Mixing
Fundamental group of the $m$-fold suspension of a finite discrete space
$begingroup$
Recall that the suspension of a topological space $X$ is the space $SX$ resulting by identifying $Xtimes{0}$ and $Xtimes{1}$ to single points of the "cylinder" $Xtimes[0,1]$. Now let $X_m$ be a discrete space consisting of $m$ points, and denote by $S^nX$ the $n$-fold suspension, i.e., $S^nX=SS^{n-1}X$.
What is the fundamental group of $S^nX_m$ for $n,mgeq1$?
Here is my attempt. The case $m=2$ is easy since $SX_2$ is a circle, so the fundamental group is $mathbb Z$, and $S^2X_2$ is the sphere, whose fundamental group is trivial. Further, the suspension of the $n$-sphere is the $(n+1)$-sphere so $pi_1(S^nX_2)$ is trivial for $ngeq2$.
Now consider $m>2$. Then $SX_m$ is homotopy equivalent to an "$(m-1)$-fold eight"
$$underbrace{bigcirchspace{-.1cm}!bigcirchspace{-.1cm}!bigcirccdotsbigcirc}_{text{$m-1$ circles}}$$
so the fundamental group is $F_{m-1}$, the free group on $m-1$ generators. If I am not mistaken two homotopy equivalent spaces have homotopy equivalent suspensions, so $S^2X_m$ is homotopy equivalent to the wedge product of $m-1$ spheres, and consequently $pi_1(S^2X_m)$ is trivial. I am having trouble, however, to visualise the space $S^nX_m$ for $n>2$. How can $pi_1(S^nX_m)$ be computed?
algebraic-topology fundamental-groups
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
$endgroup$
add a comment |
$begingroup$
Recall that the suspension of a topological space $X$ is the space $SX$ resulting by identifying $Xtimes{0}$ and $Xtimes{1}$ to single points of the "cylinder" $Xtimes[0,1]$. Now let $X_m$ be a discrete space consisting of $m$ points, and denote by $S^nX$ the $n$-fold suspension, i.e., $S^nX=SS^{n-1}X$.
What is the fundamental group of $S^nX_m$ for $n,mgeq1$?
Here is my attempt. The case $m=2$ is easy since $SX_2$ is a circle, so the fundamental group is $mathbb Z$, and $S^2X_2$ is the sphere, whose fundamental group is trivial. Further, the suspension of the $n$-sphere is the $(n+1)$-sphere so $pi_1(S^nX_2)$ is trivial for $ngeq2$.
Now consider $m>2$. Then $SX_m$ is homotopy equivalent to an "$(m-1)$-fold eight"
$$underbrace{bigcirchspace{-.1cm}!bigcirchspace{-.1cm}!bigcirccdotsbigcirc}_{text{$m-1$ circles}}$$
so the fundamental group is $F_{m-1}$, the free group on $m-1$ generators. If I am not mistaken two homotopy equivalent spaces have homotopy equivalent suspensions, so $S^2X_m$ is homotopy equivalent to the wedge product of $m-1$ spheres, and consequently $pi_1(S^2X_m)$ is trivial. I am having trouble, however, to visualise the space $S^nX_m$ for $n>2$. How can $pi_1(S^nX_m)$ be computed?
algebraic-topology fundamental-groups
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
$endgroup$
$begingroup$
I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument)
$endgroup$
– Max
Jan 11 at 13:44
$begingroup$
Ok, but thanks anyway!
$endgroup$
– iqcd
Jan 11 at 14:17
add a comment |
$begingroup$
Recall that the suspension of a topological space $X$ is the space $SX$ resulting by identifying $Xtimes{0}$ and $Xtimes{1}$ to single points of the "cylinder" $Xtimes[0,1]$. Now let $X_m$ be a discrete space consisting of $m$ points, and denote by $S^nX$ the $n$-fold suspension, i.e., $S^nX=SS^{n-1}X$.
What is the fundamental group of $S^nX_m$ for $n,mgeq1$?
Here is my attempt. The case $m=2$ is easy since $SX_2$ is a circle, so the fundamental group is $mathbb Z$, and $S^2X_2$ is the sphere, whose fundamental group is trivial. Further, the suspension of the $n$-sphere is the $(n+1)$-sphere so $pi_1(S^nX_2)$ is trivial for $ngeq2$.
