Mixing
Is an inclusion always an homeomorphism?
$begingroup$
Is the inclusion always an homeomorphism?
If not, which conditions must we have for such an assertion to be true?
What's the relationship between the image of the inclusion having the subspace topology, and the inclusion being a homeomorphism?
Any help would be appreciated
general-topology
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
$endgroup$
add a comment |
$begingroup$
Is the inclusion always an homeomorphism?
If not, which conditions must we have for such an assertion to be true?
What's the relationship between the image of the inclusion having the subspace topology, and the inclusion being a homeomorphism?
Any help would be appreciated
general-topology
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
$endgroup$
1
$begingroup$
Do you mean homeomorphism onto its image?
$endgroup$
– Randall
Jan 21 at 19:58
add a comment |
$begingroup$
Is the inclusion always an homeomorphism?
If not, which conditions must we have for such an assertion to be true?
What's the relationship between the image of the inclusion having the subspace topology, and the inclusion being a homeomorphism?
Any help would be appreciated
general-topology
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
$endgroup$
Is the inclusion always an homeomorphism?
If not, which conditions must we have for such an assertion to be true?
What's the relationship between the image of the inclusion having the subspace topology, and the inclusion being a homeomorphism?
Any help would be appreciated
general-topology
general-topology
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
asked Jan 21 at 19:50
An old man in the sea.An old man in the sea.
1,65511135
1,65511135
1
$begingroup$
Do you mean homeomorphism onto its image?
$endgroup$
– Randall
Jan 21 at 19:58
add a comment |
1
$begingroup$
Do you mean homeomorphism onto its image?
$endgroup$
– Randall
Jan 21 at 19:58
1
1
$begingroup$
Do you mean homeomorphism onto its image?
$endgroup$
– Randall
Jan 21 at 19:58
$begingroup$
Do you mean homeomorphism onto its image?
$endgroup$
– Randall
Jan 21 at 19:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An inclusion map is any function $j:Ato B$, where $A$ is a subset of $B$ and $j(a)=a$ for all $ain A$. (Note that $A$ and $B$ need not have any particular structure in this case.)
A homeomorphism is any map $u:Ato B$, where $A$ and $B$ are topological spaces, $u$ is bijective, and both $u$ and $u^{-1}$ are continuous.
The only kind of map which is both an inclusion and a homeomorphism, therefore, is the identity map from a topological space onto itself.
EDIT: However, it is worth noting that every inclusion map $f:Ato B$, where $B$ is a topological space and $A$ is a subspace (endowed with the subspace topology) is a homeomorphism when viewed as a map $f:Ato f(A)$ (where $f(A)$ is also endowed with the subspace topology).
EDIT #2: Also, in the context of topological spaces, if $C$ and $B$ are topological spaces, $A$ is a subspace of $B$ (endowed with the subspace topology), and $u:Cto A$ is homeomorphism, then we may "abuse notation" and write that $u$ is an inclusion map when viewed as a function from $C$ into $B$.
answered Jan 21 at 19:59
Ben WBen W
2,324615
$endgroup$
$begingroup$
One needs that the subset has the subspace topology.
$endgroup$
– mfl
Jan 21 at 20:03
$begingroup$
@mfl You mean for an inclusion map? Actually, you can have an inclusion between sets with no topological structure at all.
$endgroup$
– Ben W
Jan 21 at 20:04
$begingroup$
Yes. Your definition of inclusion map is between sets.
$endgroup$
– mfl
Jan 21 at 20:05
$begingroup$
Ben,Thanks for your answer. I'm interested in your first Edit. How do you prove that? I know that the inclusion is continuous. But how do you prove that the inverse is also continuous?
$endgroup$
– An old man in the sea.
Jan 21 at 21:12
$begingroup$
@Anoldmaninthesea. It's pretty much by definition.
$endgroup$
– Ben W
Jan 21 at 21:16
|
show 4 more comments
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
An inclusion map is any function $j:Ato B$, where $A$ is a subset of $B$ and $j(a)=a$ for all $ain A$. (Note that $A$ and $B$ need not have any particular structure in this case.)
A homeomorphism is any map $u:Ato B$, where $A$ and $B$ are topological spaces, $u$ is bijective, and both $u$ and $u^{-1}$ are continuous.
The only kind of map which is both an inclusion and a homeomorphism, therefore, is the identity map from a topological space onto itself.
EDIT: However, it is worth noting that every inclusion map $f:Ato B$, where $B$ is a topological space and $A$ is a subspace (endowed with the subspace topology) is a homeomorphism when viewed as a map $f:Ato f(A)$ (where $f(A)$ is also endowed with the subspace topology).
EDIT #2: Also, in the context of topological spaces, if $C$ and $B$ are topological spaces, $A$ is a subspace of $B$ (endowed with the subspace topology), and $u:Cto A$ is homeomorphism, then we may "abuse notation" and write that $u$ is an inclusion map when viewed as a function from $C$ into $B$.
answered Jan 21 at 19:59
Ben WBen W
2,324615
$endgroup$
$begingroup$
One needs that the subset has the subspace topology.
$endgroup$
– mfl
Jan 21 at 20:03
$begingroup$
@mfl You mean for an inclusion map? Actually, you can have an inclusion between sets with no topological structure at all.
$endgroup$
– Ben W
Jan 21 at 20:04
$begingroup$
Yes. Your definition of inclusion map is between sets.
