Why is a space homeomorphic to the orbit space of its universal cover by its deck transformations?












2














Below is an excerpt from Hatcher's Algebraic Topology, page 72:



enter image description here



I don't understand why the part above highlighted in red is generally true. I can relate it to an example where it is true though: We can consider $S^1$ covered by $mathbb{R}$. Then the deck transformations of $mathbb{R}$ correspond to the action of $mathbb{Z}$ on $mathbb{R}$ via translation by an integer. Hence, indeed we have $mathbb{R} / mathbb{Z} cong S^1$. But I don't know why the statement $tilde{X} / G(tilde{X}) cong X$ is true in general.






general-topology algebraic-topology







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    2














    Below is an excerpt from Hatcher's Algebraic Topology, page 72:



    enter image description here



    I don't understand why the part above highlighted in red is generally true. I can relate it to an example where it is true though: We can consider $S^1$ covered by $mathbb{R}$. Then the deck transformations of $mathbb{R}$ correspond to the action of $mathbb{Z}$ on $mathbb{R}$ via translation by an integer. Hence, indeed we have $mathbb{R} / mathbb{Z} cong S^1$. But I don't know why the statement $tilde{X} / G(tilde{X}) cong X$ is true in general.






    general-topology algebraic-topology







    share|cite|improve this question



























      2












      2








      2







      Below is an excerpt from Hatcher's Algebraic Topology, page 72:



      enter image description here



      I don't understand why the part above highlighted in red is generally true. I can relate it to an example where it is true though: We can consider $S^1$ covered by $mathbb{R}$. Then the deck transformations of $mathbb{R}$ correspond to the action of $mathbb{Z}$ on $mathbb{R}$ via translation by an integer. Hence, indeed we have $mathbb{R} / mathbb{Z} cong S^1$. But I don't know why the statement $tilde{X} / G(tilde{X}) cong X$ is true in general.






      general-topology algebraic-topology







      share|cite|improve this question















      Below is an excerpt from Hatcher's Algebraic Topology, page 72:



      enter image description here



      I don't understand why the part above highlighted in red is generally true. I can relate it to an example where it is true though: We can consider $S^1$ covered by $mathbb{R}$. Then the deck transformations of $mathbb{R}$ correspond to the action of $mathbb{Z}$ on $mathbb{R}$ via translation by an integer. Hence, indeed we have $mathbb{R} / mathbb{Z} cong S^1$. But I don't know why the statement $tilde{X} / G(tilde{X}) cong X$ is true in general.






      general-topology algebraic-topology




      general-topology algebraic-topology






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      share|cite|improve this question













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      edited Nov 27 '18 at 3:27

























      asked Nov 27 '18 at 3:11









      Frederic Chopin

      321111




      321111






















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          Here's the idea: given a "normal" covering space $widetilde{X}xrightarrow{p} X$, a deck transformation is an automorphism $phiin operatorname{Aut}(widetilde{X})$ that preserves fibres, so that $ p=pcirc phi.$ It can be shown (and is in Hatcher) that given $xin X$, and any pair of points $y,zin p^{-1}(x)$ there exists a deck transformation $phi in G(widetilde{X})$ with $phi(y)=z$. That is, the action is transitive.



          So, the intuition here is that when we quotient by $G(widetilde{X})$ action on $widetilde{X}$, we think of points as being "the same" if they can be related by the action of an element of $G(widetilde{X})$. Over each $x$ there is a fibre $p^{-1}(x)$ which corresponds to a single orbit $mathcal{O}_x$ under the $G(widetilde{X})-$action. After quotienting, we are left with the space ${mathcal{O}_x:xin X}$. You can check that its quotient topology of $X/G(widetilde{X})$ is precisely the topology of $X$ so that $widetilde{X}/G(widetilde{X})cong X$ by the obvious map $mathcal{O}_xmapsto x$. (This isn't too hard to see if you leverage the local triviality of the covering space.)






