Mixing
What is the difference between CW-complex and Cellular complex? [closed]
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Is every CW-complex is a Cellular space? Is its converse true?
If it is true then what is the difference between them?
We include the definition of CW-complex in algebraic topology given by Whitehed in 1949:
Definition. A CW complex is a Hausdorff space $X$ together with a partition of $X$ into open cells (varying dimension) that satisfies two additional properties:
- For each $n$-dimensional open cell $C$ in the partition of $X$, there exists a continuous map $f$ from the $n$-dimensional closed ball to $X$ such that
- the restriction of $f$ to the interior of the closed ball is a homeomorphism onto the cell $C$, and
- the image of the boundary of the closed ball is contained in the union of a finite number of elements of the partition, each having cell dimension less than $n$.
- A subset of $X$ is closed if and only if it meets the closure of each cell in a closed set.
Definition. A cellular space is a topological space $X$, with a sequence of subspaces
$$X^0subset X^1subset X^2subset cdots subset X,$$
such that $X=bigcuplimits_{n=0} X^n$, with the following properties:
- CS(1) $X^0$ is a discrete space.
- CS(2) for each positive integer $n$, there is an index set $A_n$, and continuous map $psi_i^n: S^{n-1} to X^{n-1}$ for each $iin A_n$ and disjoint copies $D^n_i$ of $D^n$ (one for each $iin A$) by identifying the points $x$ and $psi_i^n(x)$ for each $xin S_i^{n-1}$ and each $iin A_n$.
- CS(3) A subset $Y$ of $X$ is closed iff $Ycap X^n$ is closed in $X^n,$ for each $ngeq 0$.
algebraic-topology definition cw-complexes
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
asked Apr 21 '15 at 9:10
MTMAMTMA
322
$endgroup$
closed as unclear what you're asking by Eric Wofsey, Paul Frost, user91500, Cesareo, Lord_Farin Jan 2 at 12:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Is every CW-complex is a Cellular space? Is its converse true?
If it is true then what is the difference between them?
We include the definition of CW-complex in algebraic topology given by Whitehed in 1949:
Definition. A CW complex is a Hausdorff space $X$ together with a partition of $X$ into open cells (varying dimension) that satisfies two additional properties:
- For each $n$-dimensional open cell $C$ in the partition of $X$, there exists a continuous map $f$ from the $n$-dimensional closed ball to $X$ such that
- the restriction of $f$ to the interior of the closed ball is a homeomorphism onto the cell $C$, and
- the image of the boundary of the closed ball is contained in the union of a finite number of elements of the partition, each having cell dimension less than $n$.
- A subset of $X$ is closed if and only if it meets the closure of each cell in a closed set.
Definition. A cellular space is a topological space $X$, with a sequence of subspaces
$$X^0subset X^1subset X^2subset cdots subset X,$$
such that $X=bigcuplimits_{n=0} X^n$, with the following properties:
- CS(1) $X^0$ is a discrete space.
- CS(2) for each positive integer $n$, there is an index set $A_n$, and continuous map $psi_i^n: S^{n-1} to X^{n-1}$ for each $iin A_n$ and disjoint copies $D^n_i$ of $D^n$ (one for each $iin A$) by identifying the points $x$ and $psi_i^n(x)$ for each $xin S_i^{n-1}$ and each $iin A_n$.
- CS(3) A subset $Y$ of $X$ is closed iff $Ycap X^n$ is closed in $X^n,$ for each $ngeq 0$.
algebraic-topology definition cw-complexes
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
asked Apr 21 '15 at 9:10
MTMAMTMA
322
$endgroup$
closed as unclear what you're asking by Eric Wofsey, Paul Frost, user91500, Cesareo, Lord_Farin Jan 2 at 12:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Why don't you include the definitions you are using, and explain where you are running into issues answering the question from these?
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– vociferous_rutabaga
Apr 21 '15 at 9:19
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@MPO my post is updated with definition of CW-complex and Cellular space(Complex)
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– MTMA
Apr 21 '15 at 9:45
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Quote: "and explain where you are running into issues answering the question from these".
