Prove that for two sets A and B if there is a bijection from A to B, then there is a bijection from pow(A) to pow(B) So I know that I need to prove that it is onto and one-to-one, but I am not sure how to go about it at all. 
 A: The first problem is where you say "it".  What is "it"?  You need to make a guess at what the bijection should be, then prove it really is a bijection.
Hint.  Suppose you have a bijection
$$f:\{\,1,2,3,4,5,\ldots\,\}\to\{\,a,b,c,d,e,\ldots\,\}$$
with $f(1)=a$, $f(2)=b$, $f(3)=c$.  To find a bijection
$$g:pow(A)\to pow(B)$$
you need a rule to define $g(S)$, where $S$ is any subset of $A$.  Can you think of a good suggestion for $g(\{1,2,3\})$?
A: Let $f\colon A\to B$ be a bijection from $A$ to $B$, and define
\begin{align*}
F\colon 2^A&\to 2^B\\
S&\mapsto \bigcup_{a\in S} \{f(a)\}
\end{align*}


*

*$F$ is onto: let $T\subseteq B$, and define $S=\bigcup_{b\in T} \{f^{-1}(b)\}$. Prove that $F(S)=T$.

*$F$ is injective: let $S,S^\prime\subseteq A$ such that $F(S)=F(S^\prime)$. Let $a\in S$: $f(a)\in F(S)=F(S^\prime)$, so by definition of $F(S^\prime)$ there exists $a^\prime$ s.t. $f(a^\prime)=f(a) )$. But as $f$ is bijective, this implies $a^\prime= a$ and thus $a\in S^\prime$: therefore $S\subseteq S^\prime$. By symmetry, $S^\prime\subseteq S$ and $S^\prime=S$.

A: Let's prove if $A\approx B$ and $X\approx Y$, then $X^A\approx Y^B$:
Let $\varphi$ be the bijection from $A$ to $B$, $\psi$ be the bijection from $X$ to $Y$. Define $H:X^A\to Y^B$ by $H(f)=\psi f\varphi^{-1}$. It's easy to verify that $H$ is bijective.
If $f\neq g$, then for some $x\in A$, $f(x)\neq g(x)$, Then $H(f)(\varphi(x))=\psi f\varphi\varphi^{-1}(x)=\psi f(x)\neq\psi g(x)=\psi g\varphi\varphi^{-1}(x)=H(g)(\varphi(x))$, i.e., $H(f)\neq H(g)$. So $H$ is injective.
If $h\in Y^B$, then $\psi^{-1}h\varphi\in X^A$. Clearly $H(\psi^{-1}h\varphi)=h$. So $H$ is surjective.
Now Apply $X=Y=2$ to your question, we see $\mathscr{P}(A)\approx 2^A\approx 2^B\approx\mathscr{P}(B)$.
