Products of quotient topology same as quotient of product topology












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Let $X$ be a topological space, $p:Xto Y$ be a quotient map, and $q:Xtimes Xto Ytimes Y$ be the quotient map defined by $q(x,y)=(p(x),p(y))$. Prove that the topologies on $Y$ is the same as the topology on $Ytimes Y$ as a quotient of the product topology on $Xtimes X$.






general-topology







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  • 1




    $begingroup$
    Welcome to MSE Heidi. Please read the FAQ. You need to say what you've done towards a problem in order to get help, not just state it.
    $endgroup$
    – Alexander Gruber♦
    Nov 4 '12 at 6:24










  • $begingroup$
    Are you sure that the last sentence is correct? I suspect that you want to prove that the topology on $Ytimes Y$ induced by $q$ is the same as the product topology on $Ytimes Y$, which isn’t what you actually wrote.
    $endgroup$
    – Brian M. Scott
    Nov 4 '12 at 6:27










  • $begingroup$
    I think you mean: Prove that the topology on $Y times Y$ is the same as .... This is answered below, and is also relevant to math.stackexcnge.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Nov 4 '12 at 17:41










  • $begingroup$
    oops! yes, i meant the topology on $Y times Y$ as a product of the quotient topologies on $Y$ is the same... I feel like it intuitively makes sense but I'm not sure how to start a formal proof of it
    $endgroup$
    – Heidi
    Nov 5 '12 at 9:32










  • $begingroup$
    My previous comment should have referenced math.stackexchange.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Dec 2 '18 at 15:37


















1












$begingroup$


Let $X$ be a topological space, $p:Xto Y$ be a quotient map, and $q:Xtimes Xto Ytimes Y$ be the quotient map defined by $q(x,y)=(p(x),p(y))$. Prove that the topologies on $Y$ is the same as the topology on $Ytimes Y$ as a quotient of the product topology on $Xtimes X$.






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to MSE Heidi. Please read the FAQ. You need to say what you've done towards a problem in order to get help, not just state it.
    $endgroup$
    – Alexander Gruber♦
    Nov 4 '12 at 6:24










  • $begingroup$
    Are you sure that the last sentence is correct? I suspect that you want to prove that the topology on $Ytimes Y$ induced by $q$ is the same as the product topology on $Ytimes Y$, which isn’t what you actually wrote.
    $endgroup$
    – Brian M. Scott
    Nov 4 '12 at 6:27










  • $begingroup$
    I think you mean: Prove that the topology on $Y times Y$ is the same as .... This is answered below, and is also relevant to math.stackexcnge.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Nov 4 '12 at 17:41










  • $begingroup$
    oops! yes, i meant the topology on $Y times Y$ as a product of the quotient topologies on $Y$ is the same... I feel like it intuitively makes sense but I'm not sure how to start a formal proof of it
    $endgroup$
    – Heidi
    Nov 5 '12 at 9:32










  • $begingroup$
    My previous comment should have referenced math.stackexchange.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Dec 2 '18 at 15:37
















1












1








1


3



$begingroup$


Let $X$ be a topological space, $p:Xto Y$ be a quotient map, and $q:Xtimes Xto Ytimes Y$ be the quotient map defined by $q(x,y)=(p(x),p(y))$. Prove that the topologies on $Y$ is the same as the topology on $Ytimes Y$ as a quotient of the product topology on $Xtimes X$.






general-topology







share|cite|improve this question











$endgroup$




Let $X$ be a topological space, $p:Xto Y$ be a quotient map, and $q:Xtimes Xto Ytimes Y$ be the quotient map defined by $q(x,y)=(p(x),p(y))$. Prove that the topologies on $Y$ is the same as the topology on $Ytimes Y$ as a quotient of the product topology on $Xtimes X$.






