Proof of Invariance of Domain in Introduction to Algebraic Topology by Rotman
$begingroup$
The following proof above is given in Introduction to Algebraic Topology by Rotman and I understand it except for the part where it's claimed that $y in h(N) setminus h(dot{N})$. Now I understand that we must have $y in h(N)$ since $h(x) = y$ and $x in N$, but I don't see why $y notin h(dot{N})$.
Why is it the case that $y notin h(dot{N})$?
general-topology algebraic-topology proof-explanation
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
$endgroup$
add a comment |
$begingroup$
The following proof above is given in Introduction to Algebraic Topology by Rotman and I understand it except for the part where it's claimed that $y in h(N) setminus h(dot{N})$. Now I understand that we must have $y in h(N)$ since $h(x) = y$ and $x in N$, but I don't see why $y notin h(dot{N})$.
Why is it the case that $y notin h(dot{N})$?
general-topology algebraic-topology proof-explanation
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
$endgroup$
add a comment |
$begingroup$
The following proof above is given in Introduction to Algebraic Topology by Rotman and I understand it except for the part where it's claimed that $y in h(N) setminus h(dot{N})$. Now I understand that we must have $y in h(N)$ since $h(x) = y$ and $x in N$, but I don't see why $y notin h(dot{N})$.
Why is it the case that $y notin h(dot{N})$?
general-topology algebraic-topology proof-explanation
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
$endgroup$
The following proof above is given in Introduction to Algebraic Topology by Rotman and I understand it except for the part where it's claimed that $y in h(N) setminus h(dot{N})$. Now I understand that we must have $y in h(N)$ since $h(x) = y$ and $x in N$, but I don't see why $y notin h(dot{N})$.
Why is it the case that $y notin h(dot{N})$?
general-topology algebraic-topology proof-explanation
general-topology algebraic-topology proof-explanation
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
asked Dec 12 '18 at 4:37
PerturbativePerturbative
4,38121553
4,38121553
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$xin N$, $xnotindot N$. As $h$ is a homeomorphism, it is bijective, so
$h(x)in h(N)$ and $h(x)notin h(dot N)$.
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
$begingroup$
I'm slightly confused, it's not assumed that $x notin dot{N}$ though.
$endgroup$
– Perturbative
Dec 12 '18 at 4:52
$begingroup$
$dot N$ is the boundary of $N$, so if $xindot N$, then $N$ is not a neighbourhood of $x$.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 4:55
$begingroup$
Ohh my bad, $N$ being a closed neighborhood of $x$ implies that $x in operatorname{Int}(N)$, I was mixing up the definition of a closed neighborhood
$endgroup$
– Perturbative
Dec 12 '18 at 5:00
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%2f3036230%2fproof-of-invariance-of-domain-in-introduction-to-algebraic-topology-by-rotman%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
$begingroup$
$xin N$, $xnotindot N$. As $h$ is a homeomorphism, it is bijective, so
$h(x)in h(N)$ and $h(x)notin h(dot N)$.
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
$begingroup$
I'm slightly confused, it's not assumed that $x notin dot{N}$ though.
$endgroup$
– Perturbative
Dec 12 '18 at 4:52
$begingroup$
$dot N$ is the boundary of $N$, so if $xindot N$, then $N$ is not a neighbourhood of $x$.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 4:55
$begingroup$
Ohh my bad, $N$ being a closed neighborhood of $x$ implies that $x in operatorname{Int}(N)$, I was mixing up the definition of a closed neighborhood
$endgroup$
– Perturbative
Dec 12 '18 at 5:00
add a comment |
$begingroup$
$xin N$, $xnotindot N$. As $h$ is a homeomorphism, it is bijective, so
$h(x)in h(N)$ and $h(x)notin h(dot N)$.
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
$begingroup$
I'm slightly confused, it's not assumed that $x notin dot{N}$ though.
$endgroup$
– Perturbative
Dec 12 '18 at 4:52
$begingroup$
$dot N$ is the boundary of $N$, so if $xindot N$, then $N$ is not a neighbourhood of $x$.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 4:55
$begingroup$
Ohh my bad, $N$ being a closed neighborhood of $x$ implies that $x in operatorname{Int}(N)$, I was mixing up the definition of a closed neighborhood
$endgroup$
– Perturbative
Dec 12 '18 at 5:00
add a comment |
$begingroup$
$xin N$, $xnotindot N$. As $h$ is a homeomorphism, it is bijective, so
$h(x)in h(N)$ and $h(x)notin h(dot N)$.
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
$xin N$, $xnotindot N$. As $h$ is a homeomorphism, it is bijective, so
$h(x)in h(N)$ and $h(x)notin h(dot N)$.
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Dec 12 '18 at 4:49
Lord Shark the UnknownLord Shark the Unknown
105k1160132
105k1160132
$begingroup$
I'm slightly confused, it's not assumed that $x notin dot{N}$ though.
$endgroup$
– Perturbative
Dec 12 '18 at 4:52
$begingroup$
$dot N$ is the boundary of $N$, so if $xindot N$, then $N$ is not a neighbourhood of $x$.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 4:55
$begingroup$
Ohh my bad, $N$ being a closed neighborhood of $x$ implies that $x in operatorname{Int}(N)$, I was mixing up the definition of a closed neighborhood
$endgroup$
– Perturbative
Dec 12 '18 at 5:00
add a comment |
$begingroup$
I'm slightly confused, it's not assumed that $x notin dot{N}$ though.
$endgroup$
– Perturbative
Dec 12 '18 at 4:52
$begingroup$
$dot N$ is the boundary of $N$, so if $xindot N$, then $N$ is not a neighbourhood of $x$.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 4:55
$begingroup$
Ohh my bad, $N$ being a closed neighborhood of $x$ implies that $x in operatorname{Int}(N)$, I was mixing up the definition of a closed neighborhood
$endgroup$
– Perturbative
Dec 12 '18 at 5:00
$begingroup$
I'm slightly confused, it's not assumed that $x notin dot{N}$ though.
$endgroup$
– Perturbative
Dec 12 '18 at 4:52
$begingroup$
I'm slightly confused, it's not assumed that $x notin dot{N}$ though.
$endgroup$
– Perturbative
Dec 12 '18 at 4:52
$begingroup$
$dot N$ is the boundary of $N$, so if $xindot N$, then $N$ is not a neighbourhood of $x$.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 4:55
$begingroup$
$dot N$ is the boundary of $N$, so if $xindot N$, then $N$ is not a neighbourhood of $x$.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 4:55
$begingroup$
Ohh my bad, $N$ being a closed neighborhood of $x$ implies that $x in operatorname{Int}(N)$, I was mixing up the definition of a closed neighborhood
$endgroup$
– Perturbative
Dec 12 '18 at 5:00
$begingroup$
Ohh my bad, $N$ being a closed neighborhood of $x$ implies that $x in operatorname{Int}(N)$, I was mixing up the definition of a closed neighborhood
$endgroup$
– Perturbative
Dec 12 '18 at 5:00
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%2f3036230%2fproof-of-invariance-of-domain-in-introduction-to-algebraic-topology-by-rotman%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