$endgroup$
– Did
Apr 21 '15 at 10:20
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This is almost identical with math.stackexchange.com/q/2946857.
$endgroup$
– Paul Frost
Dec 19 '18 at 8:57
1
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The statement of CS(2) makes no sense and is surely not the intended statement (but I'm not sure what the intended statement is). In particular, you have not said what these "disjoint copies $D_i^n$" are supposed to have to do with the space $X$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:05
add a comment |
$begingroup$
Is every CW-complex is a Cellular space? Is its converse true?
If it is true then what is the difference between them?
We include the definition of CW-complex in algebraic topology given by Whitehed in 1949:
Definition. A CW complex is a Hausdorff space $X$ together with a partition of $X$ into open cells (varying dimension) that satisfies two additional properties:
- For each $n$-dimensional open cell $C$ in the partition of $X$, there exists a continuous map $f$ from the $n$-dimensional closed ball to $X$ such that
- the restriction of $f$ to the interior of the closed ball is a homeomorphism onto the cell $C$, and
- the image of the boundary of the closed ball is contained in the union of a finite number of elements of the partition, each having cell dimension less than $n$.
- A subset of $X$ is closed if and only if it meets the closure of each cell in a closed set.
Definition. A cellular space is a topological space $X$, with a sequence of subspaces
$$X^0subset X^1subset X^2subset cdots subset X,$$
such that $X=bigcuplimits_{n=0} X^n$, with the following properties:
- CS(1) $X^0$ is a discrete space.
- CS(2) for each positive integer $n$, there is an index set $A_n$, and continuous map $psi_i^n: S^{n-1} to X^{n-1}$ for each $iin A_n$ and disjoint copies $D^n_i$ of $D^n$ (one for each $iin A$) by identifying the points $x$ and $psi_i^n(x)$ for each $xin S_i^{n-1}$ and each $iin A_n$.
- CS(3) A subset $Y$ of $X$ is closed iff $Ycap X^n$ is closed in $X^n,$ for each $ngeq 0$.
algebraic-topology definition cw-complexes
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
asked Apr 21 '15 at 9:10
MTMAMTMA
322
$endgroup$
Is every CW-complex is a Cellular space? Is its converse true?
If it is true then what is the difference between them?
We include the definition of CW-complex in algebraic topology given by Whitehed in 1949:
Definition. A CW complex is a Hausdorff space $X$ together with a partition of $X$ into open cells (varying dimension) that satisfies two additional properties:
- For each $n$-dimensional open cell $C$ in the partition of $X$, there exists a continuous map $f$ from the $n$-dimensional closed ball to $X$ such that
- the restriction of $f$ to the interior of the closed ball is a homeomorphism onto the cell $C$, and
- the image of the boundary of the closed ball is contained in the union of a finite number of elements of the partition, each having cell dimension less than $n$.
- A subset of $X$ is closed if and only if it meets the closure of each cell in a closed set.
Definition. A cellular space is a topological space $X$, with a sequence of subspaces
$$X^0subset X^1subset X^2subset cdots subset X,$$
such that $X=bigcuplimits_{n=0} X^n$, with the following properties:
- CS(1) $X^0$ is a discrete space.
- CS(2) for each positive integer $n$, there is an index set $A_n$, and continuous map $psi_i^n: S^{n-1} to X^{n-1}$ for each $iin A_n$ and disjoint copies $D^n_i$ of $D^n$ (one for each $iin A$) by identifying the points $x$ and $psi_i^n(x)$ for each $xin S_i^{n-1}$ and each $iin A_n$.
