Associativty of smash product on compact spaces
$begingroup$
Let $Y,Z$ be compact or $X,Z$ locally compact. Then the canonical bijection
$$ (X wedge Y) wedge Z rightarrow X wedge (Y wedge Z) $$
is a homeomoprhism.
I can prove the case when $X,Z$ are locally compact using exponential law. But I don't see how one approaches the case when $Y,Z$ compact. Hints?
general-topology algebraic-topology homotopy-theory
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
$endgroup$
add a comment |
$begingroup$
Let $Y,Z$ be compact or $X,Z$ locally compact. Then the canonical bijection
$$ (X wedge Y) wedge Z rightarrow X wedge (Y wedge Z) $$
is a homeomoprhism.
I can prove the case when $X,Z$ are locally compact using exponential law. But I don't see how one approaches the case when $Y,Z$ compact. Hints?
general-topology algebraic-topology homotopy-theory
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
$endgroup$
add a comment |
$begingroup$
Let $Y,Z$ be compact or $X,Z$ locally compact. Then the canonical bijection
$$ (X wedge Y) wedge Z rightarrow X wedge (Y wedge Z) $$
is a homeomoprhism.
I can prove the case when $X,Z$ are locally compact using exponential law. But I don't see how one approaches the case when $Y,Z$ compact. Hints?
general-topology algebraic-topology homotopy-theory
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
$endgroup$
Let $Y,Z$ be compact or $X,Z$ locally compact. Then the canonical bijection
$$ (X wedge Y) wedge Z rightarrow X wedge (Y wedge Z) $$
is a homeomoprhism.
I can prove the case when $X,Z$ are locally compact using exponential law. But I don't see how one approaches the case when $Y,Z$ compact. Hints?
general-topology algebraic-topology homotopy-theory
general-topology algebraic-topology homotopy-theory
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
asked Aug 30 '18 at 16:42
CL.CL.
2,3332925
2,3332925
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is clear that we have surjections
$$p_{X;Y,Z} : X times Y times Z stackrel{id_X times q_{Y,Z}}{longrightarrow} X times (Y wedge Z) stackrel{q_{X,Y wedge Z}}{longrightarrow} X wedge (Y wedge Z)$$
$$p_{X,Y,Z} : X times Y times Z stackrel{q_{X,Y} times id_Z}{longrightarrow} (X wedge Y) times Z stackrel{q_{X wedge Y,Z}}{longrightarrow} (X wedge Y) wedge Z$$
where the $q$-maps are quotient maps. Moreover, there is a bijection $tau : X wedge (Y wedge Z) to (X wedge Y) wedge Z$ such that $tau circ p_{X;Y,Z} = p_{X,Y;Z}$.
It therefore suffices to show that both $p$-maps are quotient maps. In turn it suffices to show that $id_X times q_{Y,Z}$ and $q_{X,Y} times id_Z$ are quotient maps.
If $X,Z$ are locally compact, this is well-known. It remains to show that if $Y,Z$ are compact, then $id_X times q_{Y,Z}$ is a quotient map. In this case we have $q_{Y,Z} : Y times Z to Y wedge Z = (Y times Z)/(Y vee Z)$ where $Y vee Z$ is compact. Now apply exercise 16 in section 2.2 in the book "Algebraic Topology" by tom Dieck.
Frankly, I have not done the exercise, but it seems to be straightforward.
See When is the product of two quotient maps a quotient map?
Noting that closed maps are quotient maps, see also Ronnie Brown's answer to https://mathoverflow.net/q/93679.
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
$endgroup$
$begingroup$
Thanks a lot Paul, here is my attempt of proof of statement: let $U subseteq X/A times Y$, such that $(q times id)^{-1}(U)$ is open. Then if $(x,y) in U$, $A times { y } subseteq (q times id)^{-1}(U)$. By tube lemma, exists open $U_y times V_y subseteq (q times id)^{-1}(U)$, hence $q(U_y) times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
However, I don't see why $q_{X,Y} times id_Z$ is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
@CyrylL. It is a quotient map because $Z$ is compact
$endgroup$
– Paul Frost
Aug 31 '18 at 17:15
$begingroup$
I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?)
