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







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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







    share|cite|improve this question









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      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Jan 13 at 17:49









      John117John117

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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.






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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.






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            • $begingroup$
              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












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            2 Answers
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            2 Answers
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            active

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            active

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            $begingroup$

            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.






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              1












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$



                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 13 at 18:19

























                answered Jan 13 at 18:13









                AweyganAweygan

                14.8k21442




                14.8k21442























                    0












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      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
















                    0












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      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














                    0












                    0








                    0





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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


















                    • $begingroup$
                      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


















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