Proof for sets and functions. 
I have been proving problems like this all day with ease, but this is is just puzzling to me. Where do I start? Also, a site with questions and answers to problems like these.
 A: You just have to go back to the definitions of the concepts.
We use that $f^{-1}(S)$ is the set of elements $x\in X$ such that $f(x)\in S$ and that $f(S)$ is the set of elements $y\in Y$ such that there exists an element $x\in S$ such that $f(x) = y$.
Assume that $y\in C\cap f(X)$, this means that $y\in f(X)$ which means there's an $x\in X$ such that $f(x) = y$. Now we have that $y\in C$ so $x\in f^{-1}(C)$. And since $x\in f^{-1}(C)$ it follows that $y=f(x) \in f(f^{-1}(C))$. From this we can conclude that $C\cap f(X)\subseteq f(f^{-1}(C))$.
Using the same kind of approach we assume $y\in f(f^{-1}(C))$ and therefore it exists an $x\in f^{-1}(C)$ such that $f(x) = y$, but since $x\in f^{-1}(C)$ we have that $y = f(x)\in C$. From this we conclude that $f(f^{-1}(C))\subseteq C$. Also quite directly from the definition we have that $f(f^{-1}(C))\subseteq f(X)$.
From this we can conclude the first equality.
The second claim is not correctly written. If we for example have $X=\{1\}$ and $Y=\{2\}$ and $f(1)=2$ and suppose $A=X$ we have that $f^{-1}(A)\emptyset$ since there's no $x\in X$ such that $f(x)\in A$ (since $f(x)=2\notin A$.
What's probably meant is that $A\subseteq f^{-1}(f(A))$ (because $A$ is part of the range and that's what $f^{-1}(S)$ is). Now it should be provable using the same techniques...
