Path component of a CW-complex is a subcomplex.
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Let $X$ be a CW-complex. I want to prove each of its path components contains a zero cell. For that, I want to prove that each of its components is again a CW-complex.
Please give me some hints.
algebraic-topology cw-complexes
asked Nov 21 at 11:49
XYZABC
297110
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Let $X$ be a CW-complex. I want to prove each of its path components contains a zero cell. For that, I want to prove that each of its components is again a CW-complex.
Please give me some hints.
algebraic-topology cw-complexes
asked Nov 21 at 11:49
XYZABC
297110
I don't know at first hand whether in a CW-complex path-components coincide with components. What I do know is that components are always closed and that any closed set of a CW-complex is a sub-complex.
– drhab
Nov 21 at 12:14
@drhab: This seems false to me, or at least I don't understand it in the usual sense of subcomplex. Consider $S^1$ represented as a CW complex with one 0-cell (at $(1, 0)$) and a single $1$-cell. The set $C = { (x, y) in S^1 mid x le 0 }$ is certainly a closed set, but not a subcomplex, in the sense that there's no subset of the cells of $S^1$ whose union is $C$.
– John Hughes
Nov 21 at 12:19
@JohnHughes You are right. The theorem states: If $(X,mathcal E)$ is a CW-complex, $mathcal E'subseteqmathcal E$ and $X'=cupmathcal E'$ then $(X',mathcal E')$ is a subcomplex iff $X'$ is closed. I overlooked the extra condition on $X'$. Here $mathcal E$ denotes the collection of cells.
– drhab
Nov 21 at 12:32
@drhab Ah...thanks. I was thinking that you might be thinking of some theorem like "there's a subdivision of the original CW complex such that $C$ is a subcomplex," but that seemed as if it might be very hard to prove. :) Indeed, I'm not even sure how I'd define "subdivision". :)
– John Hughes
Nov 21 at 13:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a CW-complex. I want to prove each of its path components contains a zero cell. For that, I want to prove that each of its components is again a CW-complex.
Please give me some hints.
algebraic-topology cw-complexes
asked Nov 21 at 11:49
XYZABC
297110
Let $X$ be a CW-complex. I want to prove each of its path components contains a zero cell. For that, I want to prove that each of its components is again a CW-complex.
Please give me some hints.
algebraic-topology cw-complexes
algebraic-topology cw-complexes
asked Nov 21 at 11:49
XYZABC
297110
asked Nov 21 at 11:49
XYZABC
297110
asked Nov 21 at 11:49
XYZABC
297110
asked Nov 21 at 11:49
XYZABC
297110
asked Nov 21 at 11:49
XYZABC
297110
297110
I don't know at first hand whether in a CW-complex path-components coincide with components. What I do know is that components are always closed and that any closed set of a CW-complex is a sub-complex.
– drhab
Nov 21 at 12:14
@drhab: This seems false to me, or at least I don't understand it in the usual sense of subcomplex. Consider $S^1$ represented as a CW complex with one 0-cell (at $(1, 0)$) and a single $1$-cell. The set $C = { (x, y) in S^1 mid x le 0 }$ is certainly a closed set, but not a subcomplex, in the sense that there's no subset of the cells of $S^1$ whose union is $C$.
– John Hughes
Nov 21 at 12:19
@JohnHughes You are right. The theorem states: If $(X,mathcal E)$ is a CW-complex, $mathcal E'subseteqmathcal E$ and $X'=cupmathcal E'$ then $(X',mathcal E')$ is a subcomplex iff $X'$ is closed. I overlooked the extra condition on $X'$. Here $mathcal E$ denotes the collection of cells.
– drhab
Nov 21 at 12:32
@drhab Ah...thanks. I was thinking that you might be thinking of some theorem like "there's a subdivision of the original CW complex such that $C$ is a subcomplex," but that seemed as if it might be very hard to prove. :) Indeed, I'm not even sure how I'd define "subdivision". :)
– John Hughes
Nov 21 at 13:00
add a comment |
I don't know at first hand whether in a CW-complex path-components coincide with components. What I do know is that components are always closed and that any closed set of a CW-complex is a sub-complex.
