Is $f^{-1}(f(A))=A$ always true?












7












$begingroup$


If we have a function $f:Xrightarrow Y$ where $Asubset X$, is it true to say that $f^{-1}(f(A))=A$?






elementary-set-theory functions







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $f^{-1}(f(A)) supseteq A$ , link
    $endgroup$
    – Peđa Terzić
    Nov 2 '11 at 7:52










  • $begingroup$
    However, we always have math.stackexchange.com/questions/746123/prove-that-ff-1-fx-fx
    $endgroup$
    – Watson
    Jan 26 '17 at 12:28
















7












$begingroup$


If we have a function $f:Xrightarrow Y$ where $Asubset X$, is it true to say that $f^{-1}(f(A))=A$?






elementary-set-theory functions







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $f^{-1}(f(A)) supseteq A$ , link
    $endgroup$
    – Peđa Terzić
    Nov 2 '11 at 7:52










  • $begingroup$
    However, we always have math.stackexchange.com/questions/746123/prove-that-ff-1-fx-fx
    $endgroup$
    – Watson
    Jan 26 '17 at 12:28














7












7








7


7



$begingroup$


If we have a function $f:Xrightarrow Y$ where $Asubset X$, is it true to say that $f^{-1}(f(A))=A$?






elementary-set-theory functions







share|cite|improve this question











$endgroup$




If we have a function $f:Xrightarrow Y$ where $Asubset X$, is it true to say that $f^{-1}(f(A))=A$?






elementary-set-theory functions




elementary-set-theory functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 22 '13 at 14:32









Martin Sleziak

45k10122277




45k10122277










asked Nov 2 '11 at 7:29









johnnymathjohnnymath

3811612




3811612








  • 4




    $begingroup$
    $f^{-1}(f(A)) supseteq A$ , link
    $endgroup$
    – Peđa Terzić
    Nov 2 '11 at 7:52










  • $begingroup$
    However, we always have math.stackexchange.com/questions/746123/prove-that-ff-1-fx-fx
    $endgroup$
    – Watson
    Jan 26 '17 at 12:28














  • 4




    $begingroup$
    $f^{-1}(f(A)) supseteq A$ , link
    $endgroup$
    – Peđa Terzić
    Nov 2 '11 at 7:52










  • $begingroup$
    However, we always have math.stackexchange.com/questions/746123/prove-that-ff-1-fx-fx
    $endgroup$
    – Watson
    Jan 26 '17 at 12:28








4




4




$begingroup$
$f^{-1}(f(A)) supseteq A$ , link
$endgroup$
– Peđa Terzić
Nov 2 '11 at 7:52




$begingroup$
$f^{-1}(f(A)) supseteq A$ , link
$endgroup$
– Peđa Terzić
Nov 2 '11 at 7:52












$begingroup$
However, we always have math.stackexchange.com/questions/746123/prove-that-ff-1-fx-fx
$endgroup$
– Watson
Jan 26 '17 at 12:28




$begingroup$
However, we always have math.stackexchange.com/questions/746123/prove-that-ff-1-fx-fx
$endgroup$
– Watson
Jan 26 '17 at 12:28










3 Answers
3






active

oldest

votes


















10












$begingroup$

As noted, the asserted equality is not true.



In general, one inclusion always holds: $$Asubseteq f^{-1}(f(A)).$$



How to see that? Remember that $xin f^{-1}(B)$ if and only if $f(x)in B$.



Now, to show $A$ is contained in $f^{-1}Bigl( f(A)Bigr)$, let $ain A$; we need to show that $ain f^{-1}Bigl( f(A)Bigr)$. But this holds if and only if $f(a)in f(A)$, which holds since $ain A$ and $f(A) = {f(x)mid xin A}$.



The other inclusion does not hold in general, but you do have the following:



Proposition. Let $fcolon Xto Y$ be a function. Then $f$ is one to one (injective) if and only if for every $Asubseteq X$, we have $A=f^{-1}(f(A))$.



Proof. Assume first that $f$ is injective, and let $Asubseteq X$. We already know that $Asubseteq f^{-1}(f(A))$, so we only need to show that $f^{-1}(f(A))subseteq A$. Let $xin f^{-1}(f(A))$; we want to prove that $xin A$. That means that $f(x)in f(A)$, so there exists $ain A$ such that $f(x)=f(a)$. But since $f$ is one-to-one, this implies $x=a$, so $xin A$, as desired.



