cell complex structure of circle
$begingroup$
I came to know that the cell complex structure of the circle $S^1$ is $e^0∪e^1$. But in my point of view it should be $D^0∪D^1$ as $e^0$ is nothing but an empty set. Can anybody make me understand where am I wrong?
algebraic-topology
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
$endgroup$
add a comment |
$begingroup$
I came to know that the cell complex structure of the circle $S^1$ is $e^0∪e^1$. But in my point of view it should be $D^0∪D^1$ as $e^0$ is nothing but an empty set. Can anybody make me understand where am I wrong?
algebraic-topology
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
$endgroup$
1
$begingroup$
I assume that by "$e$" you mean open cells and by "$D$" closed cells? Anyway $e^0$ is not an empty set - its a singleton. $0$-dimensional open and closed cells coincide.
$endgroup$
– freakish
Jan 10 at 13:12
$begingroup$
Yeah I was exactly looking for this answer. ThankYou!
$endgroup$
– Prince Thomas
Jan 10 at 13:25
1
$begingroup$
In your first sentence, you should say a cell structure rather than the cell structure. There are many cell structures for the circle: n vertices and n edges for any positive integer n.
$endgroup$
– John Palmieri
Jan 10 at 16:25
add a comment |
$begingroup$
I came to know that the cell complex structure of the circle $S^1$ is $e^0∪e^1$. But in my point of view it should be $D^0∪D^1$ as $e^0$ is nothing but an empty set. Can anybody make me understand where am I wrong?
algebraic-topology
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
$endgroup$
I came to know that the cell complex structure of the circle $S^1$ is $e^0∪e^1$. But in my point of view it should be $D^0∪D^1$ as $e^0$ is nothing but an empty set. Can anybody make me understand where am I wrong?
algebraic-topology
algebraic-topology
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
615210
615210
1
$begingroup$
I assume that by "$e$" you mean open cells and by "$D$" closed cells? Anyway $e^0$ is not an empty set - its a singleton. $0$-dimensional open and closed cells coincide.
$endgroup$
– freakish
Jan 10 at 13:12
$begingroup$
Yeah I was exactly looking for this answer. ThankYou!
$endgroup$
– Prince Thomas
Jan 10 at 13:25
1
$begingroup$
In your first sentence, you should say a cell structure rather than the cell structure. There are many cell structures for the circle: n vertices and n edges for any positive integer n.
$endgroup$
– John Palmieri
Jan 10 at 16:25
add a comment |
1
$begingroup$
I assume that by "$e$" you mean open cells and by "$D$" closed cells? Anyway $e^0$ is not an empty set - its a singleton. $0$-dimensional open and closed cells coincide.
$endgroup$
– freakish
Jan 10 at 13:12
$begingroup$
Yeah I was exactly looking for this answer. ThankYou!
$endgroup$
– Prince Thomas
Jan 10 at 13:25
1
$begingroup$
In your first sentence, you should say a cell structure rather than the cell structure. There are many cell structures for the circle: n vertices and n edges for any positive integer n.
$endgroup$
– John Palmieri
Jan 10 at 16:25
1
1
$begingroup$
I assume that by "$e$" you mean open cells and by "$D$" closed cells? Anyway $e^0$ is not an empty set - its a singleton. $0$-dimensional open and closed cells coincide.
$endgroup$
– freakish
Jan 10 at 13:12
$begingroup$
I assume that by "$e$" you mean open cells and by "$D$" closed cells? Anyway $e^0$ is not an empty set - its a singleton. $0$-dimensional open and closed cells coincide.
$endgroup$
– freakish
Jan 10 at 13:12
$begingroup$
Yeah I was exactly looking for this answer. ThankYou!
$endgroup$
– Prince Thomas
Jan 10 at 13:25
$begingroup$
Yeah I was exactly looking for this answer. ThankYou!
$endgroup$
– Prince Thomas
Jan 10 at 13:25
1
1
$begingroup$
In your first sentence, you should say a cell structure rather than the cell structure. There are many cell structures for the circle: n vertices and n edges for any positive integer n.
$endgroup$
– John Palmieri
Jan 10 at 16:25
$begingroup$
In your first sentence, you should say a cell structure rather than the cell structure. There are many cell structures for the circle: n vertices and n edges for any positive integer n.
