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
    10 hours ago








  • 1




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






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    10 hours ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    10 hours 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
    10 hours ago








  • 1




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






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    10 hours ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    10 hours 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 10 hours ago







LBJFS

















asked 10 hours ago









LBJFSLBJFS

28310




28310








  • 1




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








  • 1




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






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    10 hours ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    10 hours ago














  • 1




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








  • 1




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






  • 1




    $begingroup$
    And this is straightforward
    $endgroup$
    – YuiTo Cheng
    10 hours ago










  • $begingroup$
    Thank you very much!
    $endgroup$
    – LBJFS
    10 hours ago








1




1




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






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






1




1




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




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




1




1




$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
10 hours ago




$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
10 hours ago












$begingroup$
Thank you very much!
$endgroup$
– LBJFS
10 hours ago




$begingroup$
Thank you very much!
$endgroup$
– LBJFS
10 hours 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
        9 hours 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
        9 hours ago










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



















      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 10 hours ago









            Thomas ShelbyThomas Shelby

            4,4092726




            4,4092726























                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 10 hours 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
                          9 hours 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
                          9 hours ago










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
















                        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
                          9 hours 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
                          9 hours ago










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














                        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 hours ago

























                        answered 10 hours ago









                        user458276user458276

                        7801314




                        7801314












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










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


















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










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
















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




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












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




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











                        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).






                            share|cite|improve this answer









                            $endgroup$



                            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 9 hours ago









                            tomasztomasz

                            23.9k23482




                            23.9k23482






























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                                Is this a new Fibonacci Identity?

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                                6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w...
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                                19世紀

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                                この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
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