Can a cell-complex have no zero cell?
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My question is very simple, but I wasn't able to find an answer in various sources. Cell-complexes are commonly presented using an inductive construction where $n$-cells are attached to $(n-1)$-cells, starting with the data of a collection $X^0$ of $0$-cells.
But sometimes, one will define a $2$-cell for instance, as being a sub-cell complex of something, raising the question: can a cell complex $0$-skeleton be empty? And more generally, can a cell-complex have empty skeletons until a dimension $k$?
It doesn't seem absurd to me that the answer should be yes, but it worries me to never see the case taken into account in the inductive construction.
algebraic-topology cw-complexes
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
$endgroup$
add a comment |
$begingroup$
My question is very simple, but I wasn't able to find an answer in various sources. Cell-complexes are commonly presented using an inductive construction where $n$-cells are attached to $(n-1)$-cells, starting with the data of a collection $X^0$ of $0$-cells.
But sometimes, one will define a $2$-cell for instance, as being a sub-cell complex of something, raising the question: can a cell complex $0$-skeleton be empty? And more generally, can a cell-complex have empty skeletons until a dimension $k$?
It doesn't seem absurd to me that the answer should be yes, but it worries me to never see the case taken into account in the inductive construction.
algebraic-topology cw-complexes
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
$endgroup$
$begingroup$
The empty space is a CW-complex with no $0$-cells.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 7:41
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What do you mean by "cell complex"? Are you using that as a synonym for CW-complex?
$endgroup$
– Eric Wofsey
Jan 2 at 7:44
$begingroup$
No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link )
$endgroup$
– TryingToGetOut
Jan 2 at 7:49
add a comment |
$begingroup$
My question is very simple, but I wasn't able to find an answer in various sources. Cell-complexes are commonly presented using an inductive construction where $n$-cells are attached to $(n-1)$-cells, starting with the data of a collection $X^0$ of $0$-cells.
But sometimes, one will define a $2$-cell for instance, as being a sub-cell complex of something, raising the question: can a cell complex $0$-skeleton be empty? And more generally, can a cell-complex have empty skeletons until a dimension $k$?
It doesn't seem absurd to me that the answer should be yes, but it worries me to never see the case taken into account in the inductive construction.
algebraic-topology cw-complexes
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
$endgroup$
My question is very simple, but I wasn't able to find an answer in various sources. Cell-complexes are commonly presented using an inductive construction where $n$-cells are attached to $(n-1)$-cells, starting with the data of a collection $X^0$ of $0$-cells.
But sometimes, one will define a $2$-cell for instance, as being a sub-cell complex of something, raising the question: can a cell complex $0$-skeleton be empty? And more generally, can a cell-complex have empty skeletons until a dimension $k$?
It doesn't seem absurd to me that the answer should be yes, but it worries me to never see the case taken into account in the inductive construction.
algebraic-topology cw-complexes
algebraic-topology cw-complexes
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
asked Jan 2 at 7:39
TryingToGetOutTryingToGetOut
488
488
$begingroup$
The empty space is a CW-complex with no $0$-cells.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 7:41
$begingroup$
What do you mean by "cell complex"? Are you using that as a synonym for CW-complex?
$endgroup$
– Eric Wofsey
Jan 2 at 7:44
$begingroup$
No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link )
$endgroup$
– TryingToGetOut
Jan 2 at 7:49
add a comment |
$begingroup$
The empty space is a CW-complex with no $0$-cells.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 7:41
$begingroup$
What do you mean by "cell complex"? Are you using that as a synonym for CW-complex?
$endgroup$
– Eric Wofsey
Jan 2 at 7:44
$begingroup$
No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link )
$endgroup$
– TryingToGetOut
Jan 2 at 7:49
$begingroup$
The empty space is a CW-complex with no $0$-cells.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 7:41
$begingroup$
The empty space is a CW-complex with no $0$-cells.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 7:41
$begingroup$
What do you mean by "cell complex"? Are you using that as a synonym for CW-complex?
$endgroup$
– Eric Wofsey
Jan 2 at 7:44
$begingroup$
What do you mean by "cell complex"? Are you using that as a synonym for CW-complex?
$endgroup$
– Eric Wofsey
Jan 2 at 7:44
$begingroup$
No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link )
$endgroup$
– TryingToGetOut
Jan 2 at 7:49
$begingroup$
No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link )
$endgroup$
– TryingToGetOut
Jan 2 at 7:49
add a comment |
1 Answer
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Suppose $X$ is a nonempty cell complex and let $n$ be minimal such that $X$ has an $n$-cell. If $n>0$, then this $n$-cell has an attaching map $S^{n-1}to X^{n-1}$ where $X^{n-1}$ is the $(n-1)$-skeleton of $X$. But by minimality of $n$, $X^{n-1}=emptyset$. Since $S^{n-1}$ is nonempty, there are no maps $S^{n-1}toemptyset$, so this is a contradiction.
So, if $X$ is any nonempty CW-complex, it must have a $0$-cell. (Of course, the empty space is a CW-complex with no cells at all!)
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
$endgroup$
$begingroup$
thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance.
$endgroup$
– TryingToGetOut
Jan 2 at 7:58
2
$begingroup$
In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut
$endgroup$
– Lord Shark the Unknown
Jan 2 at 8:01
$begingroup$
Yes of course.. Thanks!
