Connectedness of the join of two spaces
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Let $X$ be an $n$-connected space and $Y$ be an $m$-connected space. How can I prove that the join $X*Y$ is $(n+m+1)$-connected?
I thought that homotopy excision would do the trick, but it does not seem so.
algebraic-topology
asked Nov 14 at 19:14
user09127
474
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show 2 more comments
up vote
0
down vote
favorite
Let $X$ be an $n$-connected space and $Y$ be an $m$-connected space. How can I prove that the join $X*Y$ is $(n+m+1)$-connected?
I thought that homotopy excision would do the trick, but it does not seem so.
algebraic-topology
asked Nov 14 at 19:14
user09127
474
If $X$, $Y$ are CW, then there is an obvious CW structure on $Xast Y$, and oce you have determined this you can use cellular methods to decide why the connectivity statement is true.
– Tyrone
Nov 15 at 11:26
I'm afraid I don't quite get your point. Wouldn't the CW structure on the join contain $X$ and $Y$ as subcomplexes? What results are you refering to?
– user09127
Nov 20 at 13:22
The join of two CW complexes $X,Y$ is a quotient of $Xtimes Itimes Y$ by a certain relation. Take the product CW structure on $Xtimes Itimes Y$ and then figure out which cells you need to quotient out. You can also use the fact that $Xast Y$ is the pushout of the inclusions $Xtimes CYleftarrow Xtimes Yrightarrow CXtimes Y$ to get a CW structure. The end result is that the cells of $Xast Y$ are the joins of the cells of $X$ and $Y$. I'll leave you to figure out what the joins $D^nast D^m$ and $S^{n-1}ast S^{m-1}$ are.
– Tyrone
Nov 20 at 13:47
If $X$ is $n$ connected, and $Y$ is $m$ connected, then you'll see that the first cell of $Xast Y$ above dimension $0$ that you need to worry about is $e^nast e^m$, so you can figure out the connectivity of $Xast Y$ from, say, cellular homology, depending on what you are confident with.
– Tyrone
Nov 20 at 13:49
To move to the general case use CW approximation and the functorality of the join construction.
– Tyrone
Nov 20 at 13:50
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be an $n$-connected space and $Y$ be an $m$-connected space. How can I prove that the join $X*Y$ is $(n+m+1)$-connected?
I thought that homotopy excision would do the trick, but it does not seem so.
algebraic-topology
asked Nov 14 at 19:14
user09127
474
Let $X$ be an $n$-connected space and $Y$ be an $m$-connected space. How can I prove that the join $X*Y$ is $(n+m+1)$-connected?
I thought that homotopy excision would do the trick, but it does not seem so.
algebraic-topology
algebraic-topology
asked Nov 14 at 19:14
user09127
474
asked Nov 14 at 19:14
user09127
474
asked Nov 14 at 19:14
user09127
474
asked Nov 14 at 19:14
user09127
474
asked Nov 14 at 19:14
user09127
474
474
If $X$, $Y$ are CW, then there is an obvious CW structure on $Xast Y$, and oce you have determined this you can use cellular methods to decide why the connectivity statement is true.
– Tyrone
Nov 15 at 11:26
I'm afraid I don't quite get your point. Wouldn't the CW structure on the join contain $X$ and $Y$ as subcomplexes? What results are you refering to?
– user09127
Nov 20 at 13:22
The join of two CW complexes $X,Y$ is a quotient of $Xtimes Itimes Y$ by a certain relation. Take the product CW structure on $Xtimes Itimes Y$ and then figure out which cells you need to quotient out. You can also use the fact that $Xast Y$ is the pushout of the inclusions $Xtimes CYleftarrow Xtimes Yrightarrow CXtimes Y$ to get a CW structure. The end result is that the cells of $Xast Y$ are the joins of the cells of $X$ and $Y$. I'll leave you to figure out what the joins $D^nast D^m$ and $S^{n-1}ast S^{m-1}$ are.
– Tyrone
Nov 20 at 13:47
If $X$ is $n$ connected, and $Y$ is $m$ connected, then you'll see that the first cell of $Xast Y$ above dimension $0$ that you need to worry about is $e^nast e^m$, so you can figure out the connectivity of $Xast Y$ from, say, cellular homology, depending on what you are confident with.
– Tyrone
Nov 20 at 13:49
To move to the general case use CW approximation and the functorality of the join construction.
– Tyrone
Nov 20 at 13:50
|
show 2 more comments
If $X$, $Y$ are CW, then there is an obvious CW structure on $Xast Y$, and oce you have determined this you can use cellular methods to decide why the connectivity statement is true.
– Tyrone
Nov 15 at 11:26
I'm afraid I don't quite get your point. Wouldn't the CW structure on the join contain $X$ and $Y$ as subcomplexes? What results are you refering to?
– user09127
Nov 20 at 13:22
The join of two CW complexes $X,Y$ is a quotient of $Xtimes Itimes Y$ by a certain relation. Take the product CW structure on $Xtimes Itimes Y$ and then figure out which cells you need to quotient out. You can also use the fact that $Xast Y$ is the pushout of the inclusions $Xtimes CYleftarrow Xtimes Yrightarrow CXtimes Y$ to get a CW structure. The end result is that the cells of $Xast Y$ are the joins of the cells of $X$ and $Y$. I'll leave you to figure out what the joins $D^nast D^m$ and $S^{n-1}ast S^{m-1}$ are.
