Slice Chart Condition Proof - Topological Embedding
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Below is some background information from Lee's Introduction To Smooth Manifolds about slice charts of embedded submanifolds. My question is at the bottom.
What I don't understand in the proof is the part highlighted in red. I don't understand how we can conclude that $S$ is a topological embedding from what we proved so far.
general-topology differential-geometry manifolds differential-topology smooth-manifolds
asked 2 days ago
Frederic Chopin
31218
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Below is some background information from Lee's Introduction To Smooth Manifolds about slice charts of embedded submanifolds. My question is at the bottom.
What I don't understand in the proof is the part highlighted in red. I don't understand how we can conclude that $S$ is a topological embedding from what we proved so far.
general-topology differential-geometry manifolds differential-topology smooth-manifolds
asked 2 days ago
Frederic Chopin
31218
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Below is some background information from Lee's Introduction To Smooth Manifolds about slice charts of embedded submanifolds. My question is at the bottom.
What I don't understand in the proof is the part highlighted in red. I don't understand how we can conclude that $S$ is a topological embedding from what we proved so far.
general-topology differential-geometry manifolds differential-topology smooth-manifolds
asked 2 days ago
Frederic Chopin
31218
Below is some background information from Lee's Introduction To Smooth Manifolds about slice charts of embedded submanifolds. My question is at the bottom.
What I don't understand in the proof is the part highlighted in red. I don't understand how we can conclude that $S$ is a topological embedding from what we proved so far.
general-topology differential-geometry manifolds differential-topology smooth-manifolds
general-topology differential-geometry manifolds differential-topology smooth-manifolds
asked 2 days ago
Frederic Chopin
31218
asked 2 days ago
Frederic Chopin
31218
edited yesterday
asked 2 days ago
Frederic Chopin
31218
asked 2 days ago
Frederic Chopin
31218
asked 2 days ago
Frederic Chopin
31218
31218
add a comment |
add a comment |
1 Answer
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Up to the point you have highlighted, we have shown that for any point $p$ in $S$, we can define a chart $(psi,V)$, where $V$ is open in $S$ and contains $p$, with $psicolon Vto widehat Vsubset Bbb R^k$ a homeomorphism. The inclusion map $iotacolon Shookrightarrow M$ is a topological embedding because $S$ is given the subspace topology, hence $iota$ is a homeomorphism onto its image.
answered yesterday
AOrtiz
10.3k21239
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Up to the point you have highlighted, we have shown that for any point $p$ in $S$, we can define a chart $(psi,V)$, where $V$ is open in $S$ and contains $p$, with $psicolon Vto widehat Vsubset Bbb R^k$ a homeomorphism. The inclusion map $iotacolon Shookrightarrow M$ is a topological embedding because $S$ is given the subspace topology, hence $iota$ is a homeomorphism onto its image.
answered yesterday
AOrtiz
10.3k21239
add a comment |
up vote
1
down vote
accepted
Up to the point you have highlighted, we have shown that for any point $p$ in $S$, we can define a chart $(psi,V)$, where $V$ is open in $S$ and contains $p$, with $psicolon Vto widehat Vsubset Bbb R^k$ a homeomorphism. The inclusion map $iotacolon Shookrightarrow M$ is a topological embedding because $S$ is given the subspace topology, hence $iota$ is a homeomorphism onto its image.
answered yesterday
AOrtiz
10.3k21239
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Up to the point you have highlighted, we have shown that for any point $p$ in $S$, we can define a chart $(psi,V)$, where $V$ is open in $S$ and contains $p$, with $psicolon Vto widehat Vsubset Bbb R^k$ a homeomorphism. The inclusion map $iotacolon Shookrightarrow M$ is a topological embedding because $S$ is given the subspace topology, hence $iota$ is a homeomorphism onto its image.
answered yesterday
AOrtiz
10.3k21239
Up to the point you have highlighted, we have shown that for any point $p$ in $S$, we can define a chart $(psi,V)$, where $V$ is open in $S$ and contains $p$, with $psicolon Vto widehat Vsubset Bbb R^k$ a homeomorphism. The inclusion map $iotacolon Shookrightarrow M$ is a topological embedding because $S$ is given the subspace topology, hence $iota$ is a homeomorphism onto its image.
answered yesterday
AOrtiz
10.3k21239
answered yesterday
AOrtiz
10.3k21239
answered yesterday
AOrtiz
10.3k21239
answered yesterday
AOrtiz
10.3k21239
10.3k21239
add a comment |
add a comment |
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