$endgroup$
– CL.
Aug 31 '18 at 17:24
$begingroup$
@CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} times id_Z$ is in general nor a quotient map or it requires an individual proof.
$endgroup$
– Paul Frost
Aug 31 '18 at 17:50
|
show 1 more comment
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$begingroup$
It is clear that we have surjections
$$p_{X;Y,Z} : X times Y times Z stackrel{id_X times q_{Y,Z}}{longrightarrow} X times (Y wedge Z) stackrel{q_{X,Y wedge Z}}{longrightarrow} X wedge (Y wedge Z)$$
$$p_{X,Y,Z} : X times Y times Z stackrel{q_{X,Y} times id_Z}{longrightarrow} (X wedge Y) times Z stackrel{q_{X wedge Y,Z}}{longrightarrow} (X wedge Y) wedge Z$$
where the $q$-maps are quotient maps. Moreover, there is a bijection $tau : X wedge (Y wedge Z) to (X wedge Y) wedge Z$ such that $tau circ p_{X;Y,Z} = p_{X,Y;Z}$.
It therefore suffices to show that both $p$-maps are quotient maps. In turn it suffices to show that $id_X times q_{Y,Z}$ and $q_{X,Y} times id_Z$ are quotient maps.
If $X,Z$ are locally compact, this is well-known. It remains to show that if $Y,Z$ are compact, then $id_X times q_{Y,Z}$ is a quotient map. In this case we have $q_{Y,Z} : Y times Z to Y wedge Z = (Y times Z)/(Y vee Z)$ where $Y vee Z$ is compact. Now apply exercise 16 in section 2.2 in the book "Algebraic Topology" by tom Dieck.
Frankly, I have not done the exercise, but it seems to be straightforward.
See When is the product of two quotient maps a quotient map?
Noting that closed maps are quotient maps, see also Ronnie Brown's answer to https://mathoverflow.net/q/93679.
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
$endgroup$
$begingroup$
Thanks a lot Paul, here is my attempt of proof of statement: let $U subseteq X/A times Y$, such that $(q times id)^{-1}(U)$ is open. Then if $(x,y) in U$, $A times { y } subseteq (q times id)^{-1}(U)$. By tube lemma, exists open $U_y times V_y subseteq (q times id)^{-1}(U)$, hence $q(U_y) times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
However, I don't see why $q_{X,Y} times id_Z$ is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
@CyrylL. It is a quotient map because $Z$ is compact
$endgroup$
– Paul Frost
Aug 31 '18 at 17:15
$begingroup$
I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?)
$endgroup$
– CL.
Aug 31 '18 at 17:24
$begingroup$
@CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} times id_Z$ is in general nor a quotient map or it requires an individual proof.
$endgroup$
– Paul Frost
Aug 31 '18 at 17:50
|
show 1 more comment
$begingroup$
It is clear that we have surjections
$$p_{X;Y,Z} : X times Y times Z stackrel{id_X times q_{Y,Z}}{longrightarrow} X times (Y wedge Z) stackrel{q_{X,Y wedge Z}}{longrightarrow} X wedge (Y wedge Z)$$
$$p_{X,Y,Z} : X times Y times Z stackrel{q_{X,Y} times id_Z}{longrightarrow} (X wedge Y) times Z stackrel{q_{X wedge Y,Z}}{longrightarrow} (X wedge Y) wedge Z$$
where the $q$-maps are quotient maps. Moreover, there is a bijection $tau : X wedge (Y wedge Z) to (X wedge Y) wedge Z$ such that $tau circ p_{X;Y,Z} = p_{X,Y;Z}$.
It therefore suffices to show that both $p$-maps are quotient maps. In turn it suffices to show that $id_X times q_{Y,Z}$ and $q_{X,Y} times id_Z$ are quotient maps.
If $X,Z$ are locally compact, this is well-known. It remains to show that if $Y,Z$ are compact, then $id_X times q_{Y,Z}$ is a quotient map. In this case we have $q_{Y,Z} : Y times Z to Y wedge Z = (Y times Z)/(Y vee Z)$ where $Y vee Z$ is compact. Now apply exercise 16 in section 2.2 in the book "Algebraic Topology" by tom Dieck.
