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Prove, disprove, or give a counterexample:

Suppose $f:X\to Y$ is onto and $A\subseteq Y$. Then $f(f^{-1}(A))=A$.

Edit: Does this work?

Suppose $f:X \to Y$ is onto and $A \subseteq Y$.

We know that for all $y\in Y$, there exists $x\in X$ such that $f(x)=y$. Let $x\in f(f^{-1}(A))$ and because $f:X\to Y$ is onto then for all $x\in A$ there exists $y\in f^{-1}(A)$ such that $f(y)= x$. Therefore, $x\in f(f^{-1}(A))$ $\to$ $A\subseteq f(f^{-1}(A))$. Thus, $f(f^{-1}(A))=A$ by definition of set equality.

There is likely a problem or two with that proof. If anyone had further input that'd be great!

Martin Sleziak
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Vincent
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1 Answers1

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The inclusion $f(f^{-1}(A))\subset A$ always holds. Conversely, let $y\in A$. Since $f$ is surjective there exists $x\in X$ such that $y=f(x)$. Hence $x\in f^{-1}(A)$ so that $y=f(x)\in f(\underbrace{f^{-1}(A)}_{\ni x})$.

PtF
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