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
620311
$endgroup$
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
620311
$endgroup$
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
620311
$endgroup$
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
620311
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
620311
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
620311
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
620311
asked Jan 10 at 13:03
Prince ThomasPrince Thomas
620311
620311
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
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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
11.4k3934
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$begingroup$
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
11.4k3934
$endgroup$
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
11.4k3934
$endgroup$
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
11.4k3934
$endgroup$
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
11.4k3934
answered Jan 10 at 13:50
Paul FrostPaul Frost
11.4k3934
answered Jan 10 at 13:50
Paul FrostPaul Frost
11.4k3934
answered Jan 10 at 13:50
Paul FrostPaul Frost
11.4k3934
11.4k3934
add a comment |
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$begingroup$
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