$endgroup$
– mfl
Jan 21 at 20:05
$begingroup$
Ben,Thanks for your answer. I'm interested in your first Edit. How do you prove that? I know that the inclusion is continuous. But how do you prove that the inverse is also continuous?
$endgroup$
– An old man in the sea.
Jan 21 at 21:12
$begingroup$
@Anoldmaninthesea. It's pretty much by definition.
$endgroup$
– Ben W
Jan 21 at 21:16
|
show 4 more comments
$begingroup$
An inclusion map is any function $j:Ato B$, where $A$ is a subset of $B$ and $j(a)=a$ for all $ain A$. (Note that $A$ and $B$ need not have any particular structure in this case.)
A homeomorphism is any map $u:Ato B$, where $A$ and $B$ are topological spaces, $u$ is bijective, and both $u$ and $u^{-1}$ are continuous.
The only kind of map which is both an inclusion and a homeomorphism, therefore, is the identity map from a topological space onto itself.
EDIT: However, it is worth noting that every inclusion map $f:Ato B$, where $B$ is a topological space and $A$ is a subspace (endowed with the subspace topology) is a homeomorphism when viewed as a map $f:Ato f(A)$ (where $f(A)$ is also endowed with the subspace topology).
EDIT #2: Also, in the context of topological spaces, if $C$ and $B$ are topological spaces, $A$ is a subspace of $B$ (endowed with the subspace topology), and $u:Cto A$ is homeomorphism, then we may "abuse notation" and write that $u$ is an inclusion map when viewed as a function from $C$ into $B$.
answered Jan 21 at 19:59
Ben WBen W
2,324615
$endgroup$
$begingroup$
One needs that the subset has the subspace topology.
$endgroup$
– mfl
Jan 21 at 20:03
$begingroup$
@mfl You mean for an inclusion map? Actually, you can have an inclusion between sets with no topological structure at all.
$endgroup$
– Ben W
Jan 21 at 20:04
$begingroup$
Yes. Your definition of inclusion map is between sets.
$endgroup$
– mfl
Jan 21 at 20:05
$begingroup$
Ben,Thanks for your answer. I'm interested in your first Edit. How do you prove that? I know that the inclusion is continuous. But how do you prove that the inverse is also continuous?
$endgroup$
– An old man in the sea.
Jan 21 at 21:12
$begingroup$
@Anoldmaninthesea. It's pretty much by definition.
$endgroup$
– Ben W
Jan 21 at 21:16
|
show 4 more comments
$begingroup$
An inclusion map is any function $j:Ato B$, where $A$ is a subset of $B$ and $j(a)=a$ for all $ain A$. (Note that $A$ and $B$ need not have any particular structure in this case.)
A homeomorphism is any map $u:Ato B$, where $A$ and $B$ are topological spaces, $u$ is bijective, and both $u$ and $u^{-1}$ are continuous.
The only kind of map which is both an inclusion and a homeomorphism, therefore, is the identity map from a topological space onto itself.
EDIT: However, it is worth noting that every inclusion map $f:Ato B$, where $B$ is a topological space and $A$ is a subspace (endowed with the subspace topology) is a homeomorphism when viewed as a map $f:Ato f(A)$ (where $f(A)$ is also endowed with the subspace topology).
EDIT #2: Also, in the context of topological spaces, if $C$ and $B$ are topological spaces, $A$ is a subspace of $B$ (endowed with the subspace topology), and $u:Cto A$ is homeomorphism, then we may "abuse notation" and write that $u$ is an inclusion map when viewed as a function from $C$ into $B$.
answered Jan 21 at 19:59
Ben WBen W
2,324615
$endgroup$
An inclusion map is any function $j:Ato B$, where $A$ is a subset of $B$ and $j(a)=a$ for all $ain A$. (Note that $A$ and $B$ need not have any particular structure in this case.)
A homeomorphism is any map $u:Ato B$, where $A$ and $B$ are topological spaces, $u$ is bijective, and both $u$ and $u^{-1}$ are continuous.
The only kind of map which is both an inclusion and a homeomorphism, therefore, is the identity map from a topological space onto itself.
EDIT: However, it is worth noting that every inclusion map $f:Ato B$, where $B$ is a topological space and $A$ is a subspace (endowed with the subspace topology) is a homeomorphism when viewed as a map $f:Ato f(A)$ (where $f(A)$ is also endowed with the subspace topology).
EDIT #2: Also, in the context of topological spaces, if $C$ and $B$ are topological spaces, $A$ is a subspace of $B$ (endowed with the subspace topology), and $u:Cto A$ is homeomorphism, then we may "abuse notation" and write that $u$ is an inclusion map when viewed as a function from $C$ into $B$.
answered Jan 21 at 19:59
Ben WBen W
2,324615
edited Jan 21 at 20:10
answered Jan 21 at 19:59
Ben WBen W
2,324615
answered Jan 21 at 19:59
Ben WBen W
2,324615
answered Jan 21 at 19:59
Ben WBen W
2,324615
2,324615
$begingroup$
One needs that the subset has the subspace topology.
$endgroup$
– mfl
Jan 21 at 20:03
$begingroup$
@mfl You mean for an inclusion map? Actually, you can have an inclusion between sets with no topological structure at all.
$endgroup$
– Ben W
Jan 21 at 20:04
$begingroup$
Yes. Your definition of inclusion map is between sets.
$endgroup$