I need help with this question:

Let $f: X \to Y$ be a function, and let $A$ be a subset of $X$. Why are the sets $A$ and $f^{-1}(f(A))$ not necessarily equal? Under what conditions are they equal?



Consider the function $f:\{1,2,3\}\rightarrow\{2,3,4\}, f(1)=2, f(2)=f(3)=3$ and let $A=\{1,2\}\subseteq\{1,2,3\}$.

$f(A)=f(\{1,2\})=\{2,3\}\implies f^{-1}(f(A))=f^{-1}(\{2,3\})=\{1,2,3\}\neq\{1,2\}=A$

We define the direct image of $A\subseteq X$ as $f(A)=\{f(x): x\in A\}$ and the inverse image of $B\subseteq Y$ as $f^{-1}(B)=\{x: x\in X, f(x)\in B \}$. Clearly, the inverse image of $f(A)$, that is $f^{-1}(f(A)),$ contains all the elements of $A$ as $f(x)\in f(A),\ \forall x\in A$. This means $A\subseteq f^{-1}(f(A))$. But this is not to say that $f^{-1}(f(A))$ contains only the elements of $A$. If $\exists\ y\in X-A$ such that $f(y)\in f(A)$, then $y\in f^{-1}(f(A))$ but $y\notin A$.

The sufficient and necessary condition for $f^{-1}(f(A))=A,\ \forall A\subseteq X$, is that $f$ is injective. You already know from above that $A\subseteq f^{-1}(f(A))$. Can you prove the converse: $f^{-1}(f(A))\subseteq A,\forall A\subseteq X$ given $f$ is injective?


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