Let $f: A \to B$ be a function, and $Y \subseteq B$. Prove or disprove: $f^{-1}(f(f^{-1}(Y))) = f^{-1}(Y)$.

My textbook has a theorem that says:

Suppose $f: A \to B$. Let $X \subseteq A$ and $Y \subseteq B$. Then:

I. $X \subseteq f^{-1}(f(X))$

II. $f(f^{-1}(Y)) \subseteq Y$

Can I combine facts (I) and (II) and apply the definition of the preimage to prove the above proposition?

  • $\begingroup$ @Jonny Suppose $f$ is a constant function and $f(x)\in Y$ for all $x$ but $Y\neq\{x\}$ then $f(f^{-1}(Y))\neq Y$. $\endgroup$ – Suzu Hirose Dec 10 '14 at 23:15
  • $\begingroup$ $f(f^{-1}(Y)) \supseteq Y$ only holds when $f$ is surjective. $\endgroup$ – Aaron Maroja Dec 10 '14 at 23:17
  • $\begingroup$ Related: math.stackexchange.com/questions/746123 $\endgroup$ – Watson Jan 26 '17 at 12:27


If $x\in f^{-1}(Y)$ then $f(x)\in f(f^{-1}(Y))$ and $x\in f^{-1}(f(f^{-1}(Y)))$, so $f^{-1}(Y)\subseteq f^{-1}(f(f^{-1}(Y)))$.

Similarly if $x\in f^{-1}(f(f^{-1}(Y)))$ then clearly $x\in f^{-1}(Y)$ so $f^{-1}(f(f^{-1}(Y)))\subseteq f^{-1}(Y)$ and the two sets are equal.

  • $\begingroup$ For $x \in f^{-1}(Y) \Rightarrow f(x) \in f(f^{-1}(Y))$ you used (I) correct? $\endgroup$ – St Vincent Dec 10 '14 at 23:42
  • $\begingroup$ I didn't use that argument explicitly, but it is clear from the definition. The above method is the standard way of proving two sets are equal. If $x$ is in one set then it must be in the other one, and vice-versa, then the two sets are equal. $\endgroup$ – Suzu Hirose Dec 10 '14 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.