- CS(3) A subset $Y$ of $X$ is closed iff $Ycap X^n$ is closed in $X^n,$ for each $ngeq 0$.
algebraic-topology definition cw-complexes
algebraic-topology definition cw-complexes
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
asked Apr 21 '15 at 9:10
MTMAMTMA
322
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
asked Apr 21 '15 at 9:10
MTMAMTMA
322
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
edited Apr 21 '15 at 11:05
Najib Idrissi
40.9k471139
40.9k471139
asked Apr 21 '15 at 9:10
MTMAMTMA
322
asked Apr 21 '15 at 9:10
MTMAMTMA
322
asked Apr 21 '15 at 9:10
MTMAMTMA
322
322
closed as unclear what you're asking by Eric Wofsey, Paul Frost, user91500, Cesareo, Lord_Farin Jan 2 at 12:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Eric Wofsey, Paul Frost, user91500, Cesareo, Lord_Farin Jan 2 at 12:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
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Why don't you include the definitions you are using, and explain where you are running into issues answering the question from these?
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– vociferous_rutabaga
Apr 21 '15 at 9:19
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@MPO my post is updated with definition of CW-complex and Cellular space(Complex)
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– MTMA
Apr 21 '15 at 9:45
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Quote: "and explain where you are running into issues answering the question from these".
$endgroup$
– Did
Apr 21 '15 at 10:20
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This is almost identical with math.stackexchange.com/q/2946857.
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– Paul Frost
Dec 19 '18 at 8:57
1
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The statement of CS(2) makes no sense and is surely not the intended statement (but I'm not sure what the intended statement is). In particular, you have not said what these "disjoint copies $D_i^n$" are supposed to have to do with the space $X$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:05
add a comment |
3
$begingroup$
Why don't you include the definitions you are using, and explain where you are running into issues answering the question from these?
$endgroup$
– vociferous_rutabaga
Apr 21 '15 at 9:19
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@MPO my post is updated with definition of CW-complex and Cellular space(Complex)
$endgroup$
– MTMA
Apr 21 '15 at 9:45
$begingroup$
Quote: "and explain where you are running into issues answering the question from these".
$endgroup$
– Did
Apr 21 '15 at 10:20
$begingroup$
This is almost identical with math.stackexchange.com/q/2946857.
$endgroup$
– Paul Frost
Dec 19 '18 at 8:57
1
$begingroup$
The statement of CS(2) makes no sense and is surely not the intended statement (but I'm not sure what the intended statement is). In particular, you have not said what these "disjoint copies $D_i^n$" are supposed to have to do with the space $X$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:05
3
3
$begingroup$
Why don't you include the definitions you are using, and explain where you are running into issues answering the question from these?
$endgroup$
– vociferous_rutabaga
Apr 21 '15 at 9:19
$begingroup$
Why don't you include the definitions you are using, and explain where you are running into issues answering the question from these?
$endgroup$
– vociferous_rutabaga
Apr 21 '15 at 9:19
$begingroup$
@MPO my post is updated with definition of CW-complex and Cellular space(Complex)
$endgroup$
– MTMA
Apr 21 '15 at 9:45
$begingroup$
@MPO my post is updated with definition of CW-complex and Cellular space(Complex)
$endgroup$
– MTMA
Apr 21 '15 at 9:45
$begingroup$
Quote: "and explain where you are running into issues answering the question from these".
$endgroup$
– Did
Apr 21 '15 at 10:20
$begingroup$
Quote: "and explain where you are running into issues answering the question from these".
$endgroup$
– Did
Apr 21 '15 at 10:20
$begingroup$
This is almost identical with math.stackexchange.com/q/2946857.
$endgroup$
– Paul Frost
Dec 19 '18 at 8:57
$begingroup$
This is almost identical with math.stackexchange.com/q/2946857.
$endgroup$
– Paul Frost
Dec 19 '18 at 8:57
1
1
$begingroup$
The statement of CS(2) makes no sense and is surely not the intended statement (but I'm not sure what the intended statement is). In particular, you have not said what these "disjoint copies $D_i^n$" are supposed to have to do with the space $X$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:05
$begingroup$
The statement of CS(2) makes no sense and is surely not the intended statement (but I'm not sure what the intended statement is). In particular, you have not said what these "disjoint copies $D_i^n$" are supposed to have to do with the space $X$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:05
add a comment |
3 Answers
3
active
oldest
votes
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the brief answer is no, every cellular complex is not a CW-complex. The inclusion goes the other way around: CW-complex are cellular complex satisfying two additional properties:
- Closure-finiteness $bf{(C)}$: The closure of each cell is covered by a finite number of cells
- Weak-topology $bf{(W)}$: a subset $F$ is closed in $X$ if and only if the intersection between $F$ and every cell of $X$ is closed.