– drhab
Nov 21 at 12:14
@drhab: This seems false to me, or at least I don't understand it in the usual sense of subcomplex. Consider $S^1$ represented as a CW complex with one 0-cell (at $(1, 0)$) and a single $1$-cell. The set $C = { (x, y) in S^1 mid x le 0 }$ is certainly a closed set, but not a subcomplex, in the sense that there's no subset of the cells of $S^1$ whose union is $C$.
– John Hughes
Nov 21 at 12:19
@JohnHughes You are right. The theorem states: If $(X,mathcal E)$ is a CW-complex, $mathcal E'subseteqmathcal E$ and $X'=cupmathcal E'$ then $(X',mathcal E')$ is a subcomplex iff $X'$ is closed. I overlooked the extra condition on $X'$. Here $mathcal E$ denotes the collection of cells.
– drhab
Nov 21 at 12:32
@drhab Ah...thanks. I was thinking that you might be thinking of some theorem like "there's a subdivision of the original CW complex such that $C$ is a subcomplex," but that seemed as if it might be very hard to prove. :) Indeed, I'm not even sure how I'd define "subdivision". :)
– John Hughes
Nov 21 at 13:00
I don't know at first hand whether in a CW-complex path-components coincide with components. What I do know is that components are always closed and that any closed set of a CW-complex is a sub-complex.
– drhab
Nov 21 at 12:14
I don't know at first hand whether in a CW-complex path-components coincide with components. What I do know is that components are always closed and that any closed set of a CW-complex is a sub-complex.
– drhab
Nov 21 at 12:14
@drhab: This seems false to me, or at least I don't understand it in the usual sense of subcomplex. Consider $S^1$ represented as a CW complex with one 0-cell (at $(1, 0)$) and a single $1$-cell. The set $C = { (x, y) in S^1 mid x le 0 }$ is certainly a closed set, but not a subcomplex, in the sense that there's no subset of the cells of $S^1$ whose union is $C$.
– John Hughes
Nov 21 at 12:19
@drhab: This seems false to me, or at least I don't understand it in the usual sense of subcomplex. Consider $S^1$ represented as a CW complex with one 0-cell (at $(1, 0)$) and a single $1$-cell. The set $C = { (x, y) in S^1 mid x le 0 }$ is certainly a closed set, but not a subcomplex, in the sense that there's no subset of the cells of $S^1$ whose union is $C$.
– John Hughes
Nov 21 at 12:19
@JohnHughes You are right. The theorem states: If $(X,mathcal E)$ is a CW-complex, $mathcal E'subseteqmathcal E$ and $X'=cupmathcal E'$ then $(X',mathcal E')$ is a subcomplex iff $X'$ is closed. I overlooked the extra condition on $X'$. Here $mathcal E$ denotes the collection of cells.
– drhab
Nov 21 at 12:32
@JohnHughes You are right. The theorem states: If $(X,mathcal E)$ is a CW-complex, $mathcal E'subseteqmathcal E$ and $X'=cupmathcal E'$ then $(X',mathcal E')$ is a subcomplex iff $X'$ is closed. I overlooked the extra condition on $X'$. Here $mathcal E$ denotes the collection of cells.
– drhab
Nov 21 at 12:32
@drhab Ah...thanks. I was thinking that you might be thinking of some theorem like "there's a subdivision of the original CW complex such that $C$ is a subcomplex," but that seemed as if it might be very hard to prove. :) Indeed, I'm not even sure how I'd define "subdivision". :)
– John Hughes
Nov 21 at 13:00
@drhab Ah...thanks. I was thinking that you might be thinking of some theorem like "there's a subdivision of the original CW complex such that $C$ is a subcomplex," but that seemed as if it might be very hard to prove. :) Indeed, I'm not even sure how I'd define "subdivision". :)
– John Hughes
Nov 21 at 13:00
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2 Answers
2
active
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up vote
1
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Let C be your pathcomponent, and take any cell $e_nto C$, which is, say, of dimension $n$. Then the boundary of $e_n$ maps to a cell of dimension ___?
