All quasi-isometry is coarsely surjective.












0












$begingroup$


Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
$f:Xto Y$ is quasi-isometry if



$d_Y(f(x),f(x'))leq Ld_X(x,x')+C$



exists coarse inverse, i.e. $overline{f}:Yto X$ such that



$d_X(overline{f}f(x),x)leq C$



$d_Y(foverline{f}(y),y)leq C$



and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$



for all $x,x'in X$, for all $y,y'in Y$



How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$






metric-spaces lipschitz-functions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
    $f:Xto Y$ is quasi-isometry if



    $d_Y(f(x),f(x'))leq Ld_X(x,x')+C$



    exists coarse inverse, i.e. $overline{f}:Yto X$ such that



    $d_X(overline{f}f(x),x)leq C$



    $d_Y(foverline{f}(y),y)leq C$



    and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$



    for all $x,x'in X$, for all $y,y'in Y$



    How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$






    metric-spaces lipschitz-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
      $f:Xto Y$ is quasi-isometry if



      $d_Y(f(x),f(x'))leq Ld_X(x,x')+C$



      exists coarse inverse, i.e. $overline{f}:Yto X$ such that



      $d_X(overline{f}f(x),x)leq C$



      $d_Y(foverline{f}(y),y)leq C$



      and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$



      for all $x,x'in X$, for all $y,y'in Y$



      How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$






      metric-spaces lipschitz-functions







      share|cite|improve this question











      $endgroup$




      Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
      $f:Xto Y$ is quasi-isometry if



      $d_Y(f(x),f(x'))leq Ld_X(x,x')+C$



      exists coarse inverse, i.e. $overline{f}:Yto X$ such that



      $d_X(overline{f}f(x),x)leq C$



      $d_Y(foverline{f}(y),y)leq C$



      and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$



      for all $x,x'in X$, for all $y,y'in Y$



      How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$






      metric-spaces lipschitz-functions




      metric-spaces lipschitz-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 7 '18 at 15:58







      eraldcoil

















      asked Nov 7 '18 at 15:53









      eraldcoileraldcoil

      388211




      388211






















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          $begingroup$

          Put $x=bar f(y)$.



          ..........



          Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.






          share|cite|improve this answer











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            active

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            active

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            1












            $begingroup$

            Put $x=bar f(y)$.



            ..........



            Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Put $x=bar f(y)$.



              ..........



              Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Put $x=bar f(y)$.



                ..........



                Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.






                share|cite|improve this answer











                $endgroup$



                Put $x=bar f(y)$.



                ..........



                Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 3 at 17:49

























                answered Dec 5 '18 at 8:11









                Alex RavskyAlex Ravsky

                39.6k32181




                39.6k32181






























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