$f$ injective iff $f(X\cap Y)=f(X)\cap f(Y)$, proof technique I am having a problem understanding what this question asks of me.
$f\colon A\to B$ is a function. Prove that the following statements are equivalent:
1) $f$ is injective
2) for all $X,Y\subseteq A\colon f(X\cap Y) = f(X)\cap f(Y)$
I do not understand what the question means by proving that the statements are equivalent. If anyone could enlighten me I would be grateful. 
 A: To prove them equivalent, show that they imply each other-- if one is provably true, this implies that the other is as well.
For example,
assume that $f$ is injective. Your job is to show that for any $X,Y \subset A$ we have that $f(X \cap Y)=f(X) \cap f(Y)$ using the assumption. 
Conversely, assume that $f(X \cap Y)=f(X) \cap f(Y)$  and prove that $f$ must be injective.
A: Showing that two statements are equivalent is showing that if you assume that the first is true, then you should be able to prove that the second is true, and viceversa. 
So in this case, the idea is to: first assume that $f$ is injective and to try to show that for all $X,Y \subset A$, $f(X \cap Y) = f(X)\cap f(Y)$. So your next question, may be, how do we prove the latter statement. Well, we have an equality between to sets, so we could show that both $f(X)\cap f(Y) \subseteq f(X \cap Y)$ and $f(X\cap Y)\subseteq f(X) \cap f(Y)$. I will do the first one:
Take $a \in f(X)\cap f(Y)$ then $a \in f(X)$ and $a \in f(Y)$, that is, $a=f(y)=f(x)$ for some $x \in X$ and some $y \in Y$. Since $f$ is injective (here we use the first statement as hyphotesis) then $x = y$ and therefore $a \in f(X \cap Y)$. Try to do the other inclusion, and to finish with showing the equivalence, assume 2) as true, and try to show that $f$ is injective. 
A: A question which asks you to prove that statements are equivalent, are most of the time solved by the principal of "ring closure". That means you proof 1)$\Rightarrow$ 2) and 2) $\Rightarrow$ 1).
1)$\Rightarrow$ 2)
So for the proof we first assume that $f$ is injective.
And we have to proof, that for all $X,Y\subseteq A$ it is $f(X\cap Y)=f(X)\cap f(Y)$
We have to show that two sets are equal. 
This is proven by showing both inclusions "$\subseteq$" and "$\supseteq$"
"$\subseteq$"
Without loss of generality we can assume $X, Y\neq\emptyset$ and $X\cap Y\neq\emptyset$. Else the statement is trivial.
Let $b\in f(X\cap Y)$. Then there is an $a\in X\cap Y$ with $f(a)=b$. Since $a\in X$ and $a\in Y$ it is $b\in f(X)$ and $b\in f(Y)$. Hence $b\in f(X)\cap f(Y)$.
Note, that we have not used that $f$ is injective yet.
"$\supseteq$"
Let $b\in f(X)\cap f(Y)$. Then there is an $a\in X$ and $a'\in Y$ such that $f(a)=b$ and $f(a')=b$. Since $f$ is injective and $f(a)=f(a')$ we get that $a=a'$!
Hence $a\in X\cap Y$ and therefore $b\in f(X\cap Y)$.
This completes the first step of the "ring closure".
2)$\Rightarrow$ 1)
Now we assume that $f(X\cap Y)=f(X)\cap f(Y)$.
We have to show, that $f$ is injective.
That means, that for $f(a)=f(a')$ we have to conclude, that $a=a'$.
Let $f(a)=f(a')=b$. Since $f(X\cap Y)=f(X)\cap f(Y)$ for EVERY $X,Y\subseteq A$ it is $\{b\}=f(\{a\})\cap f(\{a'\})=f(\{a\}\cap\{a'\})$. Hence $\{a\}\cap\{a'\}\neq\emptyset$. Hence $a=a'$!
We conclude, that $f$ is injective.
