# Proving that the power sets of bijective sets are also bijective

How can I prove that if $$f: A \rightarrow B$$ is bijective then $$g: P(A) \rightarrow P(B)$$ is also bijective. $$P(A)$$ and $$P(B)$$ are power sets.

For injectivity, is it a sufficient proof to say that: Suppose $$X_1,X_2\in P(B)$$ and $$f(X_1)=f(X_2)$$. Since $$f$$ is injective since it is bijective we can see that $$X_1=X_2$$?

Don't know how to start surjectivity proof.

• What is $g$ more specifically? You can have a bijective function $f$ between $A$ and $B$ and a function which is not bijective between $P(A)$ and $P(B)$... take for instance a constant function... – JMoravitz Nov 14 '19 at 17:57

I assume you intend for $$g$$ to be $$f_*:P(A)\to P(B)$$ that maps the set $$C$$ to $$f(C)$$.
Suppose $$C,D\in P(A)$$ are such that $$C\neq D$$. Then there exists some element $$x\in (C\setminus D)\cup (D\setminus C)$$. Suppose without loss of generality that $$x\in C\setminus D$$. Then $$f_*(C)$$ contains $$f(x)$$ and, since $$f$$ is injective, $$D$$ does not. Hence $$f_*$$ is injective.
Suppose now that $$D\in P(B)$$. Then $$f^{-1}(D)\in P(A)$$ and, since $$f$$ is surjective, $$f_*(f^{-1}(D))=D$$, hence $$f_*$$ is surjective.