# Prove or disprove: $f^{-1}(f(f^{-1}(Y))) = f^{-1}(Y)$.

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?

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

## 1 Answer

Yes.

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.

• For $x \in f^{-1}(Y) \Rightarrow f(x) \in f(f^{-1}(Y))$ you used (I) correct? – St Vincent Dec 10 '14 at 23:42
• 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. – Suzu Hirose Dec 10 '14 at 23:44