Let $qcolon Eto X$ be a covering map. If $E$ is compact then $X$ is compact and $q$ is a finite-sheeted...
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Here's what I have so far: Since $q$ is continuous and surjective then $X$ is compact. For all $xin X$ there exists an evenly covered neighborhood $U_x$, so ${U_x colon xin X}$ is an open cover of $X$. So there is a finite subcover ${ U_{x_1},dots, U_{x_n}}$.
For a contradiction, suppose $q$ is not finite-sheeted. Now for for each $x_i$, we have that $q^{-1}(U_{x_i}) = bigsqcup_{alpha in I_i} V_{x_i,alpha}$ is disjoint union of open sets of $E$. The sets ${V_{x_1,alpha}: alpha in I_1},dots,{V_{x_n,alpha}: alpha in I_n}$ form an open cover of $E$, call it $mathcal U$. So by compactness have a finite subcover $mathcal U'subseteq mathcal U$. This where I'm having trouble completing the argument. Basically I want to argue that, for say $x_1$, no finite collection of $mathcal U$ could cover the fiber $q^{-1}(x_1)$ since we assumed the fiber is infinite. But I'm having trouble how to show this.
general-topology compactness covering-spaces
asked Jan 13 at 17:49
John117John117
376
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Here's what I have so far: Since $q$ is continuous and surjective then $X$ is compact. For all $xin X$ there exists an evenly covered neighborhood $U_x$, so ${U_x colon xin X}$ is an open cover of $X$. So there is a finite subcover ${ U_{x_1},dots, U_{x_n}}$.
For a contradiction, suppose $q$ is not finite-sheeted. Now for for each $x_i$, we have that $q^{-1}(U_{x_i}) = bigsqcup_{alpha in I_i} V_{x_i,alpha}$ is disjoint union of open sets of $E$. The sets ${V_{x_1,alpha}: alpha in I_1},dots,{V_{x_n,alpha}: alpha in I_n}$ form an open cover of $E$, call it $mathcal U$. So by compactness have a finite subcover $mathcal U'subseteq mathcal U$. This where I'm having trouble completing the argument. Basically I want to argue that, for say $x_1$, no finite collection of $mathcal U$ could cover the fiber $q^{-1}(x_1)$ since we assumed the fiber is infinite. But I'm having trouble how to show this.
general-topology compactness covering-spaces
asked Jan 13 at 17:49
John117John117
376
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add a comment |
$begingroup$
Here's what I have so far: Since $q$ is continuous and surjective then $X$ is compact. For all $xin X$ there exists an evenly covered neighborhood $U_x$, so ${U_x colon xin X}$ is an open cover of $X$. So there is a finite subcover ${ U_{x_1},dots, U_{x_n}}$.
For a contradiction, suppose $q$ is not finite-sheeted. Now for for each $x_i$, we have that $q^{-1}(U_{x_i}) = bigsqcup_{alpha in I_i} V_{x_i,alpha}$ is disjoint union of open sets of $E$. The sets ${V_{x_1,alpha}: alpha in I_1},dots,{V_{x_n,alpha}: alpha in I_n}$ form an open cover of $E$, call it $mathcal U$. So by compactness have a finite subcover $mathcal U'subseteq mathcal U$. This where I'm having trouble completing the argument. Basically I want to argue that, for say $x_1$, no finite collection of $mathcal U$ could cover the fiber $q^{-1}(x_1)$ since we assumed the fiber is infinite. But I'm having trouble how to show this.
general-topology compactness covering-spaces
asked Jan 13 at 17:49
John117John117
376
$endgroup$
Here's what I have so far: Since $q$ is continuous and surjective then $X$ is compact. For all $xin X$ there exists an evenly covered neighborhood $U_x$, so ${U_x colon xin X}$ is an open cover of $X$. So there is a finite subcover ${ U_{x_1},dots, U_{x_n}}$.
For a contradiction, suppose $q$ is not finite-sheeted. Now for for each $x_i$, we have that $q^{-1}(U_{x_i}) = bigsqcup_{alpha in I_i} V_{x_i,alpha}$ is disjoint union of open sets of $E$. The sets ${V_{x_1,alpha}: alpha in I_1},dots,{V_{x_n,alpha}: alpha in I_n}$ form an open cover of $E$, call it $mathcal U$. So by compactness have a finite subcover $mathcal U'subseteq mathcal U$. This where I'm having trouble completing the argument. Basically I want to argue that, for say $x_1$, no finite collection of $mathcal U$ could cover the fiber $q^{-1}(x_1)$ since we assumed the fiber is infinite. But I'm having trouble how to show this.
general-topology compactness covering-spaces
general-topology compactness covering-spaces
asked Jan 13 at 17:49
John117John117
376
asked Jan 13 at 17:49
John117John117
376
asked Jan 13 at 17:49
John117John117
376
asked Jan 13 at 17:49
John117John117
376
asked Jan 13 at 17:49
John117John117
376
376
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2 Answers
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For convenience, I write $U_k=U_{x_k}$. Without loss of generality, choose the collection ${U_1,ldots,U_n}$ such that $n$ is minimal, i.e., no collection of $n-1$ evenly covered open subsets of $X$ covers it. Let $mathcal U$ be as you have described. I claim that $mathcal U$ is finite. Otherwise, some ${V_{k,alpha}:alphain I_k}$ must be infinite. By compactness of $E$, for some $alphain I_k$ we have $V_{k,alpha}subset cup{V_{j,beta}:jneq k}$. Applying $q$, we see that $U_ksubsetcup{U_j:jneq k}$, contradicting our assumption.