          share|cite|improve this answer





















          • To create $phi$ I would pick a small (simply connected) neighborhood $U$ of $y$ such that $p|_U$ is injective, then find in $p^{-1}(p(U)) = bigcup_j V_j$ the $V_j$ containing $z$, for $u in U$ set $phi(u) = p^{-1}(p(u)) cap V_j$. Doing that repeatedly with $u,phi(u)$ instead of $y,z$ will extend $phi$ to the whole $tilde{X}$
            – reuns
            Nov 27 '18 at 3:59













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          1 Answer
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          active

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          2














          Here's the idea: given a "normal" covering space $widetilde{X}xrightarrow{p} X$, a deck transformation is an automorphism $phiin operatorname{Aut}(widetilde{X})$ that preserves fibres, so that $ p=pcirc phi.$ It can be shown (and is in Hatcher) that given $xin X$, and any pair of points $y,zin p^{-1}(x)$ there exists a deck transformation $phi in G(widetilde{X})$ with $phi(y)=z$. That is, the action is transitive.



          So, the intuition here is that when we quotient by $G(widetilde{X})$ action on $widetilde{X}$, we think of points as being "the same" if they can be related by the action of an element of $G(widetilde{X})$. Over each $x$ there is a fibre $p^{-1}(x)$ which corresponds to a single orbit $mathcal{O}_x$ under the $G(widetilde{X})-$action. After quotienting, we are left with the space ${mathcal{O}_x:xin X}$. You can check that its quotient topology of $X/G(widetilde{X})$ is precisely the topology of $X$ so that $widetilde{X}/G(widetilde{X})cong X$ by the obvious map $mathcal{O}_xmapsto x$. (This isn't too hard to see if you leverage the local triviality of the covering space.)






          share|cite|improve this answer





















          • To create $phi$ I would pick a small (simply connected) neighborhood $U$ of $y$ such that $p|_U$ is injective, then find in $p^{-1}(p(U)) = bigcup_j V_j$ the $V_j$ containing $z$, for $u in U$ set $phi(u) = p^{-1}(p(u)) cap V_j$. Doing that repeatedly with $u,phi(u)$ instead of $y,z$ will extend $phi$ to the whole $tilde{X}$
            – reuns
            Nov 27 '18 at 3:59


















          2














          Here's the idea: given a "normal" covering space $widetilde{X}xrightarrow{p} X$, a deck transformation is an automorphism $phiin operatorname{Aut}(widetilde{X})$ that preserves fibres, so that $ p=pcirc phi.$ It can be shown (and is in Hatcher) that given $xin X$, and any pair of points $y,zin p^{-1}(x)$ there exists a deck transformation $phi in G(widetilde{X})$ with $phi(y)=z$. That is, the action is transitive.



          So, the intuition here is that when we quotient by $G(widetilde{X})$ action on $widetilde{X}$, we think of points as being "the same" if they can be related by the action of an element of $G(widetilde{X})$. Over each $x$ there is a fibre $p^{-1}(x)$ which corresponds to a single orbit $mathcal{O}_x$ under the $G(widetilde{X})-$action. After quotienting, we are left with the space ${mathcal{O}_x:xin X}$. You can check that its quotient topology of $X/G(widetilde{X})$ is precisely the topology of $X$ so that $widetilde{X}/G(widetilde{X})cong X$ by the obvious map $mathcal{O}_xmapsto x$. (This isn't too hard to see if you leverage the local triviality of the covering space.)






          share|cite|improve this answer





















          • To create $phi$ I would pick a small (simply connected) neighborhood $U$ of $y$ such that $p|_U$ is injective, then find in $p^{-1}(p(U)) = bigcup_j V_j$ the $V_j$ containing $z$, for $u in U$ set $phi(u) = p^{-1}(p(u)) cap V_j$. Doing that repeatedly with $u,phi(u)$ instead of $y,z$ will extend $phi$ to the whole $tilde{X}$
            – reuns
            Nov 27 '18 at 3:59
















          2












          2








          2






          Here's the idea: given a "normal" covering space $widetilde{X}xrightarrow{p} X$, a deck transformation is an automorphism $phiin operatorname{Aut}(widetilde{X})$ that preserves fibres, so that $ p=pcirc phi.$ It can be shown (and is in Hatcher) that given $xin X$, and any pair of points $y,zin p^{-1}(x)$ there exists a deck transformation $phi in G(widetilde{X})$ with $phi(y)=z$. That is, the action is transitive.



