If $f$ is 1-1, prove that $f(A\setminus B) = f(A)\setminus f(B)$

Solution 1:

Choose $y \in f(A \setminus B)$ then there exists $x \in A \setminus B$ such that $y = f(x)$. Observe that $f(x) \in f(A)$ since $A \setminus B \subseteq A$. Now why is $y \notin f(B)$? If it were, say $y \in f(B)$ then $y = f(x')$ for some $x' \in B$. But then we have $f(x) = y = f(x')$ and by injectivity we have $x = x'$. This means that $x' \notin B$, a contradiction. Conclude that $y \notin f(B)$ and hence $y \in f(A) \setminus f(B)$.

Solution 2:

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.