Given $f$ is one to one, prove $f(A \backslash B)=f(A)\backslash f(B)$

Question

Given $f:X\rightarrow Y$, $A, B \subset X$, and $f$ is one to one, prove that $f(A \backslash B)=f(A)\backslash f(B)$.

I have a start, but I'm not really sure where to go from here:

Answer 1

Since $x\in B^c$, we cannot have that $f(x)\in f(B)$, else we would have that $f(x)\in f(B^c)$ (as you already found) and $f(x)\in f(B)$, which implies that $x\in B^c$ and $x\in B$ by injectivity, which is a contradiction. Therefore $f(x)\in f(B)^c$.

Now you can finish the proof.