Now consider $m>2$. Then $SX_m$ is homotopy equivalent to an "$(m-1)$-fold eight"
$$underbrace{bigcirchspace{-.1cm}!bigcirchspace{-.1cm}!bigcirccdotsbigcirc}_{text{$m-1$ circles}}$$
so the fundamental group is $F_{m-1}$, the free group on $m-1$ generators. If I am not mistaken two homotopy equivalent spaces have homotopy equivalent suspensions, so $S^2X_m$ is homotopy equivalent to the wedge product of $m-1$ spheres, and consequently $pi_1(S^2X_m)$ is trivial. I am having trouble, however, to visualise the space $S^nX_m$ for $n>2$. How can $pi_1(S^nX_m)$ be computed?
algebraic-topology fundamental-groups
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
$endgroup$
Recall that the suspension of a topological space $X$ is the space $SX$ resulting by identifying $Xtimes{0}$ and $Xtimes{1}$ to single points of the "cylinder" $Xtimes[0,1]$. Now let $X_m$ be a discrete space consisting of $m$ points, and denote by $S^nX$ the $n$-fold suspension, i.e., $S^nX=SS^{n-1}X$.
What is the fundamental group of $S^nX_m$ for $n,mgeq1$?
Here is my attempt. The case $m=2$ is easy since $SX_2$ is a circle, so the fundamental group is $mathbb Z$, and $S^2X_2$ is the sphere, whose fundamental group is trivial. Further, the suspension of the $n$-sphere is the $(n+1)$-sphere so $pi_1(S^nX_2)$ is trivial for $ngeq2$.
Now consider $m>2$. Then $SX_m$ is homotopy equivalent to an "$(m-1)$-fold eight"
$$underbrace{bigcirchspace{-.1cm}!bigcirchspace{-.1cm}!bigcirccdotsbigcirc}_{text{$m-1$ circles}}$$
so the fundamental group is $F_{m-1}$, the free group on $m-1$ generators. If I am not mistaken two homotopy equivalent spaces have homotopy equivalent suspensions, so $S^2X_m$ is homotopy equivalent to the wedge product of $m-1$ spheres, and consequently $pi_1(S^2X_m)$ is trivial. I am having trouble, however, to visualise the space $S^nX_m$ for $n>2$. How can $pi_1(S^nX_m)$ be computed?
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
edited Jan 11 at 12:57
iqcd
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
asked Jan 11 at 11:28
iqcdiqcd
2,5222926
2,5222926
$begingroup$
I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument)
$endgroup$
– Max
Jan 11 at 13:44
$begingroup$
Ok, but thanks anyway!
$endgroup$
– iqcd
Jan 11 at 14:17
add a comment |
$begingroup$
I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument)
$endgroup$
– Max
Jan 11 at 13:44
$begingroup$
Ok, but thanks anyway!
$endgroup$
– iqcd
Jan 11 at 14:17
$begingroup$
I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument)
$endgroup$
– Max
Jan 11 at 13:44
$begingroup$
I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument)
$endgroup$
– Max
Jan 11 at 13:44
$begingroup$
Ok, but thanks anyway!
$endgroup$
– iqcd
Jan 11 at 14:17
$begingroup$
Ok, but thanks anyway!
$endgroup$
– iqcd
Jan 11 at 14:17
add a comment |
1 Answer
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$begingroup$
$m = 1$: All suspensions of $X_1$ are contractible.
$m > 1$:
2.1. $n = 1$: As you stated correctly, $pi_1(SX_m) = F_{m-1}$.
2.2. $n > 1$: Let us prove the following theorem.
If $Y$ is path connected, then $SY$ is simply connected.
This applies to $Y = S^{n-1}X_m$.
Let $p : Y times I to SY$ denote the quotient map. Since $Y$ is path connected, so are $Y times I$ and $SY$. As a basepoint for $SY$ choose $z_0 = p(y_0,1/2)$, where $y_0 in Y$ is an arbitrary point. The sets $U = p(Y times (0,1])$ and $V = p(Y times [0,1))$ are open in $SY$, and we have $z_0 in U cap V = p(Y times (0,1)$ which is pathwise connected. Since $U$ and $V$ are contractible, we get $pi_1(U,z_0) = pi_1(V,z_0) = 0$.
Now apply the Seifert-van Kampen-theorem to see that $pi_1(SY,z_0) = 0$.
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
$endgroup$
$begingroup$
That's a very useful result, thanks!
$endgroup$
– iqcd
Jan 11 at 14:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$m = 1$: All suspensions of $X_1$ are contractible.
$m > 1$:
2.1. $n = 1$: As you stated correctly, $pi_1(SX_m) = F_{m-1}$.
2.2. $n > 1$: Let us prove the following theorem.
If $Y$ is path connected, then $SY$ is simply connected.
This applies to $Y = S^{n-1}X_m$.