Conversely, assume that for every $Asubseteq X$, $A=f^{-1}(f(A))$. Let $x,x’in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. Let $A={x}$; then $f(x')in f(A)$, so $x'in f^{-1}(f(A))$. By assumption, $f^{-1}(f(A))=A={x}$, so we can conclude that $x'in {x}$; but this means $x'=x$, which is what we needed to prove. $Box$






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    No. Not in general. Note that if you take the constant map $xmapsto 1$ mapping $mathbb{R}tomathbb{R}$ then $f^{-1}(f({0}))=mathbb{R}$. In fact, the equality you wrote holds true for all subsets of $X$ precisely when $f$ is injective.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      No. This need not be true.



      For example, $X=Y={0,1}$, $f(x)=0$ and $A={1}$.



      $f(A) = {0}$ and $f^{-1}({0})=X$, so $f^{-1}(f(A))=Xneq A$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I think you have some typos here, but I don't want to edit your work.
        $endgroup$
        – Squirtle
        Sep 10 '12 at 15:12










      • $begingroup$
        @dustanalysis: Why do you think I have typos here?
        $endgroup$
        – Asaf Karagila♦
        Sep 10 '12 at 15:26










      • $begingroup$
        Why do you use f(x), do you mean f(X)?
        $endgroup$
        – Squirtle
        Sep 10 '12 at 15:28










      • $begingroup$
        @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $xinoperatorname{dom}(f)$ we have that $f(x)=0$.
        $endgroup$
        – Asaf Karagila♦
        Sep 10 '12 at 15:31












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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      10












      $begingroup$

      As noted, the asserted equality is not true.



      In general, one inclusion always holds: $$Asubseteq f^{-1}(f(A)).$$



      How to see that? Remember that $xin f^{-1}(B)$ if and only if $f(x)in B$.



      Now, to show $A$ is contained in $f^{-1}Bigl( f(A)Bigr)$, let $ain A$; we need to show that $ain f^{-1}Bigl( f(A)Bigr)$. But this holds if and only if $f(a)in f(A)$, which holds since $ain A$ and $f(A) = {f(x)mid xin A}$.



      The other inclusion does not hold in general, but you do have the following:



      Proposition. Let $fcolon Xto Y$ be a function. Then $f$ is one to one (injective) if and only if for every $Asubseteq X$, we have $A=f^{-1}(f(A))$.



      Proof. Assume first that $f$ is injective, and let $Asubseteq X$. We already know that $Asubseteq f^{-1}(f(A))$, so we only need to show that $f^{-1}(f(A))subseteq A$. Let $xin f^{-1}(f(A))$; we want to prove that $xin A$. That means that $f(x)in f(A)$, so there exists $ain A$ such that $f(x)=f(a)$. But since $f$ is one-to-one, this implies $x=a$, so $xin A$, as desired.



      Conversely, assume that for every $Asubseteq X$, $A=f^{-1}(f(A))$. Let $x,x’in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. Let $A={x}$; then $f(x')in f(A)$, so $x'in f^{-1}(f(A))$. By assumption, $f^{-1}(f(A))=A={x}$, so we can conclude that $x'in {x}$; but this means $x'=x$, which is what we needed to prove. $Box$






      share|cite|improve this answer











      $endgroup$


















        10












        $begingroup$

        As noted, the asserted equality is not true.



        In general, one inclusion always holds: $$Asubseteq f^{-1}(f(A)).$$



        How to see that? Remember that $xin f^{-1}(B)$ if and only if $f(x)in B$.



        Now, to show $A$ is contained in $f^{-1}Bigl( f(A)Bigr)$, let $ain A$; we need to show that $ain f^{-1}Bigl( f(A)Bigr)$. But this holds if and only if $f(a)in f(A)$, which holds since $ain A$ and $f(A) = {f(x)mid xin A}$.



        The other inclusion does not hold in general, but you do have the following:



        Proposition. Let $fcolon Xto Y$ be a function. Then $f$ is one to one (injective) if and only if for every $Asubseteq X$, we have $A=f^{-1}(f(A))$.



        Proof. Assume first that $f$ is injective, and let $Asubseteq X$. We already know that $Asubseteq f^{-1}(f(A))$, so we only need to show that $f^{-1}(f(A))subseteq A$. Let $xin f^{-1}(f(A))$; we want to prove that $xin A$. That means that $f(x)in f(A)$, so there exists $ain A$ such that $f(x)=f(a)$. But since $f$ is one-to-one, this implies $x=a$, so $xin A$, as desired.



        Conversely, assume that for every $Asubseteq X$, $A=f^{-1}(f(A))$. Let $x,x’in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. Let $A={x}$; then $f(x')in f(A)$, so $x'in f^{-1}(f(A))$. By assumption, $f^{-1}(f(A))=A={x}$, so we can conclude that $x'in {x}$; but this means $x'=x$, which is what we needed to prove. $Box$






        share|cite|improve this answer











        $endgroup$
















          10












          10








          10





          $begingroup$

          As noted, the asserted equality is not true.