A cellular complex not satisfying $bf{(C)}$ is for instance the 2-disk made of a 2-cell attached to an infinity of $0$-cell (one for each point of its boundary).
An example of a cellular complex not satisfying $bf{(W)}$ is:
$$ { 1/n ; n leq 1 } cup { 0 } subset mathbb{R} $$.
To know more:
- http://www.profmath.uqam.ca/~powell/Expose-C.W-Complex.pdf
- https://is.muni.cz/el/1431/jaro2015/M8130/um/39015882/at02n.pdf
They both adress this specific question.
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
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1
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Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete.
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– Eric Wofsey
Jan 2 at 8:02
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Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one...
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– TryingToGetOut
Jan 2 at 8:05
add a comment |
$begingroup$
Both concepts as defined in the question agree. This is proved in the appendix of Hatcher's book "Algebraic Topology". See Proposition A.2.
See also the references given by Jeanne Lefevre.
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
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1
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There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$.
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– Eric Wofsey
Jan 2 at 7:58
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Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$.
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– Eric Wofsey
Jan 2 at 8:07
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@EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" .
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– Paul Frost
Jan 2 at 11:07
add a comment |
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A CW-complex is built inductively, with cells of dimension $n$ only allowed to be attached in the $n$-th step. A cell complex is similar but cells of any dimension may be attached in each step, so there exist cell complexes that are not CW-complexes.
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
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My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible.
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– Autolatry
Apr 21 '15 at 11:12
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@ Autolatry Please give an example of a Cell complex which is not CW-complex
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– MTMA
Apr 21 '15 at 11:44
1
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Read the definition given in the question -- the cells complexes are built inductively here.
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– Najib Idrissi
Apr 21 '15 at 11:48
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@MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex.
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– Autolatry
Apr 21 '15 at 11:49
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@Autolatry You should give us your definition of a cell complex.
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– Paul Frost
Jan 3 at 23:22
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
the brief answer is no, every cellular complex is not a CW-complex. The inclusion goes the other way around: CW-complex are cellular complex satisfying two additional properties:
- Closure-finiteness $bf{(C)}$: The closure of each cell is covered by a finite number of cells
- Weak-topology $bf{(W)}$: a subset $F$ is closed in $X$ if and only if the intersection between $F$ and every cell of $X$ is closed.
A cellular complex not satisfying $bf{(C)}$ is for instance the 2-disk made of a 2-cell attached to an infinity of $0$-cell (one for each point of its boundary).
An example of a cellular complex not satisfying $bf{(W)}$ is:
$$ { 1/n ; n leq 1 } cup { 0 } subset mathbb{R} $$.
To know more:
- http://www.profmath.uqam.ca/~powell/Expose-C.W-Complex.pdf
- https://is.muni.cz/el/1431/jaro2015/M8130/um/39015882/at02n.pdf
They both adress this specific question.
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
$endgroup$
1
$begingroup$
Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete.
$endgroup$
– Eric Wofsey
Jan 2 at 8:02
$begingroup$
Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one...
$endgroup$
– TryingToGetOut
Jan 2 at 8:05
add a comment |
$begingroup$
the brief answer is no, every cellular complex is not a CW-complex. The inclusion goes the other way around: CW-complex are cellular complex satisfying two additional properties:
- Closure-finiteness $bf{(C)}$: The closure of each cell is covered by a finite number of cells
- Weak-topology $bf{(W)}$: a subset $F$ is closed in $X$ if and only if the intersection between $F$ and every cell of $X$ is closed.
