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Let $X,Y$ be sets and $f: X \to Y$. If $A,B \subseteq Y$, is it true that $$f^{-1}(A) \subseteq f^{-1}(B) \qquad \Rightarrow \qquad A \subseteq B$$

I know this is a stupid question, but at the moment I cannot come up with either a counterexample or a proof. I am sorry to ask this with that much reputation, but at the moment I do not really see if this is right or wrong.

Edit. It is wrong. For example consider $f = x^2$. Then $$f^{-1}([0,1]) = [0,1] = f^{-1}([-1,1])$$ but clearly $$[-1,1] \nsubseteq [0,1]$$ Thanks for the answers until now. May this implication be fixed with either assuming $f$ to be injective or surjective?

Ethan Bolker
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TheGeekGreek
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3 Answers3

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It's false. Let $X = Y = \{0,1\}$. Choose $A = \{0,1\}$, $B = \{1\}$, and $\forall x. f(x) = 1$, then $f^{-1}(A) = f^{-1}(B) = X$.

Uwe
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Let $X=\{a,b\}$, $Y=\{c,d\}$, $A=\{c\}$, $B=\{d\}$ and $f(a)=f(b)=d$. Then $$f^{-1}(A)=\emptyset\subseteq X=f^{-1}(B),$$ but $B\not\subseteq A$.

Rasmus Erlemann
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What are $f$, $A$ and $B$? If there is no assumption we take $f=c\in Y$ a constant function, $B=\{c\}$ and $A=Y$ then $f^{-1}(A)=f^{-1}(B)=X$ but $A\not\subset B$.

Kara
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  • $f,A,B$ are arbitrary only depending on the choosen sets $X$ and $Y$. This is an abstract question on functions in general. Thanks for your counterexample. – TheGeekGreek Jan 22 '17 at 14:43