TITLE: Preimage of Intersection of Two Sets = Intersection of Preimage of Each Set : $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$
QUESTION [6 upvotes]: Prove If $f$ is a function , $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$. 
Proof attempt
I am guessing here $A$ and $B$ are sets in the range of $f$. Let's assume $x$ belongs to both $A$ and $B$ and $f^{-1}$ exists for both $A$ and $B$. 
Then there must exist a $y$ such that $y = f^{-1}(x)$.
Now by our assumptions $x$ is in intersection of $A$ and $B$ and since $f^{-1}(A)$ and $f^{-1}(B)$ exist then $f^{-1}(A) \cap f^{-1}(B)$ must also exist. Also since $A$ and $B$ exist and are not equal to null set so $A \cap B$ exists and $f^{-1}(A\cap B)$ also must exist and contain our $y$?? Not sure about the ending in this attempt. Any help would be much appreciated. 
Sources : ♦ 2nd Ed $\;$  P219 9.60(e) $\;$  Mathematical Proofs by Gary Chartrand,
♦ P214 $\;$ Theorem 12.4.#3 $\;$  Book of Proof by Richard Hammack,
♦ P257-258 $\;$  Theorem 5.4.2.#2(a) $\;$   How to Prove It by D Velleman.

REPLY [19 votes]: Note that by definition of inverse image, we have $$x\in f^{-1}(A \cap B)$$
$$\Leftrightarrow f(x)\in A\cap B$$
$$\Leftrightarrow f(x)\in A\mbox{ and }f(x)\in B$$
$$\Leftrightarrow x\in f^{-1}(A)\mbox{ and }x\in f^{-1}(B)$$
$$\Leftrightarrow x\in f^{-1}(A)\cap f^{-1}(B).$$
Therefore, we have $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$.