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?
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?
The inverse image of a function exists if it is bijective.
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.
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 $A=f^{-1}(f(A))$.