          In general, one inclusion always holds: $$Asubseteq f^{-1}(f(A)).$$



          How to see that? Remember that $xin f^{-1}(B)$ if and only if $f(x)in B$.



          Now, to show $A$ is contained in $f^{-1}Bigl( f(A)Bigr)$, let $ain A$; we need to show that $ain f^{-1}Bigl( f(A)Bigr)$. But this holds if and only if $f(a)in f(A)$, which holds since $ain A$ and $f(A) = {f(x)mid xin A}$.



          The other inclusion does not hold in general, but you do have the following:



          Proposition. Let $fcolon Xto Y$ be a function. Then $f$ is one to one (injective) if and only if for every $Asubseteq X$, we have $A=f^{-1}(f(A))$.



          Proof. Assume first that $f$ is injective, and let $Asubseteq X$. We already know that $Asubseteq f^{-1}(f(A))$, so we only need to show that $f^{-1}(f(A))subseteq A$. Let $xin f^{-1}(f(A))$; we want to prove that $xin A$. That means that $f(x)in f(A)$, so there exists $ain A$ such that $f(x)=f(a)$. But since $f$ is one-to-one, this implies $x=a$, so $xin A$, as desired.



          Conversely, assume that for every $Asubseteq X$, $A=f^{-1}(f(A))$. Let $x,x’in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. Let $A={x}$; then $f(x')in f(A)$, so $x'in f^{-1}(f(A))$. By assumption, $f^{-1}(f(A))=A={x}$, so we can conclude that $x'in {x}$; but this means $x'=x$, which is what we needed to prove. $Box$






          share|cite|improve this answer











          $endgroup$



          As noted, the asserted equality is not true.



          In general, one inclusion always holds: $$Asubseteq f^{-1}(f(A)).$$



          How to see that? Remember that $xin f^{-1}(B)$ if and only if $f(x)in B$.



          Now, to show $A$ is contained in $f^{-1}Bigl( f(A)Bigr)$, let $ain A$; we need to show that $ain f^{-1}Bigl( f(A)Bigr)$. But this holds if and only if $f(a)in f(A)$, which holds since $ain A$ and $f(A) = {f(x)mid xin A}$.



          The other inclusion does not hold in general, but you do have the following:



          Proposition. Let $fcolon Xto Y$ be a function. Then $f$ is one to one (injective) if and only if for every $Asubseteq X$, we have $A=f^{-1}(f(A))$.



          Proof. Assume first that $f$ is injective, and let $Asubseteq X$. We already know that $Asubseteq f^{-1}(f(A))$, so we only need to show that $f^{-1}(f(A))subseteq A$. Let $xin f^{-1}(f(A))$; we want to prove that $xin A$. That means that $f(x)in f(A)$, so there exists $ain A$ such that $f(x)=f(a)$. But since $f$ is one-to-one, this implies $x=a$, so $xin A$, as desired.



          Conversely, assume that for every $Asubseteq X$, $A=f^{-1}(f(A))$. Let $x,x’in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. Let $A={x}$; then $f(x')in f(A)$, so $x'in f^{-1}(f(A))$. By assumption, $f^{-1}(f(A))=A={x}$, so we can conclude that $x'in {x}$; but this means $x'=x$, which is what we needed to prove. $Box$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 15:36

























          answered Nov 2 '11 at 16:33









          Arturo MagidinArturo Magidin

          266k34590921




          266k34590921























              3












              $begingroup$

              No. Not in general. Note that if you take the constant map $xmapsto 1$ mapping $mathbb{R}tomathbb{R}$ then $f^{-1}(f({0}))=mathbb{R}$. In fact, the equality you wrote holds true for all subsets of $X$ precisely when $f$ is injective.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                No. Not in general. Note that if you take the constant map $xmapsto 1$ mapping $mathbb{R}tomathbb{R}$ then $f^{-1}(f({0}))=mathbb{R}$. In fact, the equality you wrote holds true for all subsets of $X$ precisely when $f$ is injective.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  No. Not in general. Note that if you take the constant map $xmapsto 1$ mapping $mathbb{R}tomathbb{R}$ then $f^{-1}(f({0}))=mathbb{R}$. In fact, the equality you wrote holds true for all subsets of $X$ precisely when $f$ is injective.






                  share|cite|improve this answer











                  $endgroup$



                  No. Not in general. Note that if you take the constant map $xmapsto 1$ mapping $mathbb{R}tomathbb{R}$ then $f^{-1}(f({0}))=mathbb{R}$. In fact, the equality you wrote holds true for all subsets of $X$ precisely when $f$ is injective.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 22 '13 at 14:43









                  Michael Albanese

                  64.7k1599315




                  64.7k1599315










                  answered Nov 2 '11 at 7:32









                  Alex YoucisAlex Youcis

                  36.3k775115




                  36.3k775115























                      2












                      $begingroup$

                      No. This need not be true.



