Let $pi: E rightarrow M$ be a fiber bundle with fiber $F$. Then $pi$ is a proper map $iff F$ is compact.
$begingroup$
Let $pi: E rightarrow M$ be a fiber bundle with fiber $F$. Then $pi$ is a proper map $iff F$ is compact.
I'm assuming everything is smooth. The "$implies$" direction is trivial but I'm struggling with the converse. I have already proved that $pi$ is an open quotient map and a smooth submersion in a previous part of the problem. I thought maybe the fact that $pi$ is an open map would be useful.
I take a compact subset $K subset M$ and look at an open cover ${U_{alpha}}_{alpha in A}$ and want to try to find a finite subcovering of that. I've tried looking at the images $V_{alpha}=pi(U_alpha)$ which are an open cover of $K$ and reducing to a finite subcover by compactness, but I can't figure out how to make this work. Obviously we need to use the assumptions on $F$ being compact so we probably need to bring in local trivializations somewhere.
differential-geometry
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
$endgroup$
add a comment |
$begingroup$
Let $pi: E rightarrow M$ be a fiber bundle with fiber $F$. Then $pi$ is a proper map $iff F$ is compact.
I'm assuming everything is smooth. The "$implies$" direction is trivial but I'm struggling with the converse. I have already proved that $pi$ is an open quotient map and a smooth submersion in a previous part of the problem. I thought maybe the fact that $pi$ is an open map would be useful.
I take a compact subset $K subset M$ and look at an open cover ${U_{alpha}}_{alpha in A}$ and want to try to find a finite subcovering of that. I've tried looking at the images $V_{alpha}=pi(U_alpha)$ which are an open cover of $K$ and reducing to a finite subcover by compactness, but I can't figure out how to make this work. Obviously we need to use the assumptions on $F$ being compact so we probably need to bring in local trivializations somewhere.
differential-geometry
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
$endgroup$
add a comment |
$begingroup$
Let $pi: E rightarrow M$ be a fiber bundle with fiber $F$. Then $pi$ is a proper map $iff F$ is compact.
I'm assuming everything is smooth. The "$implies$" direction is trivial but I'm struggling with the converse. I have already proved that $pi$ is an open quotient map and a smooth submersion in a previous part of the problem. I thought maybe the fact that $pi$ is an open map would be useful.
I take a compact subset $K subset M$ and look at an open cover ${U_{alpha}}_{alpha in A}$ and want to try to find a finite subcovering of that. I've tried looking at the images $V_{alpha}=pi(U_alpha)$ which are an open cover of $K$ and reducing to a finite subcover by compactness, but I can't figure out how to make this work. Obviously we need to use the assumptions on $F$ being compact so we probably need to bring in local trivializations somewhere.
differential-geometry
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
$endgroup$
Let $pi: E rightarrow M$ be a fiber bundle with fiber $F$. Then $pi$ is a proper map $iff F$ is compact.
I'm assuming everything is smooth. The "$implies$" direction is trivial but I'm struggling with the converse. I have already proved that $pi$ is an open quotient map and a smooth submersion in a previous part of the problem. I thought maybe the fact that $pi$ is an open map would be useful.
I take a compact subset $K subset M$ and look at an open cover ${U_{alpha}}_{alpha in A}$ and want to try to find a finite subcovering of that. I've tried looking at the images $V_{alpha}=pi(U_alpha)$ which are an open cover of $K$ and reducing to a finite subcover by compactness, but I can't figure out how to make this work. Obviously we need to use the assumptions on $F$ being compact so we probably need to bring in local trivializations somewhere.
differential-geometry
differential-geometry
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
asked Dec 3 '18 at 1:11
TuoTuoTuoTuo
1,748516
1,748516
add a comment |
add a comment |
1 Answer
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$begingroup$
Consider the family $mathcal F$ of the open subsets $Usubseteq M$ such that:
$overline U$ is homeomorphic to the closed unit ball;
there is some trivializing open set $V$ such that $Vsupseteq overline U$.
$mathcal F$ is a basis for the topology of $M$. Therefore, there is a finite cover of $K$ by elements of $mathcal F$. Now, $pi^{-1}[K]$ is a closed subset of $bigcup_{j=1}^m pi^{-1}left[overline U_jright]$. By definition there are homeomorphisms $chi_j$ defined on neighbourhoods of $pi^{-1}left[overline U_jright]$ that send $pi^{-1}left[overline U_jright]$ to the compact topological space $overline U_jtimes F$. Therefore, $pi^{-1}[K]$ is a closed subset of a finite union of compact sets, and thus compact.
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
$endgroup$
$begingroup$
Thanks, this is quite tricky.
$endgroup$
– TuoTuo
Dec 4 '18 at 0:57
$begingroup$
It's a rather common idea. Once seen, it can be used almost all the time.
$endgroup$
– Saucy O'Path
Dec 4 '18 at 1:48
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Consider the family $mathcal F$ of the open subsets $Usubseteq M$ such that:
$overline U$ is homeomorphic to the closed unit ball;
there is some trivializing open set $V$ such that $Vsupseteq overline U$.
