Preserving Bijectivity under pointwise limit












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The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.




Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.







real-analysis convergence pointwise-convergence







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    0












    $begingroup$


    The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.




    Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.







    real-analysis convergence pointwise-convergence







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.




      Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.







      real-analysis convergence pointwise-convergence







      share|cite|improve this question









      $endgroup$




      The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.




      Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.







      real-analysis convergence pointwise-convergence




      real-analysis convergence pointwise-convergence






      share|cite|improve this question













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      asked Nov 8 '18 at 4:07









      Better2BLuckyBetter2BLucky

      586




      586






















          2 Answers
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          $begingroup$

          Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define



          $$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$



          Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,



          $$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$



          Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Typo: $g circ f = 0$ on $(0,1)$.
            $endgroup$
            – Paul Frost
            Dec 2 '18 at 9:31










          • $begingroup$
            @PaulFrost Indeed. Thanks. Corrected.
            $endgroup$
            – zhw.
            Dec 2 '18 at 16:57



















          0












          $begingroup$

          It is not true.



          Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
          $$h_i(t) =
          begin{cases}
          frac{t}{i} & t le frac{1}{2} \
          frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
          end{cases}
          $$
          The inverse homeomorphisms are given by
          $$h_i^{-1} =
          begin{cases}
          ti & t le frac{1}{2i} \
          frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
          end{cases}
          $$
          Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
          $$h : I to I, h(t) =
          begin{cases}
          0 & t le frac{1}{2} \
          2(t-frac{1}{2}) & t ge frac{1}{2}
          end{cases}
          $$
          The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
          $$h^- : I to I, h^-(t) =
          begin{cases}
          0 & t = 0 \
          frac{1}{2}+frac{1}{2}t & t > 0
          end{cases}
          $$
          This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.



          To get examples on $mathbb{R}^n$, define
          $$f_i(x) =
          begin{cases}
          x & lVert x rVert ge 1 \
          h_i(lVert x rVert)x & lVert x rVert le 1
          end{cases}
          $$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2












            $begingroup$

            Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define



            $$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$



            Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,



            $$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$



            Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Typo: $g circ f = 0$ on $(0,1)$.
              $endgroup$
              – Paul Frost
              Dec 2 '18 at 9:31










            • $begingroup$
              @PaulFrost Indeed. Thanks. Corrected.
              $endgroup$
              – zhw.
              Dec 2 '18 at 16:57
















            2












            $begingroup$

            Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define



            $$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$



            Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,



            $$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$



            Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Typo: $g circ f = 0$ on $(0,1)$.
              $endgroup$
              – Paul Frost
              Dec 2 '18 at 9:31










            • $begingroup$
              @PaulFrost Indeed. Thanks. Corrected.
              $endgroup$
              – zhw.
              Dec 2 '18 at 16:57














            2












            2








            2





            $begingroup$

            Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define



            $$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$



            Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,



            $$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$



            Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.






            share|cite|improve this answer











            $endgroup$



            Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define



            $$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$



            Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,



            $$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$



            Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 '18 at 19:21

























            answered Dec 1 '18 at 18:57









            zhw.zhw.

            71.8k43075




            71.8k43075












            • $begingroup$
              Typo: $g circ f = 0$ on $(0,1)$.
              $endgroup$
              – Paul Frost
              Dec 2 '18 at 9:31










            • $begingroup$
              @PaulFrost Indeed. Thanks. Corrected.
              $endgroup$
              – zhw.
              Dec 2 '18 at 16:57


















            • $begingroup$
              Typo: $g circ f = 0$ on $(0,1)$.
              $endgroup$
              – Paul Frost
              Dec 2 '18 at 9:31










            • $begingroup$
              @PaulFrost Indeed. Thanks. Corrected.
              $endgroup$
              – zhw.
              Dec 2 '18 at 16:57
















            $begingroup$
            Typo: $g circ f = 0$ on $(0,1)$.
            $endgroup$
            – Paul Frost
            Dec 2 '18 at 9:31




            $begingroup$
            Typo: $g circ f = 0$ on $(0,1)$.
            $endgroup$
            – Paul Frost
            Dec 2 '18 at 9:31












            $begingroup$
            @PaulFrost Indeed. Thanks. Corrected.
            $endgroup$
            – zhw.
            Dec 2 '18 at 16:57




            $begingroup$
            @PaulFrost Indeed. Thanks. Corrected.
            $endgroup$
            – zhw.
            Dec 2 '18 at 16:57











            0












            $begingroup$

            It is not true.



            Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
            $$h_i(t) =
            begin{cases}
            frac{t}{i} & t le frac{1}{2} \
            frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
            end{cases}
            $$
            The inverse homeomorphisms are given by
            $$h_i^{-1} =
            begin{cases}
            ti & t le frac{1}{2i} \
            frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
            end{cases}
            $$
            Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
            $$h : I to I, h(t) =
            begin{cases}
            0 & t le frac{1}{2} \
            2(t-frac{1}{2}) & t ge frac{1}{2}
            end{cases}
            $$
            The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
            $$h^- : I to I, h^-(t) =
            begin{cases}
            0 & t = 0 \
            frac{1}{2}+frac{1}{2}t & t > 0
            end{cases}
            $$
            This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.



            To get examples on $mathbb{R}^n$, define
            $$f_i(x) =
            begin{cases}
            x & lVert x rVert ge 1 \
            h_i(lVert x rVert)x & lVert x rVert le 1
            end{cases}
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              It is not true.



              Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
              $$h_i(t) =
              begin{cases}
              frac{t}{i} & t le frac{1}{2} \
              frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
              end{cases}
              $$
              The inverse homeomorphisms are given by
              $$h_i^{-1} =
              begin{cases}
              ti & t le frac{1}{2i} \
              frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
              end{cases}
              $$
              Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
              $$h : I to I, h(t) =
              begin{cases}
              0 & t le frac{1}{2} \
              2(t-frac{1}{2}) & t ge frac{1}{2}
              end{cases}
              $$
              The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
              $$h^- : I to I, h^-(t) =
              begin{cases}
              0 & t = 0 \
              frac{1}{2}+frac{1}{2}t & t > 0
              end{cases}
              $$
              This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.



              To get examples on $mathbb{R}^n$, define
              $$f_i(x) =
              begin{cases}
              x & lVert x rVert ge 1 \
              h_i(lVert x rVert)x & lVert x rVert le 1
              end{cases}
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                It is not true.



                Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
                $$h_i(t) =
                begin{cases}
                frac{t}{i} & t le frac{1}{2} \
                frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
                end{cases}
                $$
                The inverse homeomorphisms are given by
                $$h_i^{-1} =
                begin{cases}
                ti & t le frac{1}{2i} \
                frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
                end{cases}
                $$
                Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
                $$h : I to I, h(t) =
                begin{cases}
                0 & t le frac{1}{2} \
                2(t-frac{1}{2}) & t ge frac{1}{2}
                end{cases}
                $$
                The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
                $$h^- : I to I, h^-(t) =
                begin{cases}
                0 & t = 0 \
                frac{1}{2}+frac{1}{2}t & t > 0
                end{cases}
                $$
                This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.



                To get examples on $mathbb{R}^n$, define
                $$f_i(x) =
                begin{cases}
                x & lVert x rVert ge 1 \
                h_i(lVert x rVert)x & lVert x rVert le 1
                end{cases}
                $$






                share|cite|improve this answer









                $endgroup$



                It is not true.



                Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
                $$h_i(t) =
                begin{cases}
                frac{t}{i} & t le frac{1}{2} \
                frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
                end{cases}
                $$
                The inverse homeomorphisms are given by
                $$h_i^{-1} =
                begin{cases}
                ti & t le frac{1}{2i} \
                frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
                end{cases}
                $$
                Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
                $$h : I to I, h(t) =
                begin{cases}
                0 & t le frac{1}{2} \
                2(t-frac{1}{2}) & t ge frac{1}{2}
                end{cases}
                $$
                The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
                $$h^- : I to I, h^-(t) =
                begin{cases}
                0 & t = 0 \
                frac{1}{2}+frac{1}{2}t & t > 0
                end{cases}
                $$
                This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.



                To get examples on $mathbb{R}^n$, define
                $$f_i(x) =
                begin{cases}
                x & lVert x rVert ge 1 \
                h_i(lVert x rVert)x & lVert x rVert le 1
                end{cases}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 16:29









                Paul FrostPaul Frost

                9,8453732




                9,8453732






























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