Direct mapping and inverse mapping: when they can be equal?












0












$begingroup$


I read the following phrase from a textbook




Bijection $ f : M rightarrow M$ known as symmetry if $ f^{-1} = f$.




I am curious, are there any examples of $ f^{-1} = f$ outside of identity relation - $ id_X $ ?






relations inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    any sort of transposition will do. For example, let f($x_1,x_2,x_3$) = ($x_3,x_2,x_1$).
    $endgroup$
    – Joel Pereira
    Jan 2 at 23:42






  • 1




    $begingroup$
    $-x$ on $mathbb R$, $frac 1 x$ on $(0,infty)$. How many millions do you want?
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 0:05
















0












$begingroup$


I read the following phrase from a textbook




Bijection $ f : M rightarrow M$ known as symmetry if $ f^{-1} = f$.




I am curious, are there any examples of $ f^{-1} = f$ outside of identity relation - $ id_X $ ?






relations inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    any sort of transposition will do. For example, let f($x_1,x_2,x_3$) = ($x_3,x_2,x_1$).
    $endgroup$
    – Joel Pereira
    Jan 2 at 23:42






  • 1




    $begingroup$
    $-x$ on $mathbb R$, $frac 1 x$ on $(0,infty)$. How many millions do you want?
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 0:05














0












0








0


1



$begingroup$


I read the following phrase from a textbook




Bijection $ f : M rightarrow M$ known as symmetry if $ f^{-1} = f$.




I am curious, are there any examples of $ f^{-1} = f$ outside of identity relation - $ id_X $ ?






relations inverse-function







share|cite|improve this question











$endgroup$




I read the following phrase from a textbook




Bijection $ f : M rightarrow M$ known as symmetry if $ f^{-1} = f$.




I am curious, are there any examples of $ f^{-1} = f$ outside of identity relation - $ id_X $ ?






relations inverse-function




relations inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 0:52









Asaf Karagila♦

305k33435766




305k33435766










asked Jan 2 at 23:40









DaddyMDaddyM

1515




1515












  • $begingroup$
    any sort of transposition will do. For example, let f($x_1,x_2,x_3$) = ($x_3,x_2,x_1$).
    $endgroup$
    – Joel Pereira
    Jan 2 at 23:42






  • 1




    $begingroup$
    $-x$ on $mathbb R$, $frac 1 x$ on $(0,infty)$. How many millions do you want?
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 0:05


















  • $begingroup$
    any sort of transposition will do. For example, let f($x_1,x_2,x_3$) = ($x_3,x_2,x_1$).
    $endgroup$
    – Joel Pereira
    Jan 2 at 23:42






  • 1




    $begingroup$
    $-x$ on $mathbb R$, $frac 1 x$ on $(0,infty)$. How many millions do you want?
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 0:05
















$begingroup$
any sort of transposition will do. For example, let f($x_1,x_2,x_3$) = ($x_3,x_2,x_1$).
$endgroup$
– Joel Pereira
Jan 2 at 23:42




$begingroup$
any sort of transposition will do. For example, let f($x_1,x_2,x_3$) = ($x_3,x_2,x_1$).
$endgroup$
– Joel Pereira
Jan 2 at 23:42




1




1




$begingroup$
$-x$ on $mathbb R$, $frac 1 x$ on $(0,infty)$. How many millions do you want?
$endgroup$
– Kavi Rama Murthy
Jan 3 at 0:05




$begingroup$
$-x$ on $mathbb R$, $frac 1 x$ on $(0,infty)$. How many millions do you want?
$endgroup$
– Kavi Rama Murthy
Jan 3 at 0:05










2 Answers
2






active

oldest

votes


















2












$begingroup$

For a set $M$, a function $f:Mto M$ such that $fcirc f=text{id}_M$ is usually known as an involution on $M$. There exists a one-to-one correspondence between the set $mathcal{I}$ of all involutions on $M$ and the set $mathcal{P}$ of partitions of $M$ into subsets of size $1$ or $2$. Such a one-to-one correspondence is given by $phi:mathcal{P}to mathcal{I}$ sending $Pinmathcal{P}$ to
$$f_P(x)=begin{cases}x&text{if }{x}in P,,\y&text{if }{x,y}in Ptext{ with }xneq y,.end{cases}$$
The inverse of $phi$ is $psi:mathcal{I}to mathcal{P}$ sending $finmathcal{I}$ to
$$P_f:=Big{{x},Big|,xin Mtext{ and }f(x)=xBig}cupBig{big{x,f(x)big},Big|,xin Mtext{ and }f(x)neq xBig},.$$
If $M$ is a finite set with $m$ elements and $a_m$ denotes the number of involutions on $M$, then $a_0=1$, $a_1=1$, and $$a_m=a_{m-1}+(m-1)a_{m-2}text{ for }m=2,3,4,ldots,.$$
If $M$ is an infinite set, then the cardinality of $mathcal{I}$ is $2^{|M|}$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Here's a really simple example:



