Products of quotient topology same as quotient of product topology
$begingroup$
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
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
asked Nov 4 '12 at 5:29
HeidiHeidi
61
$endgroup$
add a comment |
$begingroup$
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
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
asked Nov 4 '12 at 5:29
HeidiHeidi
61
$endgroup$
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
add a comment |
$begingroup$
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
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
asked Nov 4 '12 at 5:29
HeidiHeidi
61
$endgroup$
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
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
asked Nov 4 '12 at 5:29
HeidiHeidi
61
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
asked Nov 4 '12 at 5:29
HeidiHeidi
61
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
edited Nov 4 '12 at 5:31
Brian M. Scott
458k38511913
458k38511913
asked Nov 4 '12 at 5:29
HeidiHeidi
61
asked Nov 4 '12 at 5:29
HeidiHeidi
61
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f228687%2fproducts-of-quotient-topology-same-as-quotient-of-product-topology%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
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.
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
edited Dec 3 '18 at 12:01
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
answered Nov 4 '12 at 17:25
Ronnie BrownRonnie Brown
12.1k12939
12.1k12939
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f228687%2fproducts-of-quotient-topology-same-as-quotient-of-product-topology%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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