Projective space, showing open map to one point compactification of $Bbb C$

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Let $Bbb P^1$ be complex projective space with the quotient topology from $Bbb C^2-0$, where we identify $(a,b)sim(c,d)$ if there exists $lambdainBbb C^times$ such that $lambda(a,b)=(c,d)$. Denote the equivalence class of $(a,b)$ by $[a:b]$.



Let $BbbCcupinfty$ be the one point compactification of $Bbb C$. I want to show that the map
$$gamma:BbbP^1toBbbCcupinfty$$
given by
$$gamma([1:b/a])=b/a$$
$$gamma([0:1])=infty,$$
is open.



I have shown that whenever $Usubset U_0subsetBbbP^1$ doesn't contain $[0:1]$ that it is open. So I am only interested in $U$ open in $BbbP^1$ containing $[0:1]$.



I denote by $U_0=[a:b]mid ane 0$.



So let $U$ be open in $BbbP^1$ such that $[0:1]in U$. Then $U=(Ucap U_0)cup [0:1]$, and thus $gamma(U)=gamma(Ucap U_0)cup gamma([0:1])=gamma(Ucap U_0)cupinfty$. Where previous calculations show $gamma(Ucap U_0)$ is open in $Bbb C$.




This means that (from the definition of the topology of the one point compactification) all that is left is to show that $gamma(Ucap U_0)^c$ is quasi-compact, but I can't seem to show this. Any help is greatly appreciated!











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    Let $Bbb P^1$ be complex projective space with the quotient topology from $Bbb C^2-0$, where we identify $(a,b)sim(c,d)$ if there exists $lambdainBbb C^times$ such that $lambda(a,b)=(c,d)$. Denote the equivalence class of $(a,b)$ by $[a:b]$.



    Let $BbbCcupinfty$ be the one point compactification of $Bbb C$. I want to show that the map
    $$gamma:BbbP^1toBbbCcupinfty$$
    given by
    $$gamma([1:b/a])=b/a$$
    $$gamma([0:1])=infty,$$
    is open.



    I have shown that whenever $Usubset U_0subsetBbbP^1$ doesn't contain $[0:1]$ that it is open. So I am only interested in $U$ open in $BbbP^1$ containing $[0:1]$.



    I denote by $U_0=[a:b]mid ane 0$.



    So let $U$ be open in $BbbP^1$ such that $[0:1]in U$. Then $U=(Ucap U_0)cup [0:1]$, and thus $gamma(U)=gamma(Ucap U_0)cup gamma([0:1])=gamma(Ucap U_0)cupinfty$. Where previous calculations show $gamma(Ucap U_0)$ is open in $Bbb C$.




    This means that (from the definition of the topology of the one point compactification) all that is left is to show that $gamma(Ucap U_0)^c$ is quasi-compact, but I can't seem to show this. Any help is greatly appreciated!











    share|cite|improve this question

























      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $Bbb P^1$ be complex projective space with the quotient topology from $Bbb C^2-0$, where we identify $(a,b)sim(c,d)$ if there exists $lambdainBbb C^times$ such that $lambda(a,b)=(c,d)$. Denote the equivalence class of $(a,b)$ by $[a:b]$.



      Let $BbbCcupinfty$ be the one point compactification of $Bbb C$. I want to show that the map
      $$gamma:BbbP^1toBbbCcupinfty$$
      given by
      $$gamma([1:b/a])=b/a$$
      $$gamma([0:1])=infty,$$
      is open.



      I have shown that whenever $Usubset U_0subsetBbbP^1$ doesn't contain $[0:1]$ that it is open. So I am only interested in $U$ open in $BbbP^1$ containing $[0:1]$.



      I denote by $U_0=[a:b]mid ane 0$.



      So let $U$ be open in $BbbP^1$ such that $[0:1]in U$. Then $U=(Ucap U_0)cup [0:1]$, and thus $gamma(U)=gamma(Ucap U_0)cup gamma([0:1])=gamma(Ucap U_0)cupinfty$. Where previous calculations show $gamma(Ucap U_0)$ is open in $Bbb C$.




      This means that (from the definition of the topology of the one point compactification) all that is left is to show that $gamma(Ucap U_0)^c$ is quasi-compact, but I can't seem to show this. Any help is greatly appreciated!











      share|cite|improve this question















      Let $Bbb P^1$ be complex projective space with the quotient topology from $Bbb C^2-0$, where we identify $(a,b)sim(c,d)$ if there exists $lambdainBbb C^times$ such that $lambda(a,b)=(c,d)$. Denote the equivalence class of $(a,b)$ by $[a:b]$.