    Let $X$ be a finite set with $vert X vert$ even; then number the elements of $X$ from $1$ to $2n$, $n in Bbb N$:



    $X = {x_1, x_2, ldots, x_n, x_{n + 1}, x_{n + 2}, ldots, x_{2n} }; tag 1$



    define



    $f:X to X tag 2$



    by



    $f(x_i) = x_{i + n}, ; 1 le i le n, tag 3$



    $f(x_i) = x_{i - n}, ; n + 1 le i le 2n; tag 4$



    then since $n ge 2$,



    $f ne text{Id}_X, tag 5$



    but



    $f(f(x_i)) = f(x_{i + n}) = x_i, ; 1 le i le n, tag 6$



    $f(f(x_i)) = f(x_{i -n}) = x_i, ; n + 1 le i le 2n, tag 7$



    so we see that



    $f^2 = text{Id}_X; tag 8$



    therefore, for $x in X$,



    $f(f(x)) = x, tag 9$



    whence



    $f^{-1}(x) = f^{-1}(f(f(x)) = f(x). tag{10}$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Or, you know, a set with at least two elements, and just switch two of them.
      $endgroup$
      – Asaf Karagila♦
      Jan 3 at 0:53










    • $begingroup$
      @AsafKaragila: Oh yeah, of course. I guess I went for the obvious generalization! :)
      $endgroup$
      – Robert Lewis
      Jan 3 at 0:58











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060104%2fdirect-mapping-and-inverse-mapping-when-they-can-be-equal%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    For a set $M$, a function $f:Mto M$ such that $fcirc f=text{id}_M$ is usually known as an involution on $M$. There exists a one-to-one correspondence between the set $mathcal{I}$ of all involutions on $M$ and the set $mathcal{P}$ of partitions of $M$ into subsets of size $1$ or $2$. Such a one-to-one correspondence is given by $phi:mathcal{P}to mathcal{I}$ sending $Pinmathcal{P}$ to
    $$f_P(x)=begin{cases}x&text{if }{x}in P,,\y&text{if }{x,y}in Ptext{ with }xneq y,.end{cases}$$
    The inverse of $phi$ is $psi:mathcal{I}to mathcal{P}$ sending $finmathcal{I}$ to
    $$P_f:=Big{{x},Big|,xin Mtext{ and }f(x)=xBig}cupBig{big{x,f(x)big},Big|,xin Mtext{ and }f(x)neq xBig},.$$
    If $M$ is a finite set with $m$ elements and $a_m$ denotes the number of involutions on $M$, then $a_0=1$, $a_1=1$, and $$a_m=a_{m-1}+(m-1)a_{m-2}text{ for }m=2,3,4,ldots,.$$
    If $M$ is an infinite set, then the cardinality of $mathcal{I}$ is $2^{|M|}$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      For a set $M$, a function $f:Mto M$ such that $fcirc f=text{id}_M$ is usually known as an involution on $M$. There exists a one-to-one correspondence between the set $mathcal{I}$ of all involutions on $M$ and the set $mathcal{P}$ of partitions of $M$ into subsets of size $1$ or $2$. Such a one-to-one correspondence is given by $phi:mathcal{P}to mathcal{I}$ sending $Pinmathcal{P}$ to
      $$f_P(x)=begin{cases}x&text{if }{x}in P,,\y&text{if }{x,y}in Ptext{ with }xneq y,.end{cases}$$
      The inverse of $phi$ is $psi:mathcal{I}to mathcal{P}$ sending $finmathcal{I}$ to
      $$P_f:=Big{{x},Big|,xin Mtext{ and }f(x)=xBig}cupBig{big{x,f(x)big},Big|,xin Mtext{ and }f(x)neq xBig},.$$
      If $M$ is a finite set with $m$ elements and $a_m$ denotes the number of involutions on $M$, then $a_0=1$, $a_1=1$, and $$a_m=a_{m-1}+(m-1)a_{m-2}text{ for }m=2,3,4,ldots,.$$
      If $M$ is an infinite set, then the cardinality of $mathcal{I}$ is $2^{|M|}$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        For a set $M$, a function $f:Mto M$ such that $fcirc f=text{id}_M$ is usually known as an involution on $M$. There exists a one-to-one correspondence between the set $mathcal{I}$ of all involutions on $M$ and the set $mathcal{P}$ of partitions of $M$ into subsets of size $1$ or $2$. Such a one-to-one correspondence is given by $phi:mathcal{P}to mathcal{I}$ sending $Pinmathcal{P}$ to
        $$f_P(x)=begin{cases}x&text{if }{x}in P,,\y&text{if }{x,y}in Ptext{ with }xneq y,.end{cases}$$
        The inverse of $phi$ is $psi:mathcal{I}to mathcal{P}$ sending $finmathcal{I}$ to
        $$P_f:=Big{{x},Big|,xin Mtext{ and }f(x)=xBig}cupBig{big{x,f(x)big},Big|,xin Mtext{ and }f(x)neq xBig},.$$
        If $M$ is a finite set with $m$ elements and $a_m$ denotes the number of involutions on $M$, then $a_0=1$, $a_1=1$, and $$a_m=a_{m-1}+(m-1)a_{m-2}text{ for }m=2,3,4,ldots,.$$
        If $M$ is an infinite set, then the cardinality of $mathcal{I}$ is $2^{|M|}$.