Frankly, I have not done the exercise, but it seems to be straightforward.
See When is the product of two quotient maps a quotient map?
Noting that closed maps are quotient maps, see also Ronnie Brown's answer to https://mathoverflow.net/q/93679.
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
$endgroup$
$begingroup$
Thanks a lot Paul, here is my attempt of proof of statement: let $U subseteq X/A times Y$, such that $(q times id)^{-1}(U)$ is open. Then if $(x,y) in U$, $A times { y } subseteq (q times id)^{-1}(U)$. By tube lemma, exists open $U_y times V_y subseteq (q times id)^{-1}(U)$, hence $q(U_y) times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
However, I don't see why $q_{X,Y} times id_Z$ is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
@CyrylL. It is a quotient map because $Z$ is compact
$endgroup$
– Paul Frost
Aug 31 '18 at 17:15
$begingroup$
I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?)
$endgroup$
– CL.
Aug 31 '18 at 17:24
$begingroup$
@CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} times id_Z$ is in general nor a quotient map or it requires an individual proof.
$endgroup$
– Paul Frost
Aug 31 '18 at 17:50
|
show 1 more comment
$begingroup$
It is clear that we have surjections
$$p_{X;Y,Z} : X times Y times Z stackrel{id_X times q_{Y,Z}}{longrightarrow} X times (Y wedge Z) stackrel{q_{X,Y wedge Z}}{longrightarrow} X wedge (Y wedge Z)$$
$$p_{X,Y,Z} : X times Y times Z stackrel{q_{X,Y} times id_Z}{longrightarrow} (X wedge Y) times Z stackrel{q_{X wedge Y,Z}}{longrightarrow} (X wedge Y) wedge Z$$
where the $q$-maps are quotient maps. Moreover, there is a bijection $tau : X wedge (Y wedge Z) to (X wedge Y) wedge Z$ such that $tau circ p_{X;Y,Z} = p_{X,Y;Z}$.
It therefore suffices to show that both $p$-maps are quotient maps. In turn it suffices to show that $id_X times q_{Y,Z}$ and $q_{X,Y} times id_Z$ are quotient maps.
If $X,Z$ are locally compact, this is well-known. It remains to show that if $Y,Z$ are compact, then $id_X times q_{Y,Z}$ is a quotient map. In this case we have $q_{Y,Z} : Y times Z to Y wedge Z = (Y times Z)/(Y vee Z)$ where $Y vee Z$ is compact. Now apply exercise 16 in section 2.2 in the book "Algebraic Topology" by tom Dieck.
Frankly, I have not done the exercise, but it seems to be straightforward.
See When is the product of two quotient maps a quotient map?
Noting that closed maps are quotient maps, see also Ronnie Brown's answer to https://mathoverflow.net/q/93679.
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
$endgroup$
It is clear that we have surjections
$$p_{X;Y,Z} : X times Y times Z stackrel{id_X times q_{Y,Z}}{longrightarrow} X times (Y wedge Z) stackrel{q_{X,Y wedge Z}}{longrightarrow} X wedge (Y wedge Z)$$
$$p_{X,Y,Z} : X times Y times Z stackrel{q_{X,Y} times id_Z}{longrightarrow} (X wedge Y) times Z stackrel{q_{X wedge Y,Z}}{longrightarrow} (X wedge Y) wedge Z$$
where the $q$-maps are quotient maps. Moreover, there is a bijection $tau : X wedge (Y wedge Z) to (X wedge Y) wedge Z$ such that $tau circ p_{X;Y,Z} = p_{X,Y;Z}$.
It therefore suffices to show that both $p$-maps are quotient maps. In turn it suffices to show that $id_X times q_{Y,Z}$ and $q_{X,Y} times id_Z$ are quotient maps.
If $X,Z$ are locally compact, this is well-known. It remains to show that if $Y,Z$ are compact, then $id_X times q_{Y,Z}$ is a quotient map. In this case we have $q_{Y,Z} : Y times Z to Y wedge Z = (Y times Z)/(Y vee Z)$ where $Y vee Z$ is compact. Now apply exercise 16 in section 2.2 in the book "Algebraic Topology" by tom Dieck.
