Surjection $f$ induces surjection $\mathcal P (f)$ on power sets Just wanted to make sure the way I approach this was correct because it seemed a bit too simple for an answer: 
Question: Let $\mathcal P \left({X}\right)$ represent the power set of $X$. Let $f: X \rightarrow Y$ be a function and this function induces this function $$\mathcal P \left({f}\right): \mathcal P \left({X}\right) \rightarrow \mathcal P \left({Y}\right)$$ given by $\mathcal P \left({f}\right)(A) = f(A)$ where $A \in \mathcal P \left({X}\right)$. 
Prove if $f$ is surjective, then $\mathcal P \left({f}\right)$ is surjective.
Proof: Since $f$ is surjective, then $\forall y \in Y, \exists x \in X$ such that $f(x) = y$. So we have that $\mathcal P \left({f}\right)(A) = f(A).$ But $f$ is surjective, so this has to be mapped to $y \in Y$, so $\mathcal P \left({f}\right)$ is surjective. 

You still need to improve your proof. Because you already have by definition of your function that $\mathcal{P}(f)(A) = f(A)$, then I didn't understand the conclusion there. It's unclear what you mean by "has to be mapped to" and also your conclusion. 
An idea would be if you take any $B \in \mathcal {P}(Y)$ then $B \subset Y$ and as $f$ is surjective you have that $f(f^{-1}(B)) = B$. So take $A = f^{-1}(B) \subset X$ and you will have 
$$\mathcal{P}(f)(A) = f(f^{-1}(B)) = B$$ 
Therefore the function $\mathcal{P}$ is in fact surjective. 