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
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In a covering map all fibres are relatively discrete by definition. If $X$ has the (usually assumed) property that all singletons are closed, all fibres are also closed (and thus compact when $E$ is). The only discrete compact spaces are finite. QED.
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
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This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$
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– John117
Jan 14 at 12:02
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2 Answers
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2 Answers
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active
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For convenience, I write $U_k=U_{x_k}$. Without loss of generality, choose the collection ${U_1,ldots,U_n}$ such that $n$ is minimal, i.e., no collection of $n-1$ evenly covered open subsets of $X$ covers it. Let $mathcal U$ be as you have described. I claim that $mathcal U$ is finite. Otherwise, some ${V_{k,alpha}:alphain I_k}$ must be infinite. By compactness of $E$, for some $alphain I_k$ we have $V_{k,alpha}subset cup{V_{j,beta}:jneq k}$. Applying $q$, we see that $U_ksubsetcup{U_j:jneq k}$, contradicting our assumption.
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
$endgroup$
add a comment |
$begingroup$
For convenience, I write $U_k=U_{x_k}$. Without loss of generality, choose the collection ${U_1,ldots,U_n}$ such that $n$ is minimal, i.e., no collection of $n-1$ evenly covered open subsets of $X$ covers it. Let $mathcal U$ be as you have described. I claim that $mathcal U$ is finite. Otherwise, some ${V_{k,alpha}:alphain I_k}$ must be infinite. By compactness of $E$, for some $alphain I_k$ we have $V_{k,alpha}subset cup{V_{j,beta}:jneq k}$. Applying $q$, we see that $U_ksubsetcup{U_j:jneq k}$, contradicting our assumption.
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
$endgroup$
add a comment |
$begingroup$
For convenience, I write $U_k=U_{x_k}$. Without loss of generality, choose the collection ${U_1,ldots,U_n}$ such that $n$ is minimal, i.e., no collection of $n-1$ evenly covered open subsets of $X$ covers it. Let $mathcal U$ be as you have described. I claim that $mathcal U$ is finite. Otherwise, some ${V_{k,alpha}:alphain I_k}$ must be infinite. By compactness of $E$, for some $alphain I_k$ we have $V_{k,alpha}subset cup{V_{j,beta}:jneq k}$. Applying $q$, we see that $U_ksubsetcup{U_j:jneq k}$, contradicting our assumption.
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
$endgroup$
For convenience, I write $U_k=U_{x_k}$. Without loss of generality, choose the collection ${U_1,ldots,U_n}$ such that $n$ is minimal, i.e., no collection of $n-1$ evenly covered open subsets of $X$ covers it. Let $mathcal U$ be as you have described. I claim that $mathcal U$ is finite. Otherwise, some ${V_{k,alpha}:alphain I_k}$ must be infinite. By compactness of $E$, for some $alphain I_k$ we have $V_{k,alpha}subset cup{V_{j,beta}:jneq k}$. Applying $q$, we see that $U_ksubsetcup{U_j:jneq k}$, contradicting our assumption.
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
edited Jan 13 at 18:19
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
answered Jan 13 at 18:13
AweyganAweygan
14.8k21442
14.8k21442
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add a comment |
$begingroup$
In a covering map all fibres are relatively discrete by definition. If $X$ has the (usually assumed) property that all singletons are closed, all fibres are also closed (and thus compact when $E$ is). The only discrete compact spaces are finite. QED.
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
$begingroup$
This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$
$endgroup$
– John117
Jan 14 at 12:02
add a comment |
$begingroup$
In a covering map all fibres are relatively discrete by definition. If $X$ has the (usually assumed) property that all singletons are closed, all fibres are also closed (and thus compact when $E$ is). The only discrete compact spaces are finite. QED.
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
$begingroup$
This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$
$endgroup$
– John117
Jan 14 at 12:02
add a comment |
$begingroup$
In a covering map all fibres are relatively discrete by definition. If $X$ has the (usually assumed) property that all singletons are closed, all fibres are also closed (and thus compact when $E$ is). The only discrete compact spaces are finite. QED.
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
In a covering map all fibres are relatively discrete by definition. If $X$ has the (usually assumed) property that all singletons are closed, all fibres are also closed (and thus compact when $E$ is). The only discrete compact spaces are finite. QED.
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
answered Jan 13 at 23:12
Henno BrandsmaHenno Brandsma
117k349127
117k349127
$begingroup$
This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$
$endgroup$
– John117
Jan 14 at 12:02
add a comment |
$begingroup$
This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$
$endgroup$
– John117
Jan 14 at 12:02
$begingroup$
This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$
$endgroup$
– John117
Jan 14 at 12:02
$begingroup$
This is how I initially thought to solve the problem but here we don't have any additional assumptions on $X$ allowing us to conclude singletons are closed in $X$
$endgroup$
– John117
Jan 14 at 12:02
add a comment |
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