Unramified map of Riemann surfaces

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Let $f:S to T$ be a surjective, unramified, holomorphic map between connected Riemann surfaces. If $S$ is not compact is it always true that $f$ is a covering?



This is of course true if $S$ is compact or, more generally, if $f$ is proper. However, I can not see why this should be true in general.




gn.general-topology cv.complex-variables riemann-surfaces




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    up vote
    4
    down vote

    favorite












    Let $f:S to T$ be a surjective, unramified, holomorphic map between connected Riemann surfaces. If $S$ is not compact is it always true that $f$ is a covering?



    This is of course true if $S$ is compact or, more generally, if $f$ is proper. However, I can not see why this should be true in general.




    gn.general-topology cv.complex-variables riemann-surfaces




    share|cite|improve this question






















      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      Let $f:S to T$ be a surjective, unramified, holomorphic map between connected Riemann surfaces. If $S$ is not compact is it always true that $f$ is a covering?



      This is of course true if $S$ is compact or, more generally, if $f$ is proper. However, I can not see why this should be true in general.




      gn.general-topology cv.complex-variables riemann-surfaces




      share|cite|improve this question












      Let $f:S to T$ be a surjective, unramified, holomorphic map between connected Riemann surfaces. If $S$ is not compact is it always true that $f$ is a covering?



      This is of course true if $S$ is compact or, more generally, if $f$ is proper. However, I can not see why this should be true in general.





      gn.general-topology cv.complex-variables riemann-surfaces





      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 2 at 8:48









      Chitrabhanu

      23216




      23216




















          2 Answers
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          The simplest "non-trivial" example is $$zmapsto int_0^ze^-zeta^2dzeta:quad Cto C.$$
          It is surjective, and not ramified. But it is certainly not a covering because every covering over a simply connected surface is a homeomorphism.



          You can make the target surface compact if you wish. Consider the map from $C$ to
          $S$, where $S$ is the Riemann sphere, $f(z)=y_1/y_0$, where
          $y_j$ are two linearly independent solutions of the Airy equation
          $$y''=zy.$$ It is also surjective and unramified: $f'=-W(y_1,y_0)/y_0^2$,
          where $W$ is the Wronskian determinant which is constant.






          share|cite|improve this answer



























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            Let $C to D$ be a nontrivial finite étale cover and $p in C$ a point. Then $C setminus p to D$ is still surjective and unramified, but it's not a covering.






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            • Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan.
              – Chitrabhanu
              Sep 2 at 10:35










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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            The simplest "non-trivial" example is $$zmapsto int_0^ze^-zeta^2dzeta:quad Cto C.$$
            It is surjective, and not ramified. But it is certainly not a covering because every covering over a simply connected surface is a homeomorphism.



            You can make the target surface compact if you wish. Consider the map from $C$ to
            $S$, where $S$ is the Riemann sphere, $f(z)=y_1/y_0$, where
            $y_j$ are two linearly independent solutions of the Airy equation
            $$y''=zy.$$ It is also surjective and unramified: $f'=-W(y_1,y_0)/y_0^2$,
            where $W$ is the Wronskian determinant which is constant.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              The simplest "non-trivial" example is $$zmapsto int_0^ze^-zeta^2dzeta:quad Cto C.$$
              It is surjective, and not ramified. But it is certainly not a covering because every covering over a simply connected surface is a homeomorphism.



              You can make the target surface compact if you wish. Consider the map from $C$ to
              $S$, where $S$ is the Riemann sphere, $f(z)=y_1/y_0$, where
              $y_j$ are two linearly independent solutions of the Airy equation
              $$y''=zy.$$ It is also surjective and unramified: $f'=-W(y_1,y_0)/y_0^2$,
              where $W$ is the Wronskian determinant which is constant.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                The simplest "non-trivial" example is $$zmapsto int_0^ze^-zeta^2dzeta:quad Cto C.$$
                It is surjective, and not ramified. But it is certainly not a covering because every covering over a simply connected surface is a homeomorphism.



                You can make the target surface compact if you wish. Consider the map from $C$ to
                $S$, where $S$ is the Riemann sphere, $f(z)=y_1/y_0$, where
                $y_j$ are two linearly independent solutions of the Airy equation
                $$y''=zy.$$ It is also surjective and unramified: $f'=-W(y_1,y_0)/y_0^2$,
                where $W$ is the Wronskian determinant which is constant.






                share|cite|improve this answer












                The simplest "non-trivial" example is $$zmapsto int_0^ze^-zeta^2dzeta:quad Cto C.$$
                It is surjective, and not ramified. But it is certainly not a covering because every covering over a simply connected surface is a homeomorphism.



                You can make the target surface compact if you wish. Consider the map from $C$ to
                $S$, where $S$ is the Riemann sphere, $f(z)=y_1/y_0$, where
                $y_j$ are two linearly independent solutions of the Airy equation
                $$y''=zy.$$ It is also surjective and unramified: $f'=-W(y_1,y_0)/y_0^2$,
                where $W$ is the Wronskian determinant which is constant.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 2 at 13:31









                Alexandre Eremenko

                47k6129244




                47k6129244




















                    up vote
                    8
                    down vote













                    Let $C to D$ be a nontrivial finite étale cover and $p in C$ a point. Then $C setminus p to D$ is still surjective and unramified, but it's not a covering.






                    share|cite|improve this answer




















                    • Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan.
                      – Chitrabhanu
                      Sep 2 at 10:35














                    up vote
                    8
                    down vote













                    Let $C to D$ be a nontrivial finite étale cover and $p in C$ a point. Then $C setminus p to D$ is still surjective and unramified, but it's not a covering.






                    share|cite|improve this answer




















                    • Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan.
                      – Chitrabhanu
                      Sep 2 at 10:35












                    up vote
                    8
                    down vote










                    up vote
                    8
                    down vote









                    Let $C to D$ be a nontrivial finite étale cover and $p in C$ a point. Then $C setminus p to D$ is still surjective and unramified, but it's not a covering.






                    share|cite|improve this answer












                    Let $C to D$ be a nontrivial finite étale cover and $p in C$ a point. Then $C setminus p to D$ is still surjective and unramified, but it's not a covering.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 2 at 8:58









                    Dan Petersen

                    24k265130




                    24k265130











                    • Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan.
                      – Chitrabhanu
                      Sep 2 at 10:35
















                    • Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan.
                      – Chitrabhanu
                      Sep 2 at 10:35















                    Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan.
                    – Chitrabhanu
                    Sep 2 at 10:35




                    Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan.
                    – Chitrabhanu
                    Sep 2 at 10:35

















                     

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