Left multiplication is homeomorphism of topological groups












2












$begingroup$


This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.



To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.



Here I stuck. Can I conclude it is open? If yes, why?



Thank you in advance for your help.






general-topology continuity topological-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Just notice left multiplication is a restiction of the multiplication function to the subspace ${a}times G$
    $endgroup$
    – YuiTo Cheng
    2 days ago








  • 1




    $begingroup$
    In the end, it sufficies to notice that the restriction of a continuous function is continuous.
    $endgroup$
    – LBJFS
    2 days ago






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    2 days ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    2 days ago
















2












$begingroup$


This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.



To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.



Here I stuck. Can I conclude it is open? If yes, why?



Thank you in advance for your help.






general-topology continuity topological-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Just notice left multiplication is a restiction of the multiplication function to the subspace ${a}times G$
    $endgroup$
    – YuiTo Cheng
    2 days ago








  • 1




    $begingroup$
    In the end, it sufficies to notice that the restriction of a continuous function is continuous.
    $endgroup$
    – LBJFS
    2 days ago






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    2 days ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    2 days ago














2












2








2





$begingroup$


This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.



To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.



Here I stuck. Can I conclude it is open? If yes, why?



Thank you in advance for your help.






general-topology continuity topological-groups







share|cite|improve this question











$endgroup$




This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.



To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.



Here I stuck. Can I conclude it is open? If yes, why?



Thank you in advance for your help.






general-topology continuity topological-groups




general-topology continuity topological-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







LBJFS

















asked 2 days ago









LBJFSLBJFS

28910




28910








  • 1




    $begingroup$
    Just notice left multiplication is a restiction of the multiplication function to the subspace ${a}times G$
    $endgroup$
    – YuiTo Cheng
    2 days ago








  • 1




    $begingroup$
    In the end, it sufficies to notice that the restriction of a continuous function is continuous.
    $endgroup$
    – LBJFS
    2 days ago






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    2 days ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    2 days ago














  • 1




    $begingroup$
    Just notice left multiplication is a restiction of the multiplication function to the subspace ${a}times G$
    $endgroup$
    – YuiTo Cheng
    2 days ago








  • 1




    $begingroup$
    In the end, it sufficies to notice that the restriction of a continuous function is continuous.
    $endgroup$
    – LBJFS
    2 days ago






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    2 days ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    2 days ago








1




1




$begingroup$
Just notice left multiplication is a restiction of the multiplication function to the subspace ${a}times G$
$endgroup$
– YuiTo Cheng
2 days ago






$begingroup$
Just notice left multiplication is a restiction of the multiplication function to the subspace ${a}times G$
$endgroup$
– YuiTo Cheng
2 days ago






1




1




$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
2 days ago




$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
2 days ago




1




1




$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
2 days ago




$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
2 days ago












$begingroup$
Thank you very much!
$endgroup$
– LBJFS
2 days ago




$begingroup$
Thank you very much!
$endgroup$
– LBJFS
2 days ago










4 Answers
4






active

oldest

votes


















3












$begingroup$

Hint:



For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to ${a}×G $. So what can you conclude?






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.



    And the inverse of $f_a$ is $f_{a^{-1}}$, which is continuous by the same reason.



    Therefore, $f_a$ is a homeomorphism.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      This is essentially rephrasing the other answer, but in a Category Theoretic perspective:



      Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^{-1}:Grightarrow G$ expressing multiplication, identity, and inversion.



      Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$



      Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.



      Together, we have the desired result.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I don't know category theory, but I love different perspectives! Thank you very much.
        $endgroup$
        – LBJFS
        2 days ago










      • $begingroup$
        I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
        $endgroup$
        – tomasz
        2 days ago










      • $begingroup$
        @tomasz I elaborated a bit. Hopefully this makes more sense.
        $endgroup$
        – user458276
        2 days ago










      • $begingroup$
        I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
        $endgroup$
        – tomasz
        yesterday





















      1












      $begingroup$

      Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).



      Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Hint:



        For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to ${a}×G $. So what can you conclude?






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Hint:



          For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to ${a}×G $. So what can you conclude?






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Hint:



            For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to ${a}×G $. So what can you conclude?






            share|cite|improve this answer









            $endgroup$



            Hint:



            For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to ${a}×G $. So what can you conclude?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Thomas ShelbyThomas Shelby

            4,4592726




            4,4592726























                2












                $begingroup$

                The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.



                And the inverse of $f_a$ is $f_{a^{-1}}$, which is continuous by the same reason.



                Therefore, $f_a$ is a homeomorphism.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.



                  And the inverse of $f_a$ is $f_{a^{-1}}$, which is continuous by the same reason.



                  Therefore, $f_a$ is a homeomorphism.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.



                    And the inverse of $f_a$ is $f_{a^{-1}}$, which is continuous by the same reason.