A cellular complex not satisfying $bf{(C)}$ is for instance the 2-disk made of a 2-cell attached to an infinity of $0$-cell (one for each point of its boundary).
An example of a cellular complex not satisfying $bf{(W)}$ is:
$$ { 1/n ; n leq 1 } cup { 0 } subset mathbb{R} $$.
To know more:
- http://www.profmath.uqam.ca/~powell/Expose-C.W-Complex.pdf
- https://is.muni.cz/el/1431/jaro2015/M8130/um/39015882/at02n.pdf
They both adress this specific question.
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
$endgroup$
1
$begingroup$
Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete.
$endgroup$
– Eric Wofsey
Jan 2 at 8:02
$begingroup$
Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one...
$endgroup$
– TryingToGetOut
Jan 2 at 8:05
add a comment |
$begingroup$
the brief answer is no, every cellular complex is not a CW-complex. The inclusion goes the other way around: CW-complex are cellular complex satisfying two additional properties:
- Closure-finiteness $bf{(C)}$: The closure of each cell is covered by a finite number of cells
- Weak-topology $bf{(W)}$: a subset $F$ is closed in $X$ if and only if the intersection between $F$ and every cell of $X$ is closed.
A cellular complex not satisfying $bf{(C)}$ is for instance the 2-disk made of a 2-cell attached to an infinity of $0$-cell (one for each point of its boundary).
An example of a cellular complex not satisfying $bf{(W)}$ is:
$$ { 1/n ; n leq 1 } cup { 0 } subset mathbb{R} $$.
To know more:
- http://www.profmath.uqam.ca/~powell/Expose-C.W-Complex.pdf
- https://is.muni.cz/el/1431/jaro2015/M8130/um/39015882/at02n.pdf
They both adress this specific question.
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
$endgroup$
the brief answer is no, every cellular complex is not a CW-complex. The inclusion goes the other way around: CW-complex are cellular complex satisfying two additional properties:
- Closure-finiteness $bf{(C)}$: The closure of each cell is covered by a finite number of cells
- Weak-topology $bf{(W)}$: a subset $F$ is closed in $X$ if and only if the intersection between $F$ and every cell of $X$ is closed.
A cellular complex not satisfying $bf{(C)}$ is for instance the 2-disk made of a 2-cell attached to an infinity of $0$-cell (one for each point of its boundary).
An example of a cellular complex not satisfying $bf{(W)}$ is:
$$ { 1/n ; n leq 1 } cup { 0 } subset mathbb{R} $$.
To know more:
- http://www.profmath.uqam.ca/~powell/Expose-C.W-Complex.pdf
- https://is.muni.cz/el/1431/jaro2015/M8130/um/39015882/at02n.pdf
They both adress this specific question.
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
answered Dec 19 '18 at 6:48
TryingToGetOutTryingToGetOut
488
488
1
$begingroup$
Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete.
$endgroup$
– Eric Wofsey
Jan 2 at 8:02
$begingroup$
Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one...
$endgroup$
– TryingToGetOut
Jan 2 at 8:05
add a comment |
1
$begingroup$
Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete.
$endgroup$
– Eric Wofsey
Jan 2 at 8:02
$begingroup$
Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one...
$endgroup$
– TryingToGetOut
Jan 2 at 8:05
1
1
$begingroup$
Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete.
$endgroup$
– Eric Wofsey
Jan 2 at 8:02
$begingroup$
Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete.
$endgroup$
– Eric Wofsey
Jan 2 at 8:02
$begingroup$
Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one...
$endgroup$
– TryingToGetOut
Jan 2 at 8:05
$begingroup$
Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one...
$endgroup$
– TryingToGetOut
Jan 2 at 8:05
add a comment |
$begingroup$
Both concepts as defined in the question agree. This is proved in the appendix of Hatcher's book "Algebraic Topology". See Proposition A.2.
See also the references given by Jeanne Lefevre.
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
$endgroup$
1
$begingroup$
There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 7:58
$begingroup$
Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:07
$begingroup$
@EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" .