$endgroup$
– TryingToGetOut
Jan 2 at 8:07
add a comment |
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$begingroup$
Suppose $X$ is a nonempty cell complex and let $n$ be minimal such that $X$ has an $n$-cell. If $n>0$, then this $n$-cell has an attaching map $S^{n-1}to X^{n-1}$ where $X^{n-1}$ is the $(n-1)$-skeleton of $X$. But by minimality of $n$, $X^{n-1}=emptyset$. Since $S^{n-1}$ is nonempty, there are no maps $S^{n-1}toemptyset$, so this is a contradiction.
So, if $X$ is any nonempty CW-complex, it must have a $0$-cell. (Of course, the empty space is a CW-complex with no cells at all!)
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
$endgroup$
$begingroup$
thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance.
$endgroup$
– TryingToGetOut
Jan 2 at 7:58
2
$begingroup$
In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut
$endgroup$
– Lord Shark the Unknown
Jan 2 at 8:01
$begingroup$
Yes of course.. Thanks!
$endgroup$
– TryingToGetOut
Jan 2 at 8:07
add a comment |
$begingroup$
Suppose $X$ is a nonempty cell complex and let $n$ be minimal such that $X$ has an $n$-cell. If $n>0$, then this $n$-cell has an attaching map $S^{n-1}to X^{n-1}$ where $X^{n-1}$ is the $(n-1)$-skeleton of $X$. But by minimality of $n$, $X^{n-1}=emptyset$. Since $S^{n-1}$ is nonempty, there are no maps $S^{n-1}toemptyset$, so this is a contradiction.
So, if $X$ is any nonempty CW-complex, it must have a $0$-cell. (Of course, the empty space is a CW-complex with no cells at all!)
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
$endgroup$
$begingroup$
thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance.
$endgroup$
– TryingToGetOut
Jan 2 at 7:58
2
$begingroup$
In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut
$endgroup$
– Lord Shark the Unknown
Jan 2 at 8:01
$begingroup$
Yes of course.. Thanks!
$endgroup$
– TryingToGetOut
Jan 2 at 8:07
add a comment |
$begingroup$
Suppose $X$ is a nonempty cell complex and let $n$ be minimal such that $X$ has an $n$-cell. If $n>0$, then this $n$-cell has an attaching map $S^{n-1}to X^{n-1}$ where $X^{n-1}$ is the $(n-1)$-skeleton of $X$. But by minimality of $n$, $X^{n-1}=emptyset$. Since $S^{n-1}$ is nonempty, there are no maps $S^{n-1}toemptyset$, so this is a contradiction.
So, if $X$ is any nonempty CW-complex, it must have a $0$-cell. (Of course, the empty space is a CW-complex with no cells at all!)
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
$endgroup$
Suppose $X$ is a nonempty cell complex and let $n$ be minimal such that $X$ has an $n$-cell. If $n>0$, then this $n$-cell has an attaching map $S^{n-1}to X^{n-1}$ where $X^{n-1}$ is the $(n-1)$-skeleton of $X$. But by minimality of $n$, $X^{n-1}=emptyset$. Since $S^{n-1}$ is nonempty, there are no maps $S^{n-1}toemptyset$, so this is a contradiction.
So, if $X$ is any nonempty CW-complex, it must have a $0$-cell. (Of course, the empty space is a CW-complex with no cells at all!)
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
answered Jan 2 at 7:54
Eric WofseyEric Wofsey
183k13211338
183k13211338
$begingroup$
thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance.
$endgroup$
– TryingToGetOut
Jan 2 at 7:58
2
$begingroup$
In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut
$endgroup$
– Lord Shark the Unknown
Jan 2 at 8:01
$begingroup$
Yes of course.. Thanks!
$endgroup$
– TryingToGetOut
Jan 2 at 8:07
add a comment |
$begingroup$
thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance.
$endgroup$
– TryingToGetOut
Jan 2 at 7:58
2
$begingroup$
In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut
$endgroup$
– Lord Shark the Unknown
Jan 2 at 8:01
$begingroup$
Yes of course.. Thanks!
$endgroup$
– TryingToGetOut
Jan 2 at 8:07
$begingroup$
thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance.
$endgroup$
– TryingToGetOut
Jan 2 at 7:58
$begingroup$
thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance.
$endgroup$
– TryingToGetOut
Jan 2 at 7:58
2
2
$begingroup$
In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut
$endgroup$
– Lord Shark the Unknown
Jan 2 at 8:01
$begingroup$
In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut
$endgroup$
– Lord Shark the Unknown
Jan 2 at 8:01
$begingroup$
Yes of course.. Thanks!
$endgroup$
– TryingToGetOut
Jan 2 at 8:07
$begingroup$
Yes of course.. Thanks!
$endgroup$
– TryingToGetOut
Jan 2 at 8:07
add a comment |
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$begingroup$
The empty space is a CW-complex with no $0$-cells.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 7:41
$begingroup$
What do you mean by "cell complex"? Are you using that as a synonym for CW-complex?
$endgroup$
– Eric Wofsey
Jan 2 at 7:44
$begingroup$
No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link )
$endgroup$
– TryingToGetOut
Jan 2 at 7:49