Frankly, I have not done the exercise, but it seems to be straightforward.
See When is the product of two quotient maps a quotient map?
Noting that closed maps are quotient maps, see also Ronnie Brown's answer to https://mathoverflow.net/q/93679.
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
edited Aug 30 '18 at 19:57
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
answered Aug 30 '18 at 19:46
Paul FrostPaul Frost
12.6k31035
12.6k31035
$begingroup$
Thanks a lot Paul, here is my attempt of proof of statement: let $U subseteq X/A times Y$, such that $(q times id)^{-1}(U)$ is open. Then if $(x,y) in U$, $A times { y } subseteq (q times id)^{-1}(U)$. By tube lemma, exists open $U_y times V_y subseteq (q times id)^{-1}(U)$, hence $q(U_y) times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
However, I don't see why $q_{X,Y} times id_Z$ is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
@CyrylL. It is a quotient map because $Z$ is compact
$endgroup$
– Paul Frost
Aug 31 '18 at 17:15
$begingroup$
I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?)
$endgroup$
– CL.
Aug 31 '18 at 17:24
$begingroup$
@CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} times id_Z$ is in general nor a quotient map or it requires an individual proof.
$endgroup$
– Paul Frost
Aug 31 '18 at 17:50
|
show 1 more comment
$begingroup$
Thanks a lot Paul, here is my attempt of proof of statement: let $U subseteq X/A times Y$, such that $(q times id)^{-1}(U)$ is open. Then if $(x,y) in U$, $A times { y } subseteq (q times id)^{-1}(U)$. By tube lemma, exists open $U_y times V_y subseteq (q times id)^{-1}(U)$, hence $q(U_y) times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
However, I don't see why $q_{X,Y} times id_Z$ is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
@CyrylL. It is a quotient map because $Z$ is compact
$endgroup$
– Paul Frost
Aug 31 '18 at 17:15
$begingroup$
I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?)
$endgroup$
– CL.
Aug 31 '18 at 17:24
$begingroup$
@CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} times id_Z$ is in general nor a quotient map or it requires an individual proof.
$endgroup$
– Paul Frost
Aug 31 '18 at 17:50
$begingroup$
Thanks a lot Paul, here is my attempt of proof of statement: let $U subseteq X/A times Y$, such that $(q times id)^{-1}(U)$ is open. Then if $(x,y) in U$, $A times { y } subseteq (q times id)^{-1}(U)$. By tube lemma, exists open $U_y times V_y subseteq (q times id)^{-1}(U)$, hence $q(U_y) times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
Thanks a lot Paul, here is my attempt of proof of statement: let $U subseteq X/A times Y$, such that $(q times id)^{-1}(U)$ is open. Then if $(x,y) in U$, $A times { y } subseteq (q times id)^{-1}(U)$. By tube lemma, exists open $U_y times V_y subseteq (q times id)^{-1}(U)$, hence $q(U_y) times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
However, I don't see why $q_{X,Y} times id_Z$ is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
However, I don't see why $q_{X,Y} times id_Z$ is a quotient map.
$endgroup$
– CL.
Aug 31 '18 at 14:35
$begingroup$
@CyrylL. It is a quotient map because $Z$ is compact
$endgroup$
– Paul Frost
Aug 31 '18 at 17:15
$begingroup$
@CyrylL. It is a quotient map because $Z$ is compact
$endgroup$
– Paul Frost
Aug 31 '18 at 17:15
$begingroup$
I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?)
$endgroup$
– CL.
Aug 31 '18 at 17:24
$begingroup$
I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?)
$endgroup$
– CL.
Aug 31 '18 at 17:24
$begingroup$
@CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} times id_Z$ is in general nor a quotient map or it requires an individual proof.
$endgroup$
– Paul Frost
Aug 31 '18 at 17:50
$begingroup$
@CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} times id_Z$ is in general nor a quotient map or it requires an individual proof.
$endgroup$
– Paul Frost
Aug 31 '18 at 17:50
|
show 1 more comment
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