Uniqueness in the Universal Property of Quotient Maps
$begingroup$
Here is Munkres' way of phrasing the universal property of quotient maps:
Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?
I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).
general-topology quotient-spaces universal-property
asked Jan 2 at 15:32
user193319user193319
2,4532927
$endgroup$
add a comment |
$begingroup$
Here is Munkres' way of phrasing the universal property of quotient maps:
Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?
I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).
general-topology quotient-spaces universal-property
asked Jan 2 at 15:32
user193319user193319
2,4532927
$endgroup$
1
$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37
add a comment |
$begingroup$
Here is Munkres' way of phrasing the universal property of quotient maps:
Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?
I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).
general-topology quotient-spaces universal-property
asked Jan 2 at 15:32
user193319user193319
2,4532927
$endgroup$
Here is Munkres' way of phrasing the universal property of quotient maps:
Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?
I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).
general-topology quotient-spaces universal-property
general-topology quotient-spaces universal-property
asked Jan 2 at 15:32
user193319user193319
2,4532927
asked Jan 2 at 15:32
user193319user193319
2,4532927
asked Jan 2 at 15:32
user193319user193319
2,4532927
asked Jan 2 at 15:32
user193319user193319
2,4532927
asked Jan 2 at 15:32
user193319user193319
2,4532927
2,4532927
1
$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37
add a comment |
1
$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37
1
1
$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37
$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
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%2f3059611%2funiqueness-in-the-universal-property-of-quotient-maps%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
$begingroup$
$p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
add a comment |
$begingroup$
$p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
add a comment |
$begingroup$
$p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
$p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
115k348124
115k348124
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%2f3059611%2funiqueness-in-the-universal-property-of-quotient-maps%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
$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37