For $f:Xto Y$ and $A$ a subset of $X$, why are $A$ and $f^{-1}(f(A))$ not necessarily equal?












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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?




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    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?




    Thanks!






    functions







    share|cite|improve this question



























      0












      0








      0


      1





      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?




      Thanks!






      functions







      share|cite|improve this question















      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?




      Thanks!






      functions




      functions






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      edited Nov 29 '18 at 19:57









      Blue

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      asked Nov 29 '18 at 14:25









      FridaFrida

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



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






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



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






            share|cite|improve this answer


























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



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






              share|cite|improve this answer
























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



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






                share|cite|improve this answer












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



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







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '18 at 15:13









                Shubham JohriShubham Johri

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                4,684717






























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