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Let $f:X\to Y$ be a function between two sets. Let $A\subset X$. Is it always possible to express $A$ as $f^{-1}(B)$ for some $B\subset Y$?

I think that the question is possibly silly. Would someone point out if that is the case?

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3 Answers 3

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This is an answer to the bonus (?) question in the comments. If $f$ is not injective, say $f(x)=f(y)$ for some $x$, $y$ with $x\ne y$, then for any set $B\subset Y$, either $x$ and $y$ are both in $f^{-1}(B)$, or none of them are. So any set $A$ that contains one but not the other is not an inverse image.

Conversely, if $f$ is injective it is clear that $A=f^{-1}(f(A))$.

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No, of course not. For example, let $f$ be a constant function, $f(x)=c$. Then for any $B\subset Y$, $f^{-1}(B)$ is either $X$ (if $c \in B$) or $\varnothing$ (if $c \not \in B$). So if $X$ has more than one element, then $f^{-1}$ is not surjective.

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I am sorry but could I ask the question again for the case when $f$ is surjective – user38404 Dec 13 '12 at 20:49
Let $Y$ be a one-point set, and my $f$ is already surjective. – Chris Eagle Dec 13 '12 at 20:54

The inverse image of a function exists if it is bijective.

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