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












1














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
























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






  • 1




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
















1














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
























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






  • 1




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














1












1








1


1





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















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♦

302k32426756




302k32426756










asked Jan 2 at 23:40









DaddyM

1514




1514












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






  • 1




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


















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






  • 1




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
















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




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




1




1




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




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










2 Answers
2






active

oldest

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3














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































    1














    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

















    • 1




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










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











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    2 Answers
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    2 Answers
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    active

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    3














    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




























      3














      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


























        3












        3








        3






        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














        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









        Batominovski

        33.9k33294




        33.9k33294























            1














            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

















            • 1




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










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
















            1














            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

















            • 1




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










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














            1












            1








            1






            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












            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 Lewis

            43.9k22963




            43.9k22963








            • 1




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










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














            • 1




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










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








            1




            1




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




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












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




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


















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