        share|cite|improve this answer











        $endgroup$



        For a set $M$, a function $f:Mto M$ such that $fcirc f=text{id}_M$ is usually known as an involution on $M$. There exists a one-to-one correspondence between the set $mathcal{I}$ of all involutions on $M$ and the set $mathcal{P}$ of partitions of $M$ into subsets of size $1$ or $2$. Such a one-to-one correspondence is given by $phi:mathcal{P}to mathcal{I}$ sending $Pinmathcal{P}$ to
        $$f_P(x)=begin{cases}x&text{if }{x}in P,,\y&text{if }{x,y}in Ptext{ with }xneq y,.end{cases}$$
        The inverse of $phi$ is $psi:mathcal{I}to mathcal{P}$ sending $finmathcal{I}$ to
        $$P_f:=Big{{x},Big|,xin Mtext{ and }f(x)=xBig}cupBig{big{x,f(x)big},Big|,xin Mtext{ and }f(x)neq xBig},.$$
        If $M$ is a finite set with $m$ elements and $a_m$ denotes the number of involutions on $M$, then $a_0=1$, $a_1=1$, and $$a_m=a_{m-1}+(m-1)a_{m-2}text{ for }m=2,3,4,ldots,.$$
        If $M$ is an infinite set, then the cardinality of $mathcal{I}$ is $2^{|M|}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 0:42

























        answered Jan 3 at 1:19









        BatominovskiBatominovski

        33.1k33293




        33.1k33293























            1












            $begingroup$

            Here's a really simple example:



            Let $X$ be a finite set with $vert X vert$ even; then number the elements of $X$ from $1$ to $2n$, $n in Bbb N$:



            $X = {x_1, x_2, ldots, x_n, x_{n + 1}, x_{n + 2}, ldots, x_{2n} }; tag 1$



            define



            $f:X to X tag 2$



            by



            $f(x_i) = x_{i + n}, ; 1 le i le n, tag 3$



            $f(x_i) = x_{i - n}, ; n + 1 le i le 2n; tag 4$



            then since $n ge 2$,



            $f ne text{Id}_X, tag 5$



            but



            $f(f(x_i)) = f(x_{i + n}) = x_i, ; 1 le i le n, tag 6$



            $f(f(x_i)) = f(x_{i -n}) = x_i, ; n + 1 le i le 2n, tag 7$



            so we see that



            $f^2 = text{Id}_X; tag 8$



            therefore, for $x in X$,



            $f(f(x)) = x, tag 9$



            whence



            $f^{-1}(x) = f^{-1}(f(f(x)) = f(x). tag{10}$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Or, you know, a set with at least two elements, and just switch two of them.
              $endgroup$
              – Asaf Karagila♦
              Jan 3 at 0:53










            • $begingroup$
              @AsafKaragila: Oh yeah, of course. I guess I went for the obvious generalization! :)
              $endgroup$
              – Robert Lewis
              Jan 3 at 0:58
















            1












            $begingroup$

            Here's a really simple example:



            Let $X$ be a finite set with $vert X vert$ even; then number the elements of $X$ from $1$ to $2n$, $n in Bbb N$:



            $X = {x_1, x_2, ldots, x_n, x_{n + 1}, x_{n + 2}, ldots, x_{2n} }; tag 1$



            define



            $f:X to X tag 2$



            by



            $f(x_i) = x_{i + n}, ; 1 le i le n, tag 3$



            $f(x_i) = x_{i - n}, ; n + 1 le i le 2n; tag 4$



            then since $n ge 2$,



            $f ne text{Id}_X, tag 5$



            but



            $f(f(x_i)) = f(x_{i + n}) = x_i, ; 1 le i le n, tag 6$



            $f(f(x_i)) = f(x_{i -n}) = x_i, ; n + 1 le i le 2n, tag 7$



            so we see that



            $f^2 = text{Id}_X; tag 8$



            therefore, for $x in X$,



            $f(f(x)) = x, tag 9$



            whence



            $f^{-1}(x) = f^{-1}(f(f(x)) = f(x). tag{10}$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Or, you know, a set with at least two elements, and just switch two of them.
              $endgroup$
              – Asaf Karagila♦
              Jan 3 at 0:53