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 4 '12 at 5:31









Brian M. Scott

458k38511913




458k38511913










asked Nov 4 '12 at 5:29









HeidiHeidi

61




61








  • 1




    $begingroup$
    Welcome to MSE Heidi. Please read the FAQ. You need to say what you've done towards a problem in order to get help, not just state it.
    $endgroup$
    – Alexander Gruber♦
    Nov 4 '12 at 6:24










  • $begingroup$
    Are you sure that the last sentence is correct? I suspect that you want to prove that the topology on $Ytimes Y$ induced by $q$ is the same as the product topology on $Ytimes Y$, which isn’t what you actually wrote.
    $endgroup$
    – Brian M. Scott
    Nov 4 '12 at 6:27










  • $begingroup$
    I think you mean: Prove that the topology on $Y times Y$ is the same as .... This is answered below, and is also relevant to math.stackexcnge.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Nov 4 '12 at 17:41










  • $begingroup$
    oops! yes, i meant the topology on $Y times Y$ as a product of the quotient topologies on $Y$ is the same... I feel like it intuitively makes sense but I'm not sure how to start a formal proof of it
    $endgroup$
    – Heidi
    Nov 5 '12 at 9:32










  • $begingroup$
    My previous comment should have referenced math.stackexchange.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Dec 2 '18 at 15:37
















  • 1




    $begingroup$
    Welcome to MSE Heidi. Please read the FAQ. You need to say what you've done towards a problem in order to get help, not just state it.
    $endgroup$
    – Alexander Gruber♦
    Nov 4 '12 at 6:24










  • $begingroup$
    Are you sure that the last sentence is correct? I suspect that you want to prove that the topology on $Ytimes Y$ induced by $q$ is the same as the product topology on $Ytimes Y$, which isn’t what you actually wrote.
    $endgroup$
    – Brian M. Scott
    Nov 4 '12 at 6:27










  • $begingroup$
    I think you mean: Prove that the topology on $Y times Y$ is the same as .... This is answered below, and is also relevant to math.stackexcnge.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Nov 4 '12 at 17:41










  • $begingroup$
    oops! yes, i meant the topology on $Y times Y$ as a product of the quotient topologies on $Y$ is the same... I feel like it intuitively makes sense but I'm not sure how to start a formal proof of it
    $endgroup$
    – Heidi
    Nov 5 '12 at 9:32










  • $begingroup$
    My previous comment should have referenced math.stackexchange.com/questions/31697
    $endgroup$
    – Ronnie Brown
    Dec 2 '18 at 15:37










1




1




$begingroup$
Welcome to MSE Heidi. Please read the FAQ. You need to say what you've done towards a problem in order to get help, not just state it.
$endgroup$
– Alexander Gruber♦
Nov 4 '12 at 6:24




$begingroup$
Welcome to MSE Heidi. Please read the FAQ. You need to say what you've done towards a problem in order to get help, not just state it.
$endgroup$
– Alexander Gruber♦
Nov 4 '12 at 6:24












$begingroup$
Are you sure that the last sentence is correct? I suspect that you want to prove that the topology on $Ytimes Y$ induced by $q$ is the same as the product topology on $Ytimes Y$, which isn’t what you actually wrote.
$endgroup$
– Brian M. Scott
Nov 4 '12 at 6:27




$begingroup$
Are you sure that the last sentence is correct? I suspect that you want to prove that the topology on $Ytimes Y$ induced by $q$ is the same as the product topology on $Ytimes Y$, which isn’t what you actually wrote.
$endgroup$
– Brian M. Scott
Nov 4 '12 at 6:27












$begingroup$
I think you mean: Prove that the topology on $Y times Y$ is the same as .... This is answered below, and is also relevant to math.stackexcnge.com/questions/31697
$endgroup$
– Ronnie Brown
Nov 4 '12 at 17:41




$begingroup$
I think you mean: Prove that the topology on $Y times Y$ is the same as .... This is answered below, and is also relevant to math.stackexcnge.com/questions/31697
$endgroup$
– Ronnie Brown
Nov 4 '12 at 17:41












$begingroup$
oops! yes, i meant the topology on $Y times Y$ as a product of the quotient topologies on $Y$ is the same... I feel like it intuitively makes sense but I'm not sure how to start a formal proof of it
$endgroup$
– Heidi
Nov 5 '12 at 9:32




$begingroup$
oops! yes, i meant the topology on $Y times Y$ as a product of the quotient topologies on $Y$ is the same... I feel like it intuitively makes sense but I'm not sure how to start a formal proof of it
$endgroup$
– Heidi
Nov 5 '12 at 9:32












$begingroup$
My previous comment should have referenced math.stackexchange.com/questions/31697
$endgroup$
– Ronnie Brown
Dec 2 '18 at 15:37






$begingroup$
My previous comment should have referenced math.stackexchange.com/questions/31697
$endgroup$
– Ronnie Brown
Dec 2 '18 at 15:37












1 Answer
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The suggested result is false: the example given in the book Topology and Groupoids,(T&G) p.$111$, is actually that if $p: mathbb Q to Y$ is the quotient map identifying all of $mathbb Z$ to a single point, then $p times 1: mathbb Q times mathbb Q to Y times mathbb Q$ is not a quotient map. So it is expected that $p times p$ is not a quotient map.