$endgroup$
– Paul Frost
Jan 2 at 11:07
add a comment |
$begingroup$
Both concepts as defined in the question agree. This is proved in the appendix of Hatcher's book "Algebraic Topology". See Proposition A.2.
See also the references given by Jeanne Lefevre.
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
$endgroup$
1
$begingroup$
There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 7:58
$begingroup$
Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:07
$begingroup$
@EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" .
$endgroup$
– Paul Frost
Jan 2 at 11:07
add a comment |
$begingroup$
Both concepts as defined in the question agree. This is proved in the appendix of Hatcher's book "Algebraic Topology". See Proposition A.2.
See also the references given by Jeanne Lefevre.
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
$endgroup$
Both concepts as defined in the question agree. This is proved in the appendix of Hatcher's book "Algebraic Topology". See Proposition A.2.
See also the references given by Jeanne Lefevre.
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
edited Jan 2 at 8:08
Eric Wofsey
181k12208336
181k12208336
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
answered Dec 19 '18 at 13:20
Paul FrostPaul Frost
9,8753732
9,8753732
1
$begingroup$
There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 7:58
$begingroup$
Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:07
$begingroup$
@EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" .
$endgroup$
– Paul Frost
Jan 2 at 11:07
add a comment |
1
$begingroup$
There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 7:58
$begingroup$
Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:07
$begingroup$
@EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" .
$endgroup$
– Paul Frost
Jan 2 at 11:07
1
1
$begingroup$
There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 7:58
$begingroup$
There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 7:58
$begingroup$
Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:07
$begingroup$
Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$.
$endgroup$
– Eric Wofsey
Jan 2 at 8:07
$begingroup$
@EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" .
$endgroup$
– Paul Frost
Jan 2 at 11:07
$begingroup$
@EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" .
$endgroup$
– Paul Frost
Jan 2 at 11:07
add a comment |
$begingroup$
A CW-complex is built inductively, with cells of dimension $n$ only allowed to be attached in the $n$-th step. A cell complex is similar but cells of any dimension may be attached in each step, so there exist cell complexes that are not CW-complexes.
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
$endgroup$
$begingroup$
My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible.
$endgroup$
– Autolatry
Apr 21 '15 at 11:12
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@ Autolatry Please give an example of a Cell complex which is not CW-complex
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– MTMA
Apr 21 '15 at 11:44
1
$begingroup$
Read the definition given in the question -- the cells complexes are built inductively here.
$endgroup$
– Najib Idrissi
Apr 21 '15 at 11:48
$begingroup$
@MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex.
$endgroup$
– Autolatry
Apr 21 '15 at 11:49
$begingroup$
@Autolatry You should give us your definition of a cell complex.
$endgroup$
– Paul Frost
Jan 3 at 23:22
add a comment |
$begingroup$
A CW-complex is built inductively, with cells of dimension $n$ only allowed to be attached in the $n$-th step. A cell complex is similar but cells of any dimension may be attached in each step, so there exist cell complexes that are not CW-complexes.
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
$endgroup$
$begingroup$
My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible.
$endgroup$
– Autolatry
Apr 21 '15 at 11:12
$begingroup$
@ Autolatry Please give an example of a Cell complex which is not CW-complex
$endgroup$
– MTMA
Apr 21 '15 at 11:44
1
$begingroup$
Read the definition given in the question -- the cells complexes are built inductively here.
$endgroup$
– Najib Idrissi
Apr 21 '15 at 11:48
$begingroup$
@MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex.
$endgroup$
– Autolatry
Apr 21 '15 at 11:49
$begingroup$
@Autolatry You should give us your definition of a cell complex.
$endgroup$
– Paul Frost
Jan 3 at 23:22
add a comment |
$begingroup$
A CW-complex is built inductively, with cells of dimension $n$ only allowed to be attached in the $n$-th step. A cell complex is similar but cells of any dimension may be attached in each step, so there exist cell complexes that are not CW-complexes.
