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.

  • 1
    $\begingroup$ You should start by taking $y \in f^{-1} (A \cap B)$. In this case, there is some $x \in A \cap B$ such that $y=f(x)$. Now you want to show that $y \in f^{-1}(A) \cap f^{-1}(B)$... Don't forget that $f^{-1}(A) = \{ x : f(x) \in A \}$. $\endgroup$ Feb 5, 2012 at 12:17

1 Answer 1


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 $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$.

  • 6
    $\begingroup$ Amusingly, this is precisely the specialisation of the usual proof that right adjoints preserve limits in category theory... $\endgroup$
    – Zhen Lin
    Feb 5, 2012 at 12:38
  • $\begingroup$ @ZhenLin: Interesting observation! $\endgroup$
    – Paul
    Feb 5, 2012 at 12:46
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    $\begingroup$ This proof has problem, for it would also be true if $f^{-1}$ is replaced by $f$. However $f(A\cap B) \subset f(A) \cap f(B)$, and not $f(A\cap B=f(A) \cap f(B)$ $\endgroup$ Mar 14, 2015 at 4:39
  • $\begingroup$ @hermes Why do you say that? $\endgroup$ Apr 26, 2020 at 11:56
  • $\begingroup$ As I explained in the example, this proof also works for $f$. However, actually we do not have $f(A∩B)=f(A)∩f(B)$, but only $f(A∩B)\subset f(A)∩f(B)$. $\endgroup$ Apr 26, 2020 at 16:52

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