elementary set theory - If $f$ is 1-1, prove that $f(Asetminus B) = f(A)setminus f(B)$

I'm having a tough time with this one. Here's the background:

Let $X$ and $Y$ be sets, let $f:X\rightarrow Y$ and let $A,B\subseteq X$. For this proof, we also assume that $f$ is 1-1.

I've already proven $f(A\setminus B)\supseteq f(A)\setminus f(B)$, but where should I start to prove the other way around?

Answer

To show that $f(A\setminus B) \subset f(A) \setminus f(B)$, show that each element of the first lies in $f(A)$, but not in $f(B)$. You need injectivity for this.