                      For example, $X=Y={0,1}$, $f(x)=0$ and $A={1}$.



                      $f(A) = {0}$ and $f^{-1}({0})=X$, so $f^{-1}(f(A))=Xneq A$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I think you have some typos here, but I don't want to edit your work.
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:12










                      • $begingroup$
                        @dustanalysis: Why do you think I have typos here?
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:26










                      • $begingroup$
                        Why do you use f(x), do you mean f(X)?
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:28










                      • $begingroup$
                        @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $xinoperatorname{dom}(f)$ we have that $f(x)=0$.
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:31
















                      2












                      $begingroup$

                      No. This need not be true.



                      For example, $X=Y={0,1}$, $f(x)=0$ and $A={1}$.



                      $f(A) = {0}$ and $f^{-1}({0})=X$, so $f^{-1}(f(A))=Xneq A$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I think you have some typos here, but I don't want to edit your work.
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:12










                      • $begingroup$
                        @dustanalysis: Why do you think I have typos here?
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:26










                      • $begingroup$
                        Why do you use f(x), do you mean f(X)?
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:28










                      • $begingroup$
                        @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $xinoperatorname{dom}(f)$ we have that $f(x)=0$.
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:31














                      2












                      2








                      2





                      $begingroup$

                      No. This need not be true.



                      For example, $X=Y={0,1}$, $f(x)=0$ and $A={1}$.



                      $f(A) = {0}$ and $f^{-1}({0})=X$, so $f^{-1}(f(A))=Xneq A$.






                      share|cite|improve this answer









                      $endgroup$



                      No. This need not be true.



                      For example, $X=Y={0,1}$, $f(x)=0$ and $A={1}$.



                      $f(A) = {0}$ and $f^{-1}({0})=X$, so $f^{-1}(f(A))=Xneq A$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 2 '11 at 7:34









                      Asaf Karagila♦Asaf Karagila

                      308k33441774




                      308k33441774












                      • $begingroup$
                        I think you have some typos here, but I don't want to edit your work.
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:12










                      • $begingroup$
                        @dustanalysis: Why do you think I have typos here?
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:26










                      • $begingroup$
                        Why do you use f(x), do you mean f(X)?
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:28










                      • $begingroup$
                        @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $xinoperatorname{dom}(f)$ we have that $f(x)=0$.
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:31


















                      • $begingroup$
                        I think you have some typos here, but I don't want to edit your work.
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:12










                      • $begingroup$
                        @dustanalysis: Why do you think I have typos here?
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:26










                      • $begingroup$
                        Why do you use f(x), do you mean f(X)?
                        $endgroup$
                        – Squirtle
                        Sep 10 '12 at 15:28










                      • $begingroup$
                        @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $xinoperatorname{dom}(f)$ we have that $f(x)=0$.
                        $endgroup$
                        – Asaf Karagila♦
                        Sep 10 '12 at 15:31
















                      $begingroup$
                      I think you have some typos here, but I don't want to edit your work.
                      $endgroup$
                      – Squirtle
                      Sep 10 '12 at 15:12




                      $begingroup$
                      I think you have some typos here, but I don't want to edit your work.
                      $endgroup$
                      – Squirtle
                      Sep 10 '12 at 15:12












                      $begingroup$
                      @dustanalysis: Why do you think I have typos here?
                      $endgroup$
                      – Asaf Karagila♦
                      Sep 10 '12 at 15:26




                      $begingroup$
                      @dustanalysis: Why do you think I have typos here?
                      $endgroup$
                      – Asaf Karagila♦
                      Sep 10 '12 at 15:26












                      $begingroup$
                      Why do you use f(x), do you mean f(X)?
                      $endgroup$
                      – Squirtle
                      Sep 10 '12 at 15:28




                      $begingroup$
                      Why do you use f(x), do you mean f(X)?
                      $endgroup$
                      – Squirtle
                      Sep 10 '12 at 15:28












                      $begingroup$
                      @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $xinoperatorname{dom}(f)$ we have that $f(x)=0$.
                      $endgroup$
                      – Asaf Karagila♦
                      Sep 10 '12 at 15:31




                      $begingroup$
                      @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $xinoperatorname{dom}(f)$ we have that $f(x)=0$.
                      $endgroup$
                      – Asaf Karagila♦
                      Sep 10 '12 at 15:31


















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