$f^{-1}(f(E))=E$ implies $f$ is one-to-one

Question

Suppose that, for a function $f:X \rightarrow Y$, it is given that the following is true: $$f^{-1}(f(E))=E$$ (Where $E\subset X$). How can I use this to prove that $f$ is one-to-one? I'm really not sure how to set this up formally, as I have an intuitive understanding as to why but simply struggle to put it into proper mathematical notation.

Answer 1

Hint:

Conider the subsets $E$ that has a single element. What implies that $f^{-1}(f(\{x\}))=\{x\}$ for every $x\in X$?

Answer 2

Suppose not. Then $f(x_1)=f(x_2)=\bar y $ for some $\bar y \in Y$, and $x_1\not = x_2$...

Let $E=\{x_1\}$. Then $x_2\in f^{-1}(f(E))$... Thus $f^{-1}(f(E))\supsetneq E$, a contradiction ...