            • $begingroup$
              @AsafKaragila: Oh yeah, of course. I guess I went for the obvious generalization! :)
              $endgroup$
              – Robert Lewis
              Jan 3 at 0:58














            1












            1








            1





            $begingroup$

            Here's a really simple example:



            Let $X$ be a finite set with $vert X vert$ even; then number the elements of $X$ from $1$ to $2n$, $n in Bbb N$:



            $X = {x_1, x_2, ldots, x_n, x_{n + 1}, x_{n + 2}, ldots, x_{2n} }; tag 1$



            define



            $f:X to X tag 2$



            by



            $f(x_i) = x_{i + n}, ; 1 le i le n, tag 3$



            $f(x_i) = x_{i - n}, ; n + 1 le i le 2n; tag 4$



            then since $n ge 2$,



            $f ne text{Id}_X, tag 5$



            but



            $f(f(x_i)) = f(x_{i + n}) = x_i, ; 1 le i le n, tag 6$



            $f(f(x_i)) = f(x_{i -n}) = x_i, ; n + 1 le i le 2n, tag 7$



            so we see that



            $f^2 = text{Id}_X; tag 8$



            therefore, for $x in X$,



            $f(f(x)) = x, tag 9$



            whence



            $f^{-1}(x) = f^{-1}(f(f(x)) = f(x). tag{10}$






            share|cite|improve this answer









            $endgroup$



            Here's a really simple example:



            Let $X$ be a finite set with $vert X vert$ even; then number the elements of $X$ from $1$ to $2n$, $n in Bbb N$:



            $X = {x_1, x_2, ldots, x_n, x_{n + 1}, x_{n + 2}, ldots, x_{2n} }; tag 1$



            define



            $f:X to X tag 2$



            by



            $f(x_i) = x_{i + n}, ; 1 le i le n, tag 3$



            $f(x_i) = x_{i - n}, ; n + 1 le i le 2n; tag 4$



            then since $n ge 2$,



            $f ne text{Id}_X, tag 5$



            but



            $f(f(x_i)) = f(x_{i + n}) = x_i, ; 1 le i le n, tag 6$



            $f(f(x_i)) = f(x_{i -n}) = x_i, ; n + 1 le i le 2n, tag 7$



            so we see that



            $f^2 = text{Id}_X; tag 8$



            therefore, for $x in X$,



            $f(f(x)) = x, tag 9$



            whence



            $f^{-1}(x) = f^{-1}(f(f(x)) = f(x). tag{10}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 0:28









            Robert LewisRobert Lewis

            47.5k23067




            47.5k23067








            • 1




              $begingroup$
              Or, you know, a set with at least two elements, and just switch two of them.
              $endgroup$
              – Asaf Karagila♦
              Jan 3 at 0:53










            • $begingroup$
              @AsafKaragila: Oh yeah, of course. I guess I went for the obvious generalization! :)
              $endgroup$
              – Robert Lewis
              Jan 3 at 0:58














            • 1




              $begingroup$
              Or, you know, a set with at least two elements, and just switch two of them.
              $endgroup$
              – Asaf Karagila♦
              Jan 3 at 0:53










            • $begingroup$
              @AsafKaragila: Oh yeah, of course. I guess I went for the obvious generalization! :)
              $endgroup$
              – Robert Lewis
              Jan 3 at 0:58








            1




            1




            $begingroup$
            Or, you know, a set with at least two elements, and just switch two of them.
            $endgroup$
            – Asaf Karagila♦
            Jan 3 at 0:53




            $begingroup$
            Or, you know, a set with at least two elements, and just switch two of them.
            $endgroup$
            – Asaf Karagila♦
            Jan 3 at 0:53












            $begingroup$
            @AsafKaragila: Oh yeah, of course. I guess I went for the obvious generalization! :)
            $endgroup$
            – Robert Lewis
            Jan 3 at 0:58




            $begingroup$
            @AsafKaragila: Oh yeah, of course. I guess I went for the obvious generalization! :)
            $endgroup$
            – Robert Lewis
            Jan 3 at 0:58


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060104%2fdirect-mapping-and-inverse-mapping-when-they-can-be-equal%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more