Validity of my proof Let $A$ and $B$ be non-empty and let $f:A\rightarrow B$ be a function. We define $P(A)=\{ X; X \subseteq A \}$ as the power set of $A$ and $P(B)= \{Y; Y \subseteq B \}$ as the power set of $B$. We define $F: P(A) \rightarrow P(B)$ as $F(X)=\{f(x); x \in X$. Prove:
a) $f$ is surjective $\Rightarrow$ $F$ is surjective
b) $f$ is injective $\Rightarrow$ F is injective
My attempt at a proof:
a) We use reductio ad absurdum. Assume that:
$$\exists \beta \in P(B): \forall \alpha \in P(A): F(\alpha) \neq \beta$$
Then no elements of $\beta$ are an image of an element of $A$, which contradicts surjectivity.
This concludes the first proof.
b) By the definition of injectivity:
$$\forall \alpha, \beta \in P(A):(\alpha \neq \beta \Rightarrow F(\alpha) \neq F(\beta))$$
Let $\alpha, \beta \in P(A)$ and assume $\alpha \neq \beta$. Let's look at their images:
$$F(\alpha)=\{ f(x); x \in \alpha \}$$
$$F(\beta)=\{f(x); x \in \beta \}$$
Since $\alpha$ and $\beta$ weren't the same:
$$\exists x:(x \in \alpha \land x \notin \beta \vee x \in \beta \land x \notin \alpha )$$
We prove only the first possibility, since the second is analogous and both cannot logically be true.
So assume $x \in A \land x \notin B$. By the definition of $F$: $f(x) \in F(\alpha) \land f(x) \notin F(\beta)$. Therefore $F(\alpha) \neq F(\beta)$.
This concludes the second proof.
I'm pretty confident about the second one but the first one seems a bit sloppy. Any thoughts? 
 A: 
$$\exists \beta\in P(B):\forall\alpha\in P(A):f(\alpha)\ne\beta$$
  Then no elements of $\beta$ are an image of an element of $A$, which contradicts surjectivity.

First, I think you intend to use $F$ instead of $f$ here.  However, the sentence you write does not logically follow.  Notice that just because  we can't find a set $\alpha$ that maps to the set $\beta$ does not mean that $\beta$ is missed completely by $f$.  Now of course, we're working with stronger hypotheses, by considering the whole power set, so in this case it does, but more reasoning is needed.
It might be easiest, and clearest, to just give a direct proof.  Consider the following set:
$$\alpha_{\beta}:=\{a\in A\mid f(a)\in\beta\}$$
Now see if you can prove that $F(\alpha_{\beta})=\beta$ for all $\beta\in P(B)$.
A: You don't have to resort to "reductio at absurdum", nor to distinguish cases.
(a) Given an arbitrary $Y\in P(B)$ put
$$X:=f^{-1}(Y):=\{x\in A\ |\ f(x)\in Y\}\ .$$
Then $F(X)\subset Y$ by definition. Conversely: Since $f$ is surjective, for any $y\in Y$ there is an $x\in A$ with $f(x)=y$. It follows that $y\in F(X)$; and as $y\in Y$ was arbitrary we conclude that $Y\subset F(X)$. Together we have established that $F(X)=Y$.
(b) Assume $F(X)=F(X')$, and consider an arbitrary $x\in X$. Then $f(x)\in F(X)=F(X')$. So there exists an $x'\in X'$ with $f(x)=f(x')$, and as $f$ is injective we conclude that $x'=x$. This proves $x\in X'$, and as $x\in X$ was arbitrary we can say that $X\subset X'$. By symmetry it follows that in fact $X=X'$.
Note that it is customary to denote your $F$ ("by abuse of language") again by $f$.