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
$endgroup$
A CW-complex is built inductively, with cells of dimension $n$ only allowed to be attached in the $n$-th step. A cell complex is similar but cells of any dimension may be attached in each step, so there exist cell complexes that are not CW-complexes.
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
edited Dec 19 '18 at 14:23
Pedro Tamaroff♦
96.4k10152297
96.4k10152297
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
answered Apr 21 '15 at 10:09
AutolatryAutolatry
2,68911015
2,68911015
$begingroup$
My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible.
$endgroup$
– Autolatry
Apr 21 '15 at 11:12
$begingroup$
@ Autolatry Please give an example of a Cell complex which is not CW-complex
$endgroup$
– MTMA
Apr 21 '15 at 11:44
1
$begingroup$
Read the definition given in the question -- the cells complexes are built inductively here.
$endgroup$
– Najib Idrissi
Apr 21 '15 at 11:48
$begingroup$
@MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex.
$endgroup$
– Autolatry
Apr 21 '15 at 11:49
$begingroup$
@Autolatry You should give us your definition of a cell complex.
$endgroup$
– Paul Frost
Jan 3 at 23:22
add a comment |
$begingroup$
My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible.
$endgroup$
– Autolatry
Apr 21 '15 at 11:12
$begingroup$
@ Autolatry Please give an example of a Cell complex which is not CW-complex
$endgroup$
– MTMA
Apr 21 '15 at 11:44
1
$begingroup$
Read the definition given in the question -- the cells complexes are built inductively here.
$endgroup$
– Najib Idrissi
Apr 21 '15 at 11:48
$begingroup$
@MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex.
$endgroup$
– Autolatry
Apr 21 '15 at 11:49
$begingroup$
@Autolatry You should give us your definition of a cell complex.
$endgroup$
– Paul Frost
Jan 3 at 23:22
$begingroup$
My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible.
$endgroup$
– Autolatry
Apr 21 '15 at 11:12
$begingroup$
My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible.
$endgroup$
– Autolatry
Apr 21 '15 at 11:12
$begingroup$
@ Autolatry Please give an example of a Cell complex which is not CW-complex
$endgroup$
– MTMA
Apr 21 '15 at 11:44
$begingroup$
@ Autolatry Please give an example of a Cell complex which is not CW-complex
$endgroup$
– MTMA
Apr 21 '15 at 11:44
1
1
$begingroup$
Read the definition given in the question -- the cells complexes are built inductively here.
$endgroup$
– Najib Idrissi
Apr 21 '15 at 11:48
$begingroup$
Read the definition given in the question -- the cells complexes are built inductively here.
$endgroup$
– Najib Idrissi
Apr 21 '15 at 11:48
$begingroup$
@MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex.
$endgroup$
– Autolatry
Apr 21 '15 at 11:49
$begingroup$
@MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex.
$endgroup$
– Autolatry
Apr 21 '15 at 11:49
$begingroup$
@Autolatry You should give us your definition of a cell complex.
$endgroup$
– Paul Frost
Jan 3 at 23:22
$begingroup$
@Autolatry You should give us your definition of a cell complex.
$endgroup$
– Paul Frost
Jan 3 at 23:22
add a comment |
3
$begingroup$
Why don't you include the definitions you are using, and explain where you are running into issues answering the question from these?
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– vociferous_rutabaga
Apr 21 '15 at 9:19
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@MPO my post is updated with definition of CW-complex and Cellular space(Complex)
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– MTMA
Apr 21 '15 at 9:45
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Quote: "and explain where you are running into issues answering the question from these".
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– Did
Apr 21 '15 at 10:20
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This is almost identical with math.stackexchange.com/q/2946857.
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– Paul Frost
Dec 19 '18 at 8:57
1
$begingroup$
The statement of CS(2) makes no sense and is surely not the intended statement (but I'm not sure what the intended statement is). In particular, you have not said what these "disjoint copies $D_i^n$" are supposed to have to do with the space $X$.
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– Eric Wofsey
Jan 2 at 8:05