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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?

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  • $\begingroup$ @Jonny Suppose $f$ is a constant function and $f(x)\in Y$ for all $x$ but $Y\neq\{x\}$ then $f(f^{-1}(Y))\neq Y$. $\endgroup$ Commented Dec 10, 2014 at 23:15
  • $\begingroup$ $f(f^{-1}(Y)) \supseteq Y$ only holds when $f$ is surjective. $\endgroup$ Commented Dec 10, 2014 at 23:17
  • $\begingroup$ Related: math.stackexchange.com/questions/746123 $\endgroup$ Commented Jan 26, 2017 at 12:27

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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.

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  • $\begingroup$ For $x \in f^{-1}(Y) \Rightarrow f(x) \in f(f^{-1}(Y))$ you used (I) correct? $\endgroup$ Commented Dec 10, 2014 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. $\endgroup$ Commented Dec 10, 2014 at 23:44

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