I am trying to prove if a function is a bijection, but am having a hard time doing so. I know to prove this, I should prove the function is injective and surjective.
The question is as follows:
Suppose A and B are any two sets, and f is a bijection from A-B. Define the function g from P(A) to P(B) as follows: g(S) = {f(x) | x in S} ... therefore, g is a bijection.
So far, I have tried using the fact that f is a bijection to show that two values, x and y, are equal, so f(x) = f(y) and x = y. But I am not sure if that would satisfy for g(S) where {f(x) | x in S} = {f(y) | y in S}.
First, let's show it's onto. Let $T \in \mathcal P(B)$. We want to show there exists a $S \in \mathcal P(A)$ such that $g(S) = T$.
Since $f: A \to B$ is a bijection, then it's inverse $f^{-1}: B \to A$ exists. Let $S$ be the following,
$$S = \{ f^{-1}(y) : y \in T\}.$$
Then, $$g(S) = \{f(x) : x \in S\} = \{f(x) : x \in \{ f^{-1}(y) : y \in T\}\} = \{f(f^{-1}(y)): y \in T\} = \{y : y \in T\} = T.$$
Thus $g$ is onto. Now let's show $g$ is one-to-one. Let $S,T \in \mathcal P(A)$. We want to show that if $g(S) = g(T)$, then $S=T$. So let's suppose $g(S) = g(T)$.
Now, consider $s \in S$. By definition, $f(s) \in g(S)$. Since $g(S) = g(T)$, then it's also the case that $f(s) \in g(T)$. That means there exists an $x\in T$ such that $f(s) = f(x)$. Since $f$ is a bijection, it's injective and so $s=x$. But $x\in T$, thus $s \in T$. Since $s \in S \implies s \in T$, then $S \subseteq T$. Similarly, $T \subseteq S$. Thus, $S = T$ showing that $g$ is injective.
Since $g$ is both onto and one-to-one, it is a bijection.