The result is true for $p: X to Y$ if $X$ and $Y$ are locally compact and Hausdorff.



This problem led me in my 1961 Oxford thesis to propose using the category of Hausdorff $k$-spaces, and this proposal has been improved by using what have been called compactly generated spaces, i.e. spaces $X$ which have the final topology with respect to (a set of) continuous maps $C to X$ for $C$ compact Hausdorff. This is explained in Section 5.9 of T&G. See also this ncatlab link on convenient categories of topological spaces.






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    $begingroup$

    The suggested result is false: the example given in the book Topology and Groupoids,(T&G) p.$111$, is actually that if $p: mathbb Q to Y$ is the quotient map identifying all of $mathbb Z$ to a single point, then $p times 1: mathbb Q times mathbb Q to Y times mathbb Q$ is not a quotient map. So it is expected that $p times p$ is not a quotient map.



    The result is true for $p: X to Y$ if $X$ and $Y$ are locally compact and Hausdorff.



    This problem led me in my 1961 Oxford thesis to propose using the category of Hausdorff $k$-spaces, and this proposal has been improved by using what have been called compactly generated spaces, i.e. spaces $X$ which have the final topology with respect to (a set of) continuous maps $C to X$ for $C$ compact Hausdorff. This is explained in Section 5.9 of T&G. See also this ncatlab link on convenient categories of topological spaces.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      The suggested result is false: the example given in the book Topology and Groupoids,(T&G) p.$111$, is actually that if $p: mathbb Q to Y$ is the quotient map identifying all of $mathbb Z$ to a single point, then $p times 1: mathbb Q times mathbb Q to Y times mathbb Q$ is not a quotient map. So it is expected that $p times p$ is not a quotient map.



      The result is true for $p: X to Y$ if $X$ and $Y$ are locally compact and Hausdorff.



      This problem led me in my 1961 Oxford thesis to propose using the category of Hausdorff $k$-spaces, and this proposal has been improved by using what have been called compactly generated spaces, i.e. spaces $X$ which have the final topology with respect to (a set of) continuous maps $C to X$ for $C$ compact Hausdorff. This is explained in Section 5.9 of T&G. See also this ncatlab link on convenient categories of topological spaces.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        The suggested result is false: the example given in the book Topology and Groupoids,(T&G) p.$111$, is actually that if $p: mathbb Q to Y$ is the quotient map identifying all of $mathbb Z$ to a single point, then $p times 1: mathbb Q times mathbb Q to Y times mathbb Q$ is not a quotient map. So it is expected that $p times p$ is not a quotient map.



        The result is true for $p: X to Y$ if $X$ and $Y$ are locally compact and Hausdorff.



        This problem led me in my 1961 Oxford thesis to propose using the category of Hausdorff $k$-spaces, and this proposal has been improved by using what have been called compactly generated spaces, i.e. spaces $X$ which have the final topology with respect to (a set of) continuous maps $C to X$ for $C$ compact Hausdorff. This is explained in Section 5.9 of T&G. See also this ncatlab link on convenient categories of topological spaces.






        share|cite|improve this answer











        $endgroup$



        The suggested result is false: the example given in the book Topology and Groupoids,(T&G) p.$111$, is actually that if $p: mathbb Q to Y$ is the quotient map identifying all of $mathbb Z$ to a single point, then $p times 1: mathbb Q times mathbb Q to Y times mathbb Q$ is not a quotient map. So it is expected that $p times p$ is not a quotient map.



        The result is true for $p: X to Y$ if $X$ and $Y$ are locally compact and Hausdorff.



        This problem led me in my 1961 Oxford thesis to propose using the category of Hausdorff $k$-spaces, and this proposal has been improved by using what have been called compactly generated spaces, i.e. spaces $X$ which have the final topology with respect to (a set of) continuous maps $C to X$ for $C$ compact Hausdorff. This is explained in Section 5.9 of T&G. See also this ncatlab link on convenient categories of topological spaces.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 12:01

























        answered Nov 4 '12 at 17:25









        Ronnie BrownRonnie Brown

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