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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161917%2fleft-multiplication-is-homeomorphism-of-topological-groups%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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).






                            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 2 days ago









                            tomasztomasz

                            24k23482




                            24k23482






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161917%2fleft-multiplication-is-homeomorphism-of-topological-groups%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                How did Captain America manage to do this?

                                Image
                                2 How does Captain America lift Thor's hammer? In Avengers: Age of Ultron , he is not worthy but how he is worthy in this film. plot-explanation marvel-cinematic-universe avengers-age-of-ultron avengers-endgame share | improve this question edited 19 mins ago A J ♦ 42.7k 16 229 246 asked 39 mins ago Andi AR Andi AR 325 1 4 14
                                Read more

                                迪纳利

                                Image
                                迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
                                Read more

                                南乌拉尔铁路局

                                Image
                                南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
                                Read more