitely in $B^C$ ). Also, all the functions in $B^C$ are covered since $\psi$ is an surjection $\forall$x$\in$$B$ $\exists$$b\in A$ such that $\psi(b)=x$ Thus, the $\phi$ generates all functions from B to C.