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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.

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  • $\begingroup$ 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... $\endgroup$ – JMoravitz Nov 14 '19 at 17:57
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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.

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