$endgroup$
– John Palmieri
Jan 10 at 16:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The are two equivalent approaches to build CW-complexes. One is based on open cells, the other on closed cells. Open cells are pairwise disjoint whereas closed cells are not pairwise disjoint (unless we only have $0$-cells). In both cases the $k$-skeleton $X^k$ of a CW-complex $X$ is defined as the union of all cells with dimension $le k$. For each $n$-cell $gamma^n_alpha$ we have a characteristic map $phi^n_alpha : D^n to X^{n-1}$. If $gamma^n_alpha = e^n_alpha$ denotes an open cell, then $e^n_alpha = phi^n_alpha(D^n setminus partial D^n)$, if $gamma^n_alpha = bar{e}^n_alpha$ denotes a closed cell, then $bar{e}^n_alpha = phi^n_alpha(D^n)$. In the first case $e^n_alpha cap X^{n-1} = emptyset$ and in second case $bar{e}^n_alpha cap X^{n-1} ne emptyset$.
It seems that the "open-cell-approach" is more popular, see for example the Appendix of Hatcher's "Algebraic Topology".
Anyway, in both approaches $0$-cells are single point spaces and the $0$-skeleton $X^0$ is a discrete space. This comes from the fact that $D^0 = { * }$ and $partial D^0 = emptyset$.
Hence for $X = S^1$ we obtain an open cell decomposition $e^0 = { 1}, e^1 = S^1 setminus { 1} approx (-1,1)$ and a closed decomposition $bar{e}^0 = { 1}, bar{e}^1 = S^1$. The characteristic map for the $1$-cell is $phi^1 : [-1,1] to { 1}, phi^1(t) = e^{pi i t}$.
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
$endgroup$
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%2f3068621%2fcell-complex-structure-of-circle%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 are two equivalent approaches to build CW-complexes. One is based on open cells, the other on closed cells. Open cells are pairwise disjoint whereas closed cells are not pairwise disjoint (unless we only have $0$-cells). In both cases the $k$-skeleton $X^k$ of a CW-complex $X$ is defined as the union of all cells with dimension $le k$. For each $n$-cell $gamma^n_alpha$ we have a characteristic map $phi^n_alpha : D^n to X^{n-1}$. If $gamma^n_alpha = e^n_alpha$ denotes an open cell, then $e^n_alpha = phi^n_alpha(D^n setminus partial D^n)$, if $gamma^n_alpha = bar{e}^n_alpha$ denotes a closed cell, then $bar{e}^n_alpha = phi^n_alpha(D^n)$. In the first case $e^n_alpha cap X^{n-1} = emptyset$ and in second case $bar{e}^n_alpha cap X^{n-1} ne emptyset$.
It seems that the "open-cell-approach" is more popular, see for example the Appendix of Hatcher's "Algebraic Topology".
Anyway, in both approaches $0$-cells are single point spaces and the $0$-skeleton $X^0$ is a discrete space. This comes from the fact that $D^0 = { * }$ and $partial D^0 = emptyset$.
Hence for $X = S^1$ we obtain an open cell decomposition $e^0 = { 1}, e^1 = S^1 setminus { 1} approx (-1,1)$ and a closed decomposition $bar{e}^0 = { 1}, bar{e}^1 = S^1$. The characteristic map for the $1$-cell is $phi^1 : [-1,1] to { 1}, phi^1(t) = e^{pi i t}$.
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
$endgroup$
add a comment |
$begingroup$
The are two equivalent approaches to build CW-complexes. One is based on open cells, the other on closed cells. Open cells are pairwise disjoint whereas closed cells are not pairwise disjoint (unless we only have $0$-cells). In both cases the $k$-skeleton $X^k$ of a CW-complex $X$ is defined as the union of all cells with dimension $le k$. For each $n$-cell $gamma^n_alpha$ we have a characteristic map $phi^n_alpha : D^n to X^{n-1}$. If $gamma^n_alpha = e^n_alpha$ denotes an open cell, then $e^n_alpha = phi^n_alpha(D^n setminus partial D^n)$, if $gamma^n_alpha = bar{e}^n_alpha$ denotes a closed cell, then $bar{e}^n_alpha = phi^n_alpha(D^n)$. In the first case $e^n_alpha cap X^{n-1} = emptyset$ and in second case $bar{e}^n_alpha cap X^{n-1} ne emptyset$.
It seems that the "open-cell-approach" is more popular, see for example the Appendix of Hatcher's "Algebraic Topology".
Anyway, in both approaches $0$-cells are single point spaces and the $0$-skeleton $X^0$ is a discrete space. This comes from the fact that $D^0 = { * }$ and $partial D^0 = emptyset$.
