0
$\begingroup$

Prove or disprove:

Let $f : X \to Y$ be an onto function and let $A$ and $B$ be subsets of $Y$. If $f^{−1}(A) \subseteq f^{−1}(B)$, then $A \subseteq B$.

Is this statement true? I try to come with counterexample but I cannot; I think it is true but I am not sure.

Could you please confirm that for me ?

$\endgroup$
  • 2
    $\begingroup$ An interesting thing to keep in mind is: let $f:X\to Y$ be a function. Then, $f$ is surjective if and only if $f(f^{-1}(S))=S$ for all $S\subseteq Y$, while $f$ is injective if and only if $f^{-1}(f(T))=T$ for all $T\subseteq X$. $\endgroup$ – user228113 Mar 10 '17 at 5:18
  • $\begingroup$ @G.Sassatelli Nice information. Thanks! $\endgroup$ – Error 404 Mar 10 '17 at 5:31
3
$\begingroup$

The claim is true. Let $a \in A$. Since $f$ is onto, there exists $x \in X$ such that $f(x)=a$. But that means that $x \in f^{-1}(A)$ and thus $x \in f^{-1}(B)$. Therefore, $a=f(x) \in B,$ as desired.

The claim is not true if $f$ is not onto. To see that, let $X=Y=\mathbb{R}$ and let $A=[-2,2],$ $B=[-1,1],$ and $f(x)=0$ for all $x.$ Then clearly $X=f^{-1}(A)=f^{-1}(B)$ but $A$ is not a subset of $B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.