If $G$ is profinite, and $A$ is discrete, $f: G to A$ is continuous $implies$ $f$ factors through a normal...












1














Let $G$ be a profinite group; that is compact, and totally disconnected.



Take $A$ a discrete space, and a continuous map $f: G to A$.




$exists N$ open and normal in $G$ and a continuous map $g: G/N to A$
with $g circ pi = f$ where $pi$ is the projection




What I understand from the proof in my book so far:



Since $f$ is continuous, the image of $f$ is compact, and hence finite, since $A$ is discrete.



We can write Im$f = {x_1,..x_n}$, and set $X_i = f^{-1}{x_i}$.



Then each $X_i$ is clopen, and so it is a finite union of cosets of the form $Vg$.



The book then says to find a normal open subgroup $N$ that is contained in each such $V$, and so each $X_i$ is a union of cosets of $N$.



Can you explain this last part?






topological-groups profinite-groups







share|cite|improve this question






















  • What is $V{{}}$?
    – Lord Shark the Unknown
    Dec 5 '18 at 16:14










  • @LordSharktheUnknown I think it is supposed to be an open normal subgroup in $G$
    – Mariah
    Dec 5 '18 at 16:31










  • I assume the $V$'s denote open neighborhood of the identity. There are finitely many $V$'s involved, so their intersection is still an open neighborhood, but can we find a normal subgroup in it?
    – Berci
    Dec 5 '18 at 16:36
















1














Let $G$ be a profinite group; that is compact, and totally disconnected.



Take $A$ a discrete space, and a continuous map $f: G to A$.




$exists N$ open and normal in $G$ and a continuous map $g: G/N to A$
with $g circ pi = f$ where $pi$ is the projection




What I understand from the proof in my book so far:



Since $f$ is continuous, the image of $f$ is compact, and hence finite, since $A$ is discrete.



We can write Im$f = {x_1,..x_n}$, and set $X_i = f^{-1}{x_i}$.



Then each $X_i$ is clopen, and so it is a finite union of cosets of the form $Vg$.



The book then says to find a normal open subgroup $N$ that is contained in each such $V$, and so each $X_i$ is a union of cosets of $N$.



Can you explain this last part?






topological-groups profinite-groups







share|cite|improve this question






















  • What is $V{{}}$?
    – Lord Shark the Unknown
    Dec 5 '18 at 16:14










  • @LordSharktheUnknown I think it is supposed to be an open normal subgroup in $G$
    – Mariah
    Dec 5 '18 at 16:31










  • I assume the $V$'s denote open neighborhood of the identity. There are finitely many $V$'s involved, so their intersection is still an open neighborhood, but can we find a normal subgroup in it?
    – Berci
    Dec 5 '18 at 16:36














1












1








1







Let $G$ be a profinite group; that is compact, and totally disconnected.



Take $A$ a discrete space, and a continuous map $f: G to A$.




$exists N$ open and normal in $G$ and a continuous map $g: G/N to A$
with $g circ pi = f$ where $pi$ is the projection




What I understand from the proof in my book so far:



Since $f$ is continuous, the image of $f$ is compact, and hence finite, since $A$ is discrete.



We can write Im$f = {x_1,..x_n}$, and set $X_i = f^{-1}{x_i}$.



Then each $X_i$ is clopen, and so it is a finite union of cosets of the form $Vg$.



The book then says to find a normal open subgroup $N$ that is contained in each such $V$, and so each $X_i$ is a union of cosets of $N$.



Can you explain this last part?






topological-groups profinite-groups







share|cite|improve this question













Let $G$ be a profinite group; that is compact, and totally disconnected.



Take $A$ a discrete space, and a continuous map $f: G to A$.




$exists N$ open and normal in $G$ and a continuous map $g: G/N to A$
with $g circ pi = f$ where $pi$ is the projection




What I understand from the proof in my book so far:



Since $f$ is continuous, the image of $f$ is compact, and hence finite, since $A$ is discrete.



We can write Im$f = {x_1,..x_n}$, and set $X_i = f^{-1}{x_i}$.



Then each $X_i$ is clopen, and so it is a finite union of cosets of the form $Vg$.



The book then says to find a normal open subgroup $N$ that is contained in each such $V$, and so each $X_i$ is a union of cosets of $N$.