          So, the intuition here is that when we quotient by $G(widetilde{X})$ action on $widetilde{X}$, we think of points as being "the same" if they can be related by the action of an element of $G(widetilde{X})$. Over each $x$ there is a fibre $p^{-1}(x)$ which corresponds to a single orbit $mathcal{O}_x$ under the $G(widetilde{X})-$action. After quotienting, we are left with the space ${mathcal{O}_x:xin X}$. You can check that its quotient topology of $X/G(widetilde{X})$ is precisely the topology of $X$ so that $widetilde{X}/G(widetilde{X})cong X$ by the obvious map $mathcal{O}_xmapsto x$. (This isn't too hard to see if you leverage the local triviality of the covering space.)






          share|cite|improve this answer












          Here's the idea: given a "normal" covering space $widetilde{X}xrightarrow{p} X$, a deck transformation is an automorphism $phiin operatorname{Aut}(widetilde{X})$ that preserves fibres, so that $ p=pcirc phi.$ It can be shown (and is in Hatcher) that given $xin X$, and any pair of points $y,zin p^{-1}(x)$ there exists a deck transformation $phi in G(widetilde{X})$ with $phi(y)=z$. That is, the action is transitive.



          So, the intuition here is that when we quotient by $G(widetilde{X})$ action on $widetilde{X}$, we think of points as being "the same" if they can be related by the action of an element of $G(widetilde{X})$. Over each $x$ there is a fibre $p^{-1}(x)$ which corresponds to a single orbit $mathcal{O}_x$ under the $G(widetilde{X})-$action. After quotienting, we are left with the space ${mathcal{O}_x:xin X}$. You can check that its quotient topology of $X/G(widetilde{X})$ is precisely the topology of $X$ so that $widetilde{X}/G(widetilde{X})cong X$ by the obvious map $mathcal{O}_xmapsto x$. (This isn't too hard to see if you leverage the local triviality of the covering space.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 3:43









          Antonios-Alexandros Robotis

          9,56241640




          9,56241640












          • To create $phi$ I would pick a small (simply connected) neighborhood $U$ of $y$ such that $p|_U$ is injective, then find in $p^{-1}(p(U)) = bigcup_j V_j$ the $V_j$ containing $z$, for $u in U$ set $phi(u) = p^{-1}(p(u)) cap V_j$. Doing that repeatedly with $u,phi(u)$ instead of $y,z$ will extend $phi$ to the whole $tilde{X}$
            – reuns
            Nov 27 '18 at 3:59




















          • To create $phi$ I would pick a small (simply connected) neighborhood $U$ of $y$ such that $p|_U$ is injective, then find in $p^{-1}(p(U)) = bigcup_j V_j$ the $V_j$ containing $z$, for $u in U$ set $phi(u) = p^{-1}(p(u)) cap V_j$. Doing that repeatedly with $u,phi(u)$ instead of $y,z$ will extend $phi$ to the whole $tilde{X}$
            – reuns
            Nov 27 '18 at 3:59


















          To create $phi$ I would pick a small (simply connected) neighborhood $U$ of $y$ such that $p|_U$ is injective, then find in $p^{-1}(p(U)) = bigcup_j V_j$ the $V_j$ containing $z$, for $u in U$ set $phi(u) = p^{-1}(p(u)) cap V_j$. Doing that repeatedly with $u,phi(u)$ instead of $y,z$ will extend $phi$ to the whole $tilde{X}$
          – reuns
          Nov 27 '18 at 3:59






          To create $phi$ I would pick a small (simply connected) neighborhood $U$ of $y$ such that $p|_U$ is injective, then find in $p^{-1}(p(U)) = bigcup_j V_j$ the $V_j$ containing $z$, for $u in U$ set $phi(u) = p^{-1}(p(u)) cap V_j$. Doing that repeatedly with $u,phi(u)$ instead of $y,z$ will extend $phi$ to the whole $tilde{X}$
          – reuns
          Nov 27 '18 at 3:59




















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