Hence for $X = S^1$ we obtain an open cell decomposition $e^0 = { 1}, e^1 = S^1 setminus { 1} approx (-1,1)$ and a closed decomposition $bar{e}^0 = { 1}, bar{e}^1 = S^1$. The characteristic map for the $1$-cell is $phi^1 : [-1,1] to { 1}, phi^1(t) = e^{pi i t}$.
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
$endgroup$
add a comment |
$begingroup$
The are two equivalent approaches to build CW-complexes. One is based on open cells, the other on closed cells. Open cells are pairwise disjoint whereas closed cells are not pairwise disjoint (unless we only have $0$-cells). In both cases the $k$-skeleton $X^k$ of a CW-complex $X$ is defined as the union of all cells with dimension $le k$. For each $n$-cell $gamma^n_alpha$ we have a characteristic map $phi^n_alpha : D^n to X^{n-1}$. If $gamma^n_alpha = e^n_alpha$ denotes an open cell, then $e^n_alpha = phi^n_alpha(D^n setminus partial D^n)$, if $gamma^n_alpha = bar{e}^n_alpha$ denotes a closed cell, then $bar{e}^n_alpha = phi^n_alpha(D^n)$. In the first case $e^n_alpha cap X^{n-1} = emptyset$ and in second case $bar{e}^n_alpha cap X^{n-1} ne emptyset$.
It seems that the "open-cell-approach" is more popular, see for example the Appendix of Hatcher's "Algebraic Topology".
Anyway, in both approaches $0$-cells are single point spaces and the $0$-skeleton $X^0$ is a discrete space. This comes from the fact that $D^0 = { * }$ and $partial D^0 = emptyset$.
Hence for $X = S^1$ we obtain an open cell decomposition $e^0 = { 1}, e^1 = S^1 setminus { 1} approx (-1,1)$ and a closed decomposition $bar{e}^0 = { 1}, bar{e}^1 = S^1$. The characteristic map for the $1$-cell is $phi^1 : [-1,1] to { 1}, phi^1(t) = e^{pi i t}$.
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
$endgroup$
The are two equivalent approaches to build CW-complexes. One is based on open cells, the other on closed cells. Open cells are pairwise disjoint whereas closed cells are not pairwise disjoint (unless we only have $0$-cells). In both cases the $k$-skeleton $X^k$ of a CW-complex $X$ is defined as the union of all cells with dimension $le k$. For each $n$-cell $gamma^n_alpha$ we have a characteristic map $phi^n_alpha : D^n to X^{n-1}$. If $gamma^n_alpha = e^n_alpha$ denotes an open cell, then $e^n_alpha = phi^n_alpha(D^n setminus partial D^n)$, if $gamma^n_alpha = bar{e}^n_alpha$ denotes a closed cell, then $bar{e}^n_alpha = phi^n_alpha(D^n)$. In the first case $e^n_alpha cap X^{n-1} = emptyset$ and in second case $bar{e}^n_alpha cap X^{n-1} ne emptyset$.
It seems that the "open-cell-approach" is more popular, see for example the Appendix of Hatcher's "Algebraic Topology".
Anyway, in both approaches $0$-cells are single point spaces and the $0$-skeleton $X^0$ is a discrete space. This comes from the fact that $D^0 = { * }$ and $partial D^0 = emptyset$.
Hence for $X = S^1$ we obtain an open cell decomposition $e^0 = { 1}, e^1 = S^1 setminus { 1} approx (-1,1)$ and a closed decomposition $bar{e}^0 = { 1}, bar{e}^1 = S^1$. The characteristic map for the $1$-cell is $phi^1 : [-1,1] to { 1}, phi^1(t) = e^{pi i t}$.
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
answered Jan 10 at 13:50
Paul FrostPaul Frost
10.1k3933
10.1k3933
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%2f3068621%2fcell-complex-structure-of-circle%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
$begingroup$
I assume that by "$e$" you mean open cells and by "$D$" closed cells? Anyway $e^0$ is not an empty set - its a singleton. $0$-dimensional open and closed cells coincide.
$endgroup$
– freakish
Jan 10 at 13:12
$begingroup$
Yeah I was exactly looking for this answer. ThankYou!
$endgroup$
– Prince Thomas
Jan 10 at 13:25
1
$begingroup$
In your first sentence, you should say a cell structure rather than the cell structure. There are many cell structures for the circle: n vertices and n edges for any positive integer n.
$endgroup$
– John Palmieri
Jan 10 at 16:25