How to show $f^{-1}(Y \backslash F) = X\backslash f^{-1}(F)$? Given a function $f: X \rightarrow Y$ and $E \subset X$ and $F \subset Y$, prove that $f^{-1}(Y \backslash F) =  X \backslash f^{-1}(F)$?
I'm assuming we just have to think of $f^{-1} (Y)$. I know the definition should be $f^{-1} (Y) = \{x \in X \ : \ f(X) \in Y\}$. But I don't see how we could end up with $f^{-1}(F)$ because that would imply $f^{-1}(F)= \{f(x): x\in X : f(x) \in F\}$ ?
 A: Let $x \in f^{-1}(Y \setminus F)$. Then $f(x) \in Y$ and  $f(x) \not\in F$. Clearly, $x \in X$ and  we need $x \not\in f^{-1}(F)$. The latter follows directly from $f(x) \not \in F$. Thus $x \in X \setminus f^{-1}(F)$.
Let $x \in X \setminus f^{-1}(F)$. Then $x \in X$ and $x \not\in f^{-1}(F)$. Clearly, $f(x) \in Y$ and we need $f(x) \not\in F$. This follows directly from $x \not\in f^{-1}(F)$. Thus $f(x) \in Y \setminus F$ and therefore $x \in f^{-1}(Y \setminus F)$.

This whole argument basically is just an exercise in the definition of pre-image.
A: The following statements are equivalent:
$x \in f^{-1}(Y \setminus F)$
$f(x) \in Y \setminus F$
$f(x) \notin F$
$x \notin f^{-1}(F)$
$x \in X \setminus f^{-1}(F)$
A: Well, by the definition, we have $$f^{-1}(Y \setminus F) = \{x\in X: f(x) \in Y\setminus F\}$$
and $$f^{-1}(F) = \{x\in X: f(x) \in F\}.$$
Then, if $x \in f^{-1}(X\setminus F)$, then $x \not\in f^{-1}(F)$, or $x \in X\setminus f^{-1}(F)$. So, $f^{-1}(X\setminus F) \subset X\setminus f^{-1}(F)$.
If $x \in X\setminus f^{-1}(F)$, then $f(x) \not\in F$, or $x \in f^{-1}(Y\setminus F)$.
So, you have the conclusion.