Let $p : Y times I to SY$ denote the quotient map. Since $Y$ is path connected, so are $Y times I$ and $SY$. As a basepoint for $SY$ choose $z_0 = p(y_0,1/2)$, where $y_0 in Y$ is an arbitrary point. The sets $U = p(Y times (0,1])$ and $V = p(Y times [0,1))$ are open in $SY$, and we have $z_0 in U cap V = p(Y times (0,1)$ which is pathwise connected. Since $U$ and $V$ are contractible, we get $pi_1(U,z_0) = pi_1(V,z_0) = 0$.
Now apply the Seifert-van Kampen-theorem to see that $pi_1(SY,z_0) = 0$.
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
$endgroup$
$begingroup$
That's a very useful result, thanks!
$endgroup$
– iqcd
Jan 11 at 14:29
add a comment |
$begingroup$
$m = 1$: All suspensions of $X_1$ are contractible.
$m > 1$:
2.1. $n = 1$: As you stated correctly, $pi_1(SX_m) = F_{m-1}$.
2.2. $n > 1$: Let us prove the following theorem.
If $Y$ is path connected, then $SY$ is simply connected.
This applies to $Y = S^{n-1}X_m$.
Let $p : Y times I to SY$ denote the quotient map. Since $Y$ is path connected, so are $Y times I$ and $SY$. As a basepoint for $SY$ choose $z_0 = p(y_0,1/2)$, where $y_0 in Y$ is an arbitrary point. The sets $U = p(Y times (0,1])$ and $V = p(Y times [0,1))$ are open in $SY$, and we have $z_0 in U cap V = p(Y times (0,1)$ which is pathwise connected. Since $U$ and $V$ are contractible, we get $pi_1(U,z_0) = pi_1(V,z_0) = 0$.
Now apply the Seifert-van Kampen-theorem to see that $pi_1(SY,z_0) = 0$.
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
$endgroup$
$begingroup$
That's a very useful result, thanks!
$endgroup$
– iqcd
Jan 11 at 14:29
add a comment |
$begingroup$
$m = 1$: All suspensions of $X_1$ are contractible.
$m > 1$:
2.1. $n = 1$: As you stated correctly, $pi_1(SX_m) = F_{m-1}$.
2.2. $n > 1$: Let us prove the following theorem.
If $Y$ is path connected, then $SY$ is simply connected.
This applies to $Y = S^{n-1}X_m$.
Let $p : Y times I to SY$ denote the quotient map. Since $Y$ is path connected, so are $Y times I$ and $SY$. As a basepoint for $SY$ choose $z_0 = p(y_0,1/2)$, where $y_0 in Y$ is an arbitrary point. The sets $U = p(Y times (0,1])$ and $V = p(Y times [0,1))$ are open in $SY$, and we have $z_0 in U cap V = p(Y times (0,1)$ which is pathwise connected. Since $U$ and $V$ are contractible, we get $pi_1(U,z_0) = pi_1(V,z_0) = 0$.
Now apply the Seifert-van Kampen-theorem to see that $pi_1(SY,z_0) = 0$.
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
$endgroup$
$m = 1$: All suspensions of $X_1$ are contractible.
$m > 1$:
2.1. $n = 1$: As you stated correctly, $pi_1(SX_m) = F_{m-1}$.
2.2. $n > 1$: Let us prove the following theorem.
If $Y$ is path connected, then $SY$ is simply connected.
This applies to $Y = S^{n-1}X_m$.
Let $p : Y times I to SY$ denote the quotient map. Since $Y$ is path connected, so are $Y times I$ and $SY$. As a basepoint for $SY$ choose $z_0 = p(y_0,1/2)$, where $y_0 in Y$ is an arbitrary point. The sets $U = p(Y times (0,1])$ and $V = p(Y times [0,1))$ are open in $SY$, and we have $z_0 in U cap V = p(Y times (0,1)$ which is pathwise connected. Since $U$ and $V$ are contractible, we get $pi_1(U,z_0) = pi_1(V,z_0) = 0$.
Now apply the Seifert-van Kampen-theorem to see that $pi_1(SY,z_0) = 0$.
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
answered Jan 11 at 14:19
Paul FrostPaul Frost
10.4k3933
10.4k3933
$begingroup$
That's a very useful result, thanks!
$endgroup$
– iqcd
Jan 11 at 14:29
add a comment |
$begingroup$
That's a very useful result, thanks!
$endgroup$
– iqcd
Jan 11 at 14:29
$begingroup$
That's a very useful result, thanks!
$endgroup$
– iqcd
Jan 11 at 14:29
$begingroup$
That's a very useful result, thanks!
$endgroup$
– iqcd
Jan 11 at 14:29
add a comment |
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$begingroup$
I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument)
$endgroup$
– Max
Jan 11 at 13:44
$begingroup$
Ok, but thanks anyway!
$endgroup$
– iqcd
Jan 11 at 14:17