                    Therefore, $f_a$ is a homeomorphism.






                    share|cite|improve this answer









                    $endgroup$



                    The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.



                    And the inverse of $f_a$ is $f_{a^{-1}}$, which is continuous by the same reason.



                    Therefore, $f_a$ is a homeomorphism.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    José Carlos SantosJosé Carlos Santos

                    170k23132238




                    170k23132238























                        2












                        $begingroup$

                        This is essentially rephrasing the other answer, but in a Category Theoretic perspective:



                        Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^{-1}:Grightarrow G$ expressing multiplication, identity, and inversion.



                        Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$



                        Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.



                        Together, we have the desired result.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          I don't know category theory, but I love different perspectives! Thank you very much.
                          $endgroup$
                          – LBJFS
                          2 days ago










                        • $begingroup$
                          I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
                          $endgroup$
                          – tomasz
                          2 days ago










                        • $begingroup$
                          @tomasz I elaborated a bit. Hopefully this makes more sense.
                          $endgroup$
                          – user458276
                          2 days ago










                        • $begingroup$
                          I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
                          $endgroup$
                          – tomasz
                          yesterday


















                        2












                        $begingroup$

                        This is essentially rephrasing the other answer, but in a Category Theoretic perspective:



                        Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^{-1}:Grightarrow G$ expressing multiplication, identity, and inversion.



                        Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$



                        Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.



                        Together, we have the desired result.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          I don't know category theory, but I love different perspectives! Thank you very much.
                          $endgroup$
                          – LBJFS
                          2 days ago










                        • $begingroup$
                          I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
                          $endgroup$
                          – tomasz
                          2 days ago










                        • $begingroup$
                          @tomasz I elaborated a bit. Hopefully this makes more sense.
                          $endgroup$
                          – user458276
                          2 days ago










                        • $begingroup$
                          I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
                          $endgroup$
                          – tomasz
                          yesterday
















                        2












                        2








                        2





                        $begingroup$

                        This is essentially rephrasing the other answer, but in a Category Theoretic perspective:



                        Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^{-1}:Grightarrow G$ expressing multiplication, identity, and inversion.



                        Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$



                        Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.



                        Together, we have the desired result.






                        share|cite|improve this answer











                        $endgroup$



                        This is essentially rephrasing the other answer, but in a Category Theoretic perspective:



                        Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^{-1}:Grightarrow G$ expressing multiplication, identity, and inversion.



                        Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$



                        Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.



                        Together, we have the desired result.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 2 days ago

























                        answered 2 days ago









                        user458276user458276

                        7801314




                        7801314












                        • $begingroup$
                          I don't know category theory, but I love different perspectives! Thank you very much.
                          $endgroup$
                          – LBJFS
                          2 days ago










                        • $begingroup$
                          I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
                          $endgroup$
                          – tomasz
                          2 days ago










                        • $begingroup$
                          @tomasz I elaborated a bit. Hopefully this makes more sense.
                          $endgroup$
                          – user458276
                          2 days ago










                        • $begingroup$
                          I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
                          $endgroup$
                          – tomasz
                          yesterday




















                        • $begingroup$
                          I don't know category theory, but I love different perspectives! Thank you very much.
                          $endgroup$
                          – LBJFS
                          2 days ago










                        • $begingroup$
                          I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
                          $endgroup$
                          – tomasz
                          2 days ago










                        • $begingroup$
                          @tomasz I elaborated a bit. Hopefully this makes more sense.
                          $endgroup$
                          – user458276
                          2 days ago










                        • $begingroup$
                          I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
                          $endgroup$
                          – tomasz
                          yesterday


















                        $begingroup$
                        I don't know category theory, but I love different perspectives! Thank you very much.
                        $endgroup$
                        – LBJFS
                        2 days ago




                        $begingroup$
                        I don't know category theory, but I love different perspectives! Thank you very much.
                        $endgroup$
                        – LBJFS
                        2 days ago












                        $begingroup$
                        I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
                        $endgroup$
                        – tomasz
                        2 days ago




                        $begingroup$
                        I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
                        $endgroup$
                        – tomasz
                        2 days ago












                        $begingroup$
                        @tomasz I elaborated a bit. Hopefully this makes more sense.
                        $endgroup$
                        – user458276
                        2 days ago




                        $begingroup$
                        @tomasz I elaborated a bit. Hopefully this makes more sense.
                        $endgroup$
                        – user458276
                        2 days ago












                        $begingroup$
                        I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
                        $endgroup$
                        – tomasz
                        yesterday






                        $begingroup$
                        I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
                        $endgroup$
                        – tomasz
                        yesterday













                        1












                        $begingroup$

                        Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).



                        Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).



                          Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).



                            Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).






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                            $endgroup$



                            Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).



                            Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            tomasztomasz

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