$mathcal F$ is a basis for the topology of $M$. Therefore, there is a finite cover of $K$ by elements of $mathcal F$. Now, $pi^{-1}[K]$ is a closed subset of $bigcup_{j=1}^m pi^{-1}left[overline U_jright]$. By definition there are homeomorphisms $chi_j$ defined on neighbourhoods of $pi^{-1}left[overline U_jright]$ that send $pi^{-1}left[overline U_jright]$ to the compact topological space $overline U_jtimes F$. Therefore, $pi^{-1}[K]$ is a closed subset of a finite union of compact sets, and thus compact.
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
$endgroup$
$begingroup$
Thanks, this is quite tricky.
$endgroup$
– TuoTuo
Dec 4 '18 at 0:57
$begingroup$
It's a rather common idea. Once seen, it can be used almost all the time.
$endgroup$
– Saucy O'Path
Dec 4 '18 at 1:48
add a comment |
$begingroup$
Consider the family $mathcal F$ of the open subsets $Usubseteq M$ such that:
$overline U$ is homeomorphic to the closed unit ball;
there is some trivializing open set $V$ such that $Vsupseteq overline U$.
$mathcal F$ is a basis for the topology of $M$. Therefore, there is a finite cover of $K$ by elements of $mathcal F$. Now, $pi^{-1}[K]$ is a closed subset of $bigcup_{j=1}^m pi^{-1}left[overline U_jright]$. By definition there are homeomorphisms $chi_j$ defined on neighbourhoods of $pi^{-1}left[overline U_jright]$ that send $pi^{-1}left[overline U_jright]$ to the compact topological space $overline U_jtimes F$. Therefore, $pi^{-1}[K]$ is a closed subset of a finite union of compact sets, and thus compact.
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
$endgroup$
$begingroup$
Thanks, this is quite tricky.
$endgroup$
– TuoTuo
Dec 4 '18 at 0:57
$begingroup$
It's a rather common idea. Once seen, it can be used almost all the time.
$endgroup$
– Saucy O'Path
Dec 4 '18 at 1:48
add a comment |
$begingroup$
Consider the family $mathcal F$ of the open subsets $Usubseteq M$ such that:
$overline U$ is homeomorphic to the closed unit ball;
there is some trivializing open set $V$ such that $Vsupseteq overline U$.
$mathcal F$ is a basis for the topology of $M$. Therefore, there is a finite cover of $K$ by elements of $mathcal F$. Now, $pi^{-1}[K]$ is a closed subset of $bigcup_{j=1}^m pi^{-1}left[overline U_jright]$. By definition there are homeomorphisms $chi_j$ defined on neighbourhoods of $pi^{-1}left[overline U_jright]$ that send $pi^{-1}left[overline U_jright]$ to the compact topological space $overline U_jtimes F$. Therefore, $pi^{-1}[K]$ is a closed subset of a finite union of compact sets, and thus compact.
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
$endgroup$
Consider the family $mathcal F$ of the open subsets $Usubseteq M$ such that:
$overline U$ is homeomorphic to the closed unit ball;
there is some trivializing open set $V$ such that $Vsupseteq overline U$.
$mathcal F$ is a basis for the topology of $M$. Therefore, there is a finite cover of $K$ by elements of $mathcal F$. Now, $pi^{-1}[K]$ is a closed subset of $bigcup_{j=1}^m pi^{-1}left[overline U_jright]$. By definition there are homeomorphisms $chi_j$ defined on neighbourhoods of $pi^{-1}left[overline U_jright]$ that send $pi^{-1}left[overline U_jright]$ to the compact topological space $overline U_jtimes F$. Therefore, $pi^{-1}[K]$ is a closed subset of a finite union of compact sets, and thus compact.
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
edited Dec 3 '18 at 1:51
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
answered Dec 3 '18 at 1:38
Saucy O'PathSaucy O'Path
5,8541626
5,8541626
$begingroup$
Thanks, this is quite tricky.
$endgroup$
– TuoTuo
Dec 4 '18 at 0:57
$begingroup$
It's a rather common idea. Once seen, it can be used almost all the time.
$endgroup$
– Saucy O'Path
Dec 4 '18 at 1:48
add a comment |
$begingroup$
Thanks, this is quite tricky.
$endgroup$
– TuoTuo
Dec 4 '18 at 0:57
$begingroup$
It's a rather common idea. Once seen, it can be used almost all the time.
$endgroup$
– Saucy O'Path
Dec 4 '18 at 1:48
$begingroup$
Thanks, this is quite tricky.
$endgroup$
– TuoTuo
Dec 4 '18 at 0:57
$begingroup$
Thanks, this is quite tricky.
$endgroup$
– TuoTuo
Dec 4 '18 at 0:57
$begingroup$
It's a rather common idea. Once seen, it can be used almost all the time.
$endgroup$
– Saucy O'Path
Dec 4 '18 at 1:48
$begingroup$
It's a rather common idea. Once seen, it can be used almost all the time.
$endgroup$
– Saucy O'Path
Dec 4 '18 at 1:48
add a comment |
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