– Tyrone
Nov 20 at 13:47
If $X$ is $n$ connected, and $Y$ is $m$ connected, then you'll see that the first cell of $Xast Y$ above dimension $0$ that you need to worry about is $e^nast e^m$, so you can figure out the connectivity of $Xast Y$ from, say, cellular homology, depending on what you are confident with.
– Tyrone
Nov 20 at 13:49
To move to the general case use CW approximation and the functorality of the join construction.
– Tyrone
Nov 20 at 13:50
If $X$, $Y$ are CW, then there is an obvious CW structure on $Xast Y$, and oce you have determined this you can use cellular methods to decide why the connectivity statement is true.
– Tyrone
Nov 15 at 11:26
If $X$, $Y$ are CW, then there is an obvious CW structure on $Xast Y$, and oce you have determined this you can use cellular methods to decide why the connectivity statement is true.
– Tyrone
Nov 15 at 11:26
I'm afraid I don't quite get your point. Wouldn't the CW structure on the join contain $X$ and $Y$ as subcomplexes? What results are you refering to?
– user09127
Nov 20 at 13:22
I'm afraid I don't quite get your point. Wouldn't the CW structure on the join contain $X$ and $Y$ as subcomplexes? What results are you refering to?
– user09127
Nov 20 at 13:22
The join of two CW complexes $X,Y$ is a quotient of $Xtimes Itimes Y$ by a certain relation. Take the product CW structure on $Xtimes Itimes Y$ and then figure out which cells you need to quotient out. You can also use the fact that $Xast Y$ is the pushout of the inclusions $Xtimes CYleftarrow Xtimes Yrightarrow CXtimes Y$ to get a CW structure. The end result is that the cells of $Xast Y$ are the joins of the cells of $X$ and $Y$. I'll leave you to figure out what the joins $D^nast D^m$ and $S^{n-1}ast S^{m-1}$ are.
– Tyrone
Nov 20 at 13:47
The join of two CW complexes $X,Y$ is a quotient of $Xtimes Itimes Y$ by a certain relation. Take the product CW structure on $Xtimes Itimes Y$ and then figure out which cells you need to quotient out. You can also use the fact that $Xast Y$ is the pushout of the inclusions $Xtimes CYleftarrow Xtimes Yrightarrow CXtimes Y$ to get a CW structure. The end result is that the cells of $Xast Y$ are the joins of the cells of $X$ and $Y$. I'll leave you to figure out what the joins $D^nast D^m$ and $S^{n-1}ast S^{m-1}$ are.
– Tyrone
Nov 20 at 13:47
If $X$ is $n$ connected, and $Y$ is $m$ connected, then you'll see that the first cell of $Xast Y$ above dimension $0$ that you need to worry about is $e^nast e^m$, so you can figure out the connectivity of $Xast Y$ from, say, cellular homology, depending on what you are confident with.
– Tyrone
Nov 20 at 13:49
If $X$ is $n$ connected, and $Y$ is $m$ connected, then you'll see that the first cell of $Xast Y$ above dimension $0$ that you need to worry about is $e^nast e^m$, so you can figure out the connectivity of $Xast Y$ from, say, cellular homology, depending on what you are confident with.
– Tyrone
Nov 20 at 13:49
To move to the general case use CW approximation and the functorality of the join construction.
– Tyrone
Nov 20 at 13:50
To move to the general case use CW approximation and the functorality of the join construction.
– Tyrone
Nov 20 at 13:50
|
show 2 more comments
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If $X$, $Y$ are CW, then there is an obvious CW structure on $Xast Y$, and oce you have determined this you can use cellular methods to decide why the connectivity statement is true.
– Tyrone
Nov 15 at 11:26
I'm afraid I don't quite get your point. Wouldn't the CW structure on the join contain $X$ and $Y$ as subcomplexes? What results are you refering to?
– user09127
Nov 20 at 13:22
The join of two CW complexes $X,Y$ is a quotient of $Xtimes Itimes Y$ by a certain relation. Take the product CW structure on $Xtimes Itimes Y$ and then figure out which cells you need to quotient out. You can also use the fact that $Xast Y$ is the pushout of the inclusions $Xtimes CYleftarrow Xtimes Yrightarrow CXtimes Y$ to get a CW structure. The end result is that the cells of $Xast Y$ are the joins of the cells of $X$ and $Y$. I'll leave you to figure out what the joins $D^nast D^m$ and $S^{n-1}ast S^{m-1}$ are.
– Tyrone
Nov 20 at 13:47
If $X$ is $n$ connected, and $Y$ is $m$ connected, then you'll see that the first cell of $Xast Y$ above dimension $0$ that you need to worry about is $e^nast e^m$, so you can figure out the connectivity of $Xast Y$ from, say, cellular homology, depending on what you are confident with.
– Tyrone
Nov 20 at 13:49
To move to the general case use CW approximation and the functorality of the join construction.
– Tyrone
Nov 20 at 13:50