Can you explain this last part?






topological-groups profinite-groups




topological-groups profinite-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 16:13









MariahMariah

1,356518




1,356518












  • What is $V{{}}$?
    – Lord Shark the Unknown
    Dec 5 '18 at 16:14










  • @LordSharktheUnknown I think it is supposed to be an open normal subgroup in $G$
    – Mariah
    Dec 5 '18 at 16:31










  • I assume the $V$'s denote open neighborhood of the identity. There are finitely many $V$'s involved, so their intersection is still an open neighborhood, but can we find a normal subgroup in it?
    – Berci
    Dec 5 '18 at 16:36


















  • What is $V{{}}$?
    – Lord Shark the Unknown
    Dec 5 '18 at 16:14










  • @LordSharktheUnknown I think it is supposed to be an open normal subgroup in $G$
    – Mariah
    Dec 5 '18 at 16:31










  • I assume the $V$'s denote open neighborhood of the identity. There are finitely many $V$'s involved, so their intersection is still an open neighborhood, but can we find a normal subgroup in it?
    – Berci
    Dec 5 '18 at 16:36
















What is $V{{}}$?
– Lord Shark the Unknown
Dec 5 '18 at 16:14




What is $V{{}}$?
– Lord Shark the Unknown
Dec 5 '18 at 16:14












@LordSharktheUnknown I think it is supposed to be an open normal subgroup in $G$
– Mariah
Dec 5 '18 at 16:31




@LordSharktheUnknown I think it is supposed to be an open normal subgroup in $G$
– Mariah
Dec 5 '18 at 16:31












I assume the $V$'s denote open neighborhood of the identity. There are finitely many $V$'s involved, so their intersection is still an open neighborhood, but can we find a normal subgroup in it?
– Berci
Dec 5 '18 at 16:36




I assume the $V$'s denote open neighborhood of the identity. There are finitely many $V$'s involved, so their intersection is still an open neighborhood, but can we find a normal subgroup in it?
– Berci
Dec 5 '18 at 16:36










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Let $V$ be an open neighbourhood of the identity. Then it contains an open subgroup
$H$. As $G$ is compact, $H$ has finite index. Therefore it has finitely
many conjugates. The intersection $K$ of the conjugates of $H$ is normal and
open, so also has finite index. Therefore $Vsupseteq K$ where $K$
is an open normal subgroup.



If you have finitely many $V$s, then intersect their corresponding $K$s.






share|cite|improve this answer





















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    Let $V$ be an open neighbourhood of the identity. Then it contains an open subgroup
    $H$. As $G$ is compact, $H$ has finite index. Therefore it has finitely
    many conjugates. The intersection $K$ of the conjugates of $H$ is normal and
    open, so also has finite index. Therefore $Vsupseteq K$ where $K$
    is an open normal subgroup.



    If you have finitely many $V$s, then intersect their corresponding $K$s.






    share|cite|improve this answer


























      0














      Let $V$ be an open neighbourhood of the identity. Then it contains an open subgroup
      $H$. As $G$ is compact, $H$ has finite index. Therefore it has finitely
      many conjugates. The intersection $K$ of the conjugates of $H$ is normal and
      open, so also has finite index. Therefore $Vsupseteq K$ where $K$
      is an open normal subgroup.



      If you have finitely many $V$s, then intersect their corresponding $K$s.






      share|cite|improve this answer
























        0












        0








        0






        Let $V$ be an open neighbourhood of the identity. Then it contains an open subgroup
        $H$. As $G$ is compact, $H$ has finite index. Therefore it has finitely
        many conjugates. The intersection $K$ of the conjugates of $H$ is normal and
        open, so also has finite index. Therefore $Vsupseteq K$ where $K$
        is an open normal subgroup.



        If you have finitely many $V$s, then intersect their corresponding $K$s.






        share|cite|improve this answer












        Let $V$ be an open neighbourhood of the identity. Then it contains an open subgroup
        $H$. As $G$ is compact, $H$ has finite index. Therefore it has finitely
        many conjugates. The intersection $K$ of the conjugates of $H$ is normal and
        open, so also has finite index. Therefore $Vsupseteq K$ where $K$
        is an open normal subgroup.



        If you have finitely many $V$s, then intersect their corresponding $K$s.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 16:45









        Lord Shark the UnknownLord Shark the Unknown

        102k959132




        102k959132






























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