How to understand this example in Do Carmo?
I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
I really don't understand what he said about why $alpha$ is not an embedding...
No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?
general-topology differential-geometry riemannian-geometry curves
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
asked Nov 12 at 11:16
user450201
718
add a comment |
I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
I really don't understand what he said about why $alpha$ is not an embedding...
No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?
general-topology differential-geometry riemannian-geometry curves
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
asked Nov 12 at 11:16
user450201
718
add a comment |
I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
I really don't understand what he said about why $alpha$ is not an embedding...
No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?
general-topology differential-geometry riemannian-geometry curves
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
asked Nov 12 at 11:16
user450201
718
I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
I really don't understand what he said about why $alpha$ is not an embedding...
No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?
general-topology differential-geometry riemannian-geometry curves
general-topology differential-geometry riemannian-geometry curves
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
asked Nov 12 at 11:16
user450201
718
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
asked Nov 12 at 11:16
user450201
718
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
edited Nov 12 at 11:38
José Carlos Santos
148k22117218
148k22117218
asked Nov 12 at 11:16
user450201
718
asked Nov 12 at 11:16
user450201
718
asked Nov 12 at 11:16
user450201
718
718
add a comment |
add a comment |
1 Answer
1
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The set $C=alphabigl((-3,0)bigr)$ has two topologies:
- the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.
- the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.
Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
1
Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
– user450201
Nov 12 at 11:27
That would be correct. On the other hand, that is what Do Carmo is claiming.
– José Carlos Santos
Nov 12 at 11:34
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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votes
The set $C=alphabigl((-3,0)bigr)$ has two topologies:
- the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.
- the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.
Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
1
Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
– user450201
Nov 12 at 11:27
That would be correct. On the other hand, that is what Do Carmo is claiming.
– José Carlos Santos
Nov 12 at 11:34
add a comment |
The set $C=alphabigl((-3,0)bigr)$ has two topologies:
- the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.
- the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.
Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
1
Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
– user450201
Nov 12 at 11:27
That would be correct. On the other hand, that is what Do Carmo is claiming.
– José Carlos Santos
Nov 12 at 11:34
add a comment |
The set $C=alphabigl((-3,0)bigr)$ has two topologies:
- the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.
- the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.
Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
The set $C=alphabigl((-3,0)bigr)$ has two topologies:
- the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.
- the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.
Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
edited Nov 12 at 13:59
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
answered Nov 12 at 11:23
José Carlos Santos
148k22117218
148k22117218
1
Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
– user450201
Nov 12 at 11:27
That would be correct. On the other hand, that is what Do Carmo is claiming.
– José Carlos Santos
Nov 12 at 11:34
add a comment |
1
Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
– user450201
Nov 12 at 11:27
That would be correct. On the other hand, that is what Do Carmo is claiming.
– José Carlos Santos
Nov 12 at 11:34
1
1
Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
– user450201
Nov 12 at 11:27
Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
– user450201
Nov 12 at 11:27
That would be correct. On the other hand, that is what Do Carmo is claiming.
– José Carlos Santos
Nov 12 at 11:34
That would be correct. On the other hand, that is what Do Carmo is claiming.
– José Carlos Santos
Nov 12 at 11:34
add a comment |
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