      Let $BbbCcupinfty$ be the one point compactification of $Bbb C$. I want to show that the map
      $$gamma:BbbP^1toBbbCcupinfty$$
      given by
      $$gamma([1:b/a])=b/a$$
      $$gamma([0:1])=infty,$$
      is open.



      I have shown that whenever $Usubset U_0subsetBbbP^1$ doesn't contain $[0:1]$ that it is open. So I am only interested in $U$ open in $BbbP^1$ containing $[0:1]$.



      I denote by $U_0=[a:b]mid ane 0$.



      So let $U$ be open in $BbbP^1$ such that $[0:1]in U$. Then $U=(Ucap U_0)cup [0:1]$, and thus $gamma(U)=gamma(Ucap U_0)cup gamma([0:1])=gamma(Ucap U_0)cupinfty$. Where previous calculations show $gamma(Ucap U_0)$ is open in $Bbb C$.




      This means that (from the definition of the topology of the one point compactification) all that is left is to show that $gamma(Ucap U_0)^c$ is quasi-compact, but I can't seem to show this. Any help is greatly appreciated!








      algebraic-geometry compactness projective-space






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      edited Sep 9 at 13:25

























      asked Sep 9 at 13:07









      user591482

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          You need to show that $gamma(U)$ contains a neighborhood of $infty$. Such a neighborhood can be of the form $inftycup mathbbC-bar B(0,r)$, where $bar B(0,r)$ is the closed ball of radius $r$.



          Since the topology of $mathbbP^1$ is the quotient topology, $U$ contains a ball $p(B((0,1),s)$ where $p:mathbbC^2-0rightarrow mathbbP^1$ is the quotient map. Let $zin mathbbC-bar B(0,s)$, $1over<s$ implies that $(1over z,1)in B((0,1);s)$, this implies that $mathbbC-bar B(0,s)subsetgamma(p(B(0,s))subsetgamma(U)$.






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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            You need to show that $gamma(U)$ contains a neighborhood of $infty$. Such a neighborhood can be of the form $inftycup mathbbC-bar B(0,r)$, where $bar B(0,r)$ is the closed ball of radius $r$.



            Since the topology of $mathbbP^1$ is the quotient topology, $U$ contains a ball $p(B((0,1),s)$ where $p:mathbbC^2-0rightarrow mathbbP^1$ is the quotient map. Let $zin mathbbC-bar B(0,s)$, $1over<s$ implies that $(1over z,1)in B((0,1);s)$, this implies that $mathbbC-bar B(0,s)subsetgamma(p(B(0,s))subsetgamma(U)$.






            share|cite|improve this answer


























              up vote
              2
              down vote



              accepted










              You need to show that $gamma(U)$ contains a neighborhood of $infty$. Such a neighborhood can be of the form $inftycup mathbbC-bar B(0,r)$, where $bar B(0,r)$ is the closed ball of radius $r$.



              Since the topology of $mathbbP^1$ is the quotient topology, $U$ contains a ball $p(B((0,1),s)$ where $p:mathbbC^2-0rightarrow mathbbP^1$ is the quotient map. Let $zin mathbbC-bar B(0,s)$, $1over<s$ implies that $(1over z,1)in B((0,1);s)$, this implies that $mathbbC-bar B(0,s)subsetgamma(p(B(0,s))subsetgamma(U)$.






              share|cite|improve this answer
























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                You need to show that $gamma(U)$ contains a neighborhood of $infty$. Such a neighborhood can be of the form $inftycup mathbbC-bar B(0,r)$, where $bar B(0,r)$ is the closed ball of radius $r$.



                Since the topology of $mathbbP^1$ is the quotient topology, $U$ contains a ball $p(B((0,1),s)$ where $p:mathbbC^2-0rightarrow mathbbP^1$ is the quotient map. Let $zin mathbbC-bar B(0,s)$, $1over<s$ implies that $(1over z,1)in B((0,1);s)$, this implies that $mathbbC-bar B(0,s)subsetgamma(p(B(0,s))subsetgamma(U)$.






                share|cite|improve this answer














                You need to show that $gamma(U)$ contains a neighborhood of $infty$. Such a neighborhood can be of the form $inftycup mathbbC-bar B(0,r)$, where $bar B(0,r)$ is the closed ball of radius $r$.



                Since the topology of $mathbbP^1$ is the quotient topology, $U$ contains a ball $p(B((0,1),s)$ where $p:mathbbC^2-0rightarrow mathbbP^1$ is the quotient map. Let $zin mathbbC-bar B(0,s)$, $1over<s$ implies that $(1over z,1)in B((0,1);s)$, this implies that $mathbbC-bar B(0,s)subsetgamma(p(B(0,s))subsetgamma(U)$.







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Sep 9 at 14:15

























                answered Sep 9 at 14:10









                Tsemo Aristide

                52.5k11244




                52.5k11244



























                     

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