Why is a space homeomorphic to the orbit space of its universal cover by its deck transformations?
Below is an excerpt from Hatcher's Algebraic Topology, page 72:
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
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
add a comment |
Below is an excerpt from Hatcher's Algebraic Topology, page 72:
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
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
add a comment |
Below is an excerpt from Hatcher's Algebraic Topology, page 72:
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
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
Below is an excerpt from Hatcher's Algebraic Topology, page 72:
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
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
edited Nov 27 '18 at 3:27
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
asked Nov 27 '18 at 3:11
Frederic Chopin
321111
321111
add a comment |
add a comment |
1 Answer
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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.)
answered Nov 27 '18 at 3:43
Antonios-Alexandros Robotis
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
add a comment |
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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.)
answered Nov 27 '18 at 3:43
Antonios-Alexandros Robotis
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
add a comment |
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.)
answered Nov 27 '18 at 3:43
Antonios-Alexandros Robotis
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
add a comment |
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.)
answered Nov 27 '18 at 3:43
Antonios-Alexandros Robotis
9,56241640
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.)
answered Nov 27 '18 at 3:43
Antonios-Alexandros Robotis
9,56241640
answered Nov 27 '18 at 3:43
Antonios-Alexandros Robotis
9,56241640
answered Nov 27 '18 at 3:43
Antonios-Alexandros Robotis
9,56241640
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
add a comment |
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
add a comment |
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