Hatcher Lemma 1A.3 - Covering Spaces Of Graphs Are Graphs
$begingroup$
Below is an excerpt from Hatcher's Algebraic Topology:
There are a few things I don't understand about the highlighted red part.
What exactly is a "basic open set"?
The highlighted sentence seems to be suggesting that because of the local homeomorphism property of the covering map, any covering space in general, not just graphs and their covering spaces, will have the same "basic open sets" as $X$. Does this mean that any covering space of $X$ "has the same topology as $X$"? That seems wrong, and what does it even mean for two spaces to "have the same topology"?
general-topology algebraic-topology
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
$endgroup$
add a comment |
$begingroup$
Below is an excerpt from Hatcher's Algebraic Topology:
There are a few things I don't understand about the highlighted red part.
What exactly is a "basic open set"?
The highlighted sentence seems to be suggesting that because of the local homeomorphism property of the covering map, any covering space in general, not just graphs and their covering spaces, will have the same "basic open sets" as $X$. Does this mean that any covering space of $X$ "has the same topology as $X$"? That seems wrong, and what does it even mean for two spaces to "have the same topology"?
general-topology algebraic-topology
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
$endgroup$
add a comment |
$begingroup$
Below is an excerpt from Hatcher's Algebraic Topology:
There are a few things I don't understand about the highlighted red part.
What exactly is a "basic open set"?
The highlighted sentence seems to be suggesting that because of the local homeomorphism property of the covering map, any covering space in general, not just graphs and their covering spaces, will have the same "basic open sets" as $X$. Does this mean that any covering space of $X$ "has the same topology as $X$"? That seems wrong, and what does it even mean for two spaces to "have the same topology"?
general-topology algebraic-topology
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
$endgroup$
Below is an excerpt from Hatcher's Algebraic Topology:
There are a few things I don't understand about the highlighted red part.
What exactly is a "basic open set"?
The highlighted sentence seems to be suggesting that because of the local homeomorphism property of the covering map, any covering space in general, not just graphs and their covering spaces, will have the same "basic open sets" as $X$. Does this mean that any covering space of $X$ "has the same topology as $X$"? That seems wrong, and what does it even mean for two spaces to "have the same topology"?
general-topology algebraic-topology
general-topology algebraic-topology
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
edited Dec 9 '18 at 4:21
Frederic Chopin
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
asked Dec 9 '18 at 4:15
Frederic ChopinFrederic Chopin
321111
321111
add a comment |
add a comment |
1 Answer
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When a topology is specified by giving a basis, an open set in the basis is called a basic open set.
That's not what he is saying. $X$ is a graph, and $tilde{X}$ is some topological covering space. He wants to prove that the graph structure on $X$ gives a graph structure on $tilde{X}$. He then does so, but then wants to show that this graph structure's topology on $tilde{X}$ is the same as the original topology on $tilde{X}$ (a set can have several different topologies: he is saying these two are the same here). Hence the graph structure respects the original topology on $tilde{X}$. This happens because the graph structure is induced by $X$ and the local covering conditions lift the right way upstairs. He is not claiming that this means that $X$ and $tilde{X}$ have the same topology.
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
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$begingroup$
I see. So why do the two topologies on $tilde{X}$ have the same basic open sets? It seems like it has something to do with the local homeomorphism, but I can't still can't see why.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 5:02
1
$begingroup$
Yeah, you need to stare at it. It's because the edges in the graph are given by lifts of edges according to the cover. A priori, there's no reason for the graph-topology to agree with the original. But, because it was built by lifts, they do.
$endgroup$
– Randall
Dec 9 '18 at 5:03
add a comment |
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1 Answer
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1 Answer
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$begingroup$
When a topology is specified by giving a basis, an open set in the basis is called a basic open set.
That's not what he is saying. $X$ is a graph, and $tilde{X}$ is some topological covering space. He wants to prove that the graph structure on $X$ gives a graph structure on $tilde{X}$. He then does so, but then wants to show that this graph structure's topology on $tilde{X}$ is the same as the original topology on $tilde{X}$ (a set can have several different topologies: he is saying these two are the same here). Hence the graph structure respects the original topology on $tilde{X}$. This happens because the graph structure is induced by $X$ and the local covering conditions lift the right way upstairs. He is not claiming that this means that $X$ and $tilde{X}$ have the same topology.
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
$endgroup$
$begingroup$
I see. So why do the two topologies on $tilde{X}$ have the same basic open sets? It seems like it has something to do with the local homeomorphism, but I can't still can't see why.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 5:02
1
$begingroup$
Yeah, you need to stare at it. It's because the edges in the graph are given by lifts of edges according to the cover. A priori, there's no reason for the graph-topology to agree with the original. But, because it was built by lifts, they do.