answered Nov 21 at 12:06
Fumera
235
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up vote
1
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A subcomplex of a CW-complex $X$ is a subspace $A subset X$ which is a union of closed cells of $X$. Recall that a closed $n$-cell $e^n_alpha$ is the image of an attaching map $phi^n_alpha : D^n to X^{n-1}$ defined on the closed unit ball in $mathbb{R}^n$. In particular, each $e^n_alpha$ is pathwise connected. Note that if $e^n_alpha subset A$, then also all $e^m_beta$ such that $mathring{e}^m_beta cap e^n_alpha ne emptyset$ must be contained in $A$. Here $mathring{e}^m_beta$ denotes the open cell associated to $e^m_beta$, i.e. the set $phi^m_beta(mathring{D}^m) subset e^m_beta$. $mathring{D}^m$ is the open unit ball in $mathbb{R}^n$.
Each path component $P$ of $X$ is a union of closed cells: If $e^n_alpha cap P ne emptyset$, then $e^n_alpha cup P$ is pathwise connected so that $e^n_alpha subset e^n_alpha cup P subset P$.
answered Nov 21 at 14:40
Paul Frost
8,1041528
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let C be your pathcomponent, and take any cell $e_nto C$, which is, say, of dimension $n$. Then the boundary of $e_n$ maps to a cell of dimension ___?
answered Nov 21 at 12:06
Fumera
235
add a comment |
up vote
1
down vote
Let C be your pathcomponent, and take any cell $e_nto C$, which is, say, of dimension $n$. Then the boundary of $e_n$ maps to a cell of dimension ___?
answered Nov 21 at 12:06
Fumera
235
add a comment |
up vote
1
down vote
up vote
1
down vote
Let C be your pathcomponent, and take any cell $e_nto C$, which is, say, of dimension $n$. Then the boundary of $e_n$ maps to a cell of dimension ___?
answered Nov 21 at 12:06
Fumera
235
Let C be your pathcomponent, and take any cell $e_nto C$, which is, say, of dimension $n$. Then the boundary of $e_n$ maps to a cell of dimension ___?
answered Nov 21 at 12:06
Fumera
235
edited Nov 21 at 12:22
answered Nov 21 at 12:06
Fumera
235
answered Nov 21 at 12:06
Fumera
235
answered Nov 21 at 12:06
Fumera
235
235
add a comment |
add a comment |
up vote
1
down vote
A subcomplex of a CW-complex $X$ is a subspace $A subset X$ which is a union of closed cells of $X$. Recall that a closed $n$-cell $e^n_alpha$ is the image of an attaching map $phi^n_alpha : D^n to X^{n-1}$ defined on the closed unit ball in $mathbb{R}^n$. In particular, each $e^n_alpha$ is pathwise connected. Note that if $e^n_alpha subset A$, then also all $e^m_beta$ such that $mathring{e}^m_beta cap e^n_alpha ne emptyset$ must be contained in $A$. Here $mathring{e}^m_beta$ denotes the open cell associated to $e^m_beta$, i.e. the set $phi^m_beta(mathring{D}^m) subset e^m_beta$. $mathring{D}^m$ is the open unit ball in $mathbb{R}^n$.
Each path component $P$ of $X$ is a union of closed cells: If $e^n_alpha cap P ne emptyset$, then $e^n_alpha cup P$ is pathwise connected so that $e^n_alpha subset e^n_alpha cup P subset P$.
answered Nov 21 at 14:40
Paul Frost
8,1041528
add a comment |
up vote
1
down vote
A subcomplex of a CW-complex $X$ is a subspace $A subset X$ which is a union of closed cells of $X$. Recall that a closed $n$-cell $e^n_alpha$ is the image of an attaching map $phi^n_alpha : D^n to X^{n-1}$ defined on the closed unit ball in $mathbb{R}^n$. In particular, each $e^n_alpha$ is pathwise connected. Note that if $e^n_alpha subset A$, then also all $e^m_beta$ such that $mathring{e}^m_beta cap e^n_alpha ne emptyset$ must be contained in $A$. Here $mathring{e}^m_beta$ denotes the open cell associated to $e^m_beta$, i.e. the set $phi^m_beta(mathring{D}^m) subset e^m_beta$. $mathring{D}^m$ is the open unit ball in $mathbb{R}^n$.
Each path component $P$ of $X$ is a union of closed cells: If $e^n_alpha cap P ne emptyset$, then $e^n_alpha cup P$ is pathwise connected so that $e^n_alpha subset e^n_alpha cup P subset P$.
answered Nov 21 at 14:40
Paul Frost
8,1041528
add a comment |
up vote
1
down vote
up vote
1
down vote
A subcomplex of a CW-complex $X$ is a subspace $A subset X$ which is a union of closed cells of $X$. Recall that a closed $n$-cell $e^n_alpha$ is the image of an attaching map $phi^n_alpha : D^n to X^{n-1}$ defined on the closed unit ball in $mathbb{R}^n$. In particular, each $e^n_alpha$ is pathwise connected. Note that if $e^n_alpha subset A$, then also all $e^m_beta$ such that $mathring{e}^m_beta cap e^n_alpha ne emptyset$ must be contained in $A$. Here $mathring{e}^m_beta$ denotes the open cell associated to $e^m_beta$, i.e. the set $phi^m_beta(mathring{D}^m) subset e^m_beta$. $mathring{D}^m$ is the open unit ball in $mathbb{R}^n$.
Each path component $P$ of $X$ is a union of closed cells: If $e^n_alpha cap P ne emptyset$, then $e^n_alpha cup P$ is pathwise connected so that $e^n_alpha subset e^n_alpha cup P subset P$.
answered Nov 21 at 14:40
Paul Frost
8,1041528
A subcomplex of a CW-complex $X$ is a subspace $A subset X$ which is a union of closed cells of $X$. Recall that a closed $n$-cell $e^n_alpha$ is the image of an attaching map $phi^n_alpha : D^n to X^{n-1}$ defined on the closed unit ball in $mathbb{R}^n$. In particular, each $e^n_alpha$ is pathwise connected. Note that if $e^n_alpha subset A$, then also all $e^m_beta$ such that $mathring{e}^m_beta cap e^n_alpha ne emptyset$ must be contained in $A$. Here $mathring{e}^m_beta$ denotes the open cell associated to $e^m_beta$, i.e. the set $phi^m_beta(mathring{D}^m) subset e^m_beta$. $mathring{D}^m$ is the open unit ball in $mathbb{R}^n$.
Each path component $P$ of $X$ is a union of closed cells: If $e^n_alpha cap P ne emptyset$, then $e^n_alpha cup P$ is pathwise connected so that $e^n_alpha subset e^n_alpha cup P subset P$.
answered Nov 21 at 14:40
Paul Frost
8,1041528
answered Nov 21 at 14:40
Paul Frost
8,1041528
answered Nov 21 at 14:40
Paul Frost
8,1041528
answered Nov 21 at 14:40
Paul Frost
8,1041528
8,1041528
add a comment |
add a comment |
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I don't know at first hand whether in a CW-complex path-components coincide with components. What I do know is that components are always closed and that any closed set of a CW-complex is a sub-complex.
– drhab
Nov 21 at 12:14
@drhab: This seems false to me, or at least I don't understand it in the usual sense of subcomplex. Consider $S^1$ represented as a CW complex with one 0-cell (at $(1, 0)$) and a single $1$-cell. The set $C = { (x, y) in S^1 mid x le 0 }$ is certainly a closed set, but not a subcomplex, in the sense that there's no subset of the cells of $S^1$ whose union is $C$.
– John Hughes
Nov 21 at 12:19
@JohnHughes You are right. The theorem states: If $(X,mathcal E)$ is a CW-complex, $mathcal E'subseteqmathcal E$ and $X'=cupmathcal E'$ then $(X',mathcal E')$ is a subcomplex iff $X'$ is closed. I overlooked the extra condition on $X'$. Here $mathcal E$ denotes the collection of cells.
– drhab
Nov 21 at 12:32
@drhab Ah...thanks. I was thinking that you might be thinking of some theorem like "there's a subdivision of the original CW complex such that $C$ is a subcomplex," but that seemed as if it might be very hard to prove. :) Indeed, I'm not even sure how I'd define "subdivision". :)
– John Hughes
Nov 21 at 13:00