Proof regarding n-connectedness
If one has to prove that $R^n - {0}$ is not $(n-1)$-connected, is it necessary to prove formally that there exists a non contractible $(n-1)$-sphere or can that simply be stated. If one must formally prove this, how may that be done? Can one, for example, consider the intersection of the $(n-1)$-sphere and $R^n$ with an arbitrary 2D plane and then say (for the sake of contradiction by the definition of 1-connectedness) that the given sphere is contractible if and only if the circle that corresponds to the intersection of that sphere is contractible in the subspace topology defined by the intersection of the plane and $R^n$?
general-topology proof-verification geometric-topology
add a comment |
If one has to prove that $R^n - {0}$ is not $(n-1)$-connected, is it necessary to prove formally that there exists a non contractible $(n-1)$-sphere or can that simply be stated. If one must formally prove this, how may that be done? Can one, for example, consider the intersection of the $(n-1)$-sphere and $R^n$ with an arbitrary 2D plane and then say (for the sake of contradiction by the definition of 1-connectedness) that the given sphere is contractible if and only if the circle that corresponds to the intersection of that sphere is contractible in the subspace topology defined by the intersection of the plane and $R^n$?
general-topology proof-verification geometric-topology
I think you have an index shift in the first line. $R^n setminus{0}$ is $n-1$-connected. $Rsetminus {0}$ is not connected, $R^2setminus {0}$ is not simply connected (i.e. not $1$-connected).
– Babelfish
Jan 4 at 8:22
add a comment |
If one has to prove that $R^n - {0}$ is not $(n-1)$-connected, is it necessary to prove formally that there exists a non contractible $(n-1)$-sphere or can that simply be stated. If one must formally prove this, how may that be done? Can one, for example, consider the intersection of the $(n-1)$-sphere and $R^n$ with an arbitrary 2D plane and then say (for the sake of contradiction by the definition of 1-connectedness) that the given sphere is contractible if and only if the circle that corresponds to the intersection of that sphere is contractible in the subspace topology defined by the intersection of the plane and $R^n$?
general-topology proof-verification geometric-topology
If one has to prove that $R^n - {0}$ is not $(n-1)$-connected, is it necessary to prove formally that there exists a non contractible $(n-1)$-sphere or can that simply be stated. If one must formally prove this, how may that be done? Can one, for example, consider the intersection of the $(n-1)$-sphere and $R^n$ with an arbitrary 2D plane and then say (for the sake of contradiction by the definition of 1-connectedness) that the given sphere is contractible if and only if the circle that corresponds to the intersection of that sphere is contractible in the subspace topology defined by the intersection of the plane and $R^n$?
general-topology proof-verification geometric-topology
general-topology proof-verification geometric-topology
edited yesterday
Paul Frost
9,5802632
9,5802632
asked Jan 4 at 8:11
Aryaman GuptaAryaman Gupta
336
336
I think you have an index shift in the first line. $R^n setminus{0}$ is $n-1$-connected. $Rsetminus {0}$ is not connected, $R^2setminus {0}$ is not simply connected (i.e. not $1$-connected).
– Babelfish
Jan 4 at 8:22
add a comment |
I think you have an index shift in the first line. $R^n setminus{0}$ is $n-1$-connected. $Rsetminus {0}$ is not connected, $R^2setminus {0}$ is not simply connected (i.e. not $1$-connected).
– Babelfish
Jan 4 at 8:22
I think you have an index shift in the first line. $R^n setminus{0}$ is $n-1$-connected. $Rsetminus {0}$ is not connected, $R^2setminus {0}$ is not simply connected (i.e. not $1$-connected).
– Babelfish
Jan 4 at 8:22
I think you have an index shift in the first line. $R^n setminus{0}$ is $n-1$-connected. $Rsetminus {0}$ is not connected, $R^2setminus {0}$ is not simply connected (i.e. not $1$-connected).
– Babelfish
Jan 4 at 8:22
add a comment |
1 Answer
1
active
oldest
votes
This is not immanent in the definitions, so there is need for a proof.
Your approach, however, is not sufficient. In general, there could be a contraction of the $n-1$-sphere outside the plane, without the existence of a contraction inside the plane. I think your approach would say something about relative homotopy groups.
The standard approach (at least I don't know an easier way) to show that $R^nsetminus {0}$ is not $n-1$-connected goes by first showing that $R^n setminus {0} simeq S^{n-1}$ (homotopy equivalence). Therefore, all homotopy groups will be isomorphic.
So it remains to show that $S^{n-1}$ is not $n-1$-connected. It is well known that $S^{n-1}$ is $n-2$-connected, so by the Hurewicz theorem, $pi_{n-1}(S^{n-1}) cong H_{n-1}(S^{n-1})$. Homology is not so hard to calculate, in particular $H_{n-1}(S^{n-1})cong mathbb{Z}neq 0$, so $S^{n-1}$ is not ${n-1}$-connected.
See also Wikipedia for this.
Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.
– Babelfish
23 hours ago
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%2f3061415%2fproof-regarding-n-connectedness%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
This is not immanent in the definitions, so there is need for a proof.
Your approach, however, is not sufficient. In general, there could be a contraction of the $n-1$-sphere outside the plane, without the existence of a contraction inside the plane. I think your approach would say something about relative homotopy groups.
The standard approach (at least I don't know an easier way) to show that $R^nsetminus {0}$ is not $n-1$-connected goes by first showing that $R^n setminus {0} simeq S^{n-1}$ (homotopy equivalence). Therefore, all homotopy groups will be isomorphic.
So it remains to show that $S^{n-1}$ is not $n-1$-connected. It is well known that $S^{n-1}$ is $n-2$-connected, so by the Hurewicz theorem, $pi_{n-1}(S^{n-1}) cong H_{n-1}(S^{n-1})$. Homology is not so hard to calculate, in particular $H_{n-1}(S^{n-1})cong mathbb{Z}neq 0$, so $S^{n-1}$ is not ${n-1}$-connected.
See also Wikipedia for this.
Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.
– Babelfish
23 hours ago
add a comment |
This is not immanent in the definitions, so there is need for a proof.
Your approach, however, is not sufficient. In general, there could be a contraction of the $n-1$-sphere outside the plane, without the existence of a contraction inside the plane. I think your approach would say something about relative homotopy groups.
The standard approach (at least I don't know an easier way) to show that $R^nsetminus {0}$ is not $n-1$-connected goes by first showing that $R^n setminus {0} simeq S^{n-1}$ (homotopy equivalence). Therefore, all homotopy groups will be isomorphic.
So it remains to show that $S^{n-1}$ is not $n-1$-connected. It is well known that $S^{n-1}$ is $n-2$-connected, so by the Hurewicz theorem, $pi_{n-1}(S^{n-1}) cong H_{n-1}(S^{n-1})$. Homology is not so hard to calculate, in particular $H_{n-1}(S^{n-1})cong mathbb{Z}neq 0$, so $S^{n-1}$ is not ${n-1}$-connected.
See also Wikipedia for this.
Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.
– Babelfish
23 hours ago
add a comment |
This is not immanent in the definitions, so there is need for a proof.
Your approach, however, is not sufficient. In general, there could be a contraction of the $n-1$-sphere outside the plane, without the existence of a contraction inside the plane. I think your approach would say something about relative homotopy groups.
The standard approach (at least I don't know an easier way) to show that $R^nsetminus {0}$ is not $n-1$-connected goes by first showing that $R^n setminus {0} simeq S^{n-1}$ (homotopy equivalence). Therefore, all homotopy groups will be isomorphic.
So it remains to show that $S^{n-1}$ is not $n-1$-connected. It is well known that $S^{n-1}$ is $n-2$-connected, so by the Hurewicz theorem, $pi_{n-1}(S^{n-1}) cong H_{n-1}(S^{n-1})$. Homology is not so hard to calculate, in particular $H_{n-1}(S^{n-1})cong mathbb{Z}neq 0$, so $S^{n-1}$ is not ${n-1}$-connected.
See also Wikipedia for this.
This is not immanent in the definitions, so there is need for a proof.
Your approach, however, is not sufficient. In general, there could be a contraction of the $n-1$-sphere outside the plane, without the existence of a contraction inside the plane. I think your approach would say something about relative homotopy groups.
The standard approach (at least I don't know an easier way) to show that $R^nsetminus {0}$ is not $n-1$-connected goes by first showing that $R^n setminus {0} simeq S^{n-1}$ (homotopy equivalence). Therefore, all homotopy groups will be isomorphic.
So it remains to show that $S^{n-1}$ is not $n-1$-connected. It is well known that $S^{n-1}$ is $n-2$-connected, so by the Hurewicz theorem, $pi_{n-1}(S^{n-1}) cong H_{n-1}(S^{n-1})$. Homology is not so hard to calculate, in particular $H_{n-1}(S^{n-1})cong mathbb{Z}neq 0$, so $S^{n-1}$ is not ${n-1}$-connected.
See also Wikipedia for this.
answered Jan 4 at 8:45
BabelfishBabelfish
1,126520
1,126520
Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.
– Babelfish
23 hours ago
add a comment |
Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.
– Babelfish
23 hours ago
Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.
– Babelfish
23 hours ago
Do you need further help with this topic? Should I extend an explanation? If you are satisfied with the answer, you might consider to accept it.
– Babelfish
23 hours ago
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3061415%2fproof-regarding-n-connectedness%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
I think you have an index shift in the first line. $R^n setminus{0}$ is $n-1$-connected. $Rsetminus {0}$ is not connected, $R^2setminus {0}$ is not simply connected (i.e. not $1$-connected).
– Babelfish
Jan 4 at 8:22