$endgroup$
– Randall
Dec 9 '18 at 5:03
add a comment |
$begingroup$
When a topology is specified by giving a basis, an open set in the basis is called a basic open set.
That's not what he is saying. $X$ is a graph, and $tilde{X}$ is some topological covering space. He wants to prove that the graph structure on $X$ gives a graph structure on $tilde{X}$. He then does so, but then wants to show that this graph structure's topology on $tilde{X}$ is the same as the original topology on $tilde{X}$ (a set can have several different topologies: he is saying these two are the same here). Hence the graph structure respects the original topology on $tilde{X}$. This happens because the graph structure is induced by $X$ and the local covering conditions lift the right way upstairs. He is not claiming that this means that $X$ and $tilde{X}$ have the same topology.
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
$endgroup$
$begingroup$
I see. So why do the two topologies on $tilde{X}$ have the same basic open sets? It seems like it has something to do with the local homeomorphism, but I can't still can't see why.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 5:02
1
$begingroup$
Yeah, you need to stare at it. It's because the edges in the graph are given by lifts of edges according to the cover. A priori, there's no reason for the graph-topology to agree with the original. But, because it was built by lifts, they do.
$endgroup$
– Randall
Dec 9 '18 at 5:03
add a comment |
$begingroup$
When a topology is specified by giving a basis, an open set in the basis is called a basic open set.
That's not what he is saying. $X$ is a graph, and $tilde{X}$ is some topological covering space. He wants to prove that the graph structure on $X$ gives a graph structure on $tilde{X}$. He then does so, but then wants to show that this graph structure's topology on $tilde{X}$ is the same as the original topology on $tilde{X}$ (a set can have several different topologies: he is saying these two are the same here). Hence the graph structure respects the original topology on $tilde{X}$. This happens because the graph structure is induced by $X$ and the local covering conditions lift the right way upstairs. He is not claiming that this means that $X$ and $tilde{X}$ have the same topology.
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
$endgroup$
When a topology is specified by giving a basis, an open set in the basis is called a basic open set.
That's not what he is saying. $X$ is a graph, and $tilde{X}$ is some topological covering space. He wants to prove that the graph structure on $X$ gives a graph structure on $tilde{X}$. He then does so, but then wants to show that this graph structure's topology on $tilde{X}$ is the same as the original topology on $tilde{X}$ (a set can have several different topologies: he is saying these two are the same here). Hence the graph structure respects the original topology on $tilde{X}$. This happens because the graph structure is induced by $X$ and the local covering conditions lift the right way upstairs. He is not claiming that this means that $X$ and $tilde{X}$ have the same topology.
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
answered Dec 9 '18 at 4:33
RandallRandall
9,72111230
9,72111230
$begingroup$
I see. So why do the two topologies on $tilde{X}$ have the same basic open sets? It seems like it has something to do with the local homeomorphism, but I can't still can't see why.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 5:02
1
$begingroup$
Yeah, you need to stare at it. It's because the edges in the graph are given by lifts of edges according to the cover. A priori, there's no reason for the graph-topology to agree with the original. But, because it was built by lifts, they do.
$endgroup$
– Randall
Dec 9 '18 at 5:03
add a comment |
$begingroup$
I see. So why do the two topologies on $tilde{X}$ have the same basic open sets? It seems like it has something to do with the local homeomorphism, but I can't still can't see why.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 5:02
1
$begingroup$
Yeah, you need to stare at it. It's because the edges in the graph are given by lifts of edges according to the cover. A priori, there's no reason for the graph-topology to agree with the original. But, because it was built by lifts, they do.
$endgroup$
– Randall
Dec 9 '18 at 5:03
$begingroup$
I see. So why do the two topologies on $tilde{X}$ have the same basic open sets? It seems like it has something to do with the local homeomorphism, but I can't still can't see why.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 5:02
$begingroup$
I see. So why do the two topologies on $tilde{X}$ have the same basic open sets? It seems like it has something to do with the local homeomorphism, but I can't still can't see why.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 5:02
1
1
$begingroup$
Yeah, you need to stare at it. It's because the edges in the graph are given by lifts of edges according to the cover. A priori, there's no reason for the graph-topology to agree with the original. But, because it was built by lifts, they do.
$endgroup$
– Randall
Dec 9 '18 at 5:03
$begingroup$
Yeah, you need to stare at it. It's because the edges in the graph are given by lifts of edges according to the cover. A priori, there's no reason for the graph-topology to agree with the original. But, because it was built by lifts, they do.
$endgroup$
– Randall
Dec 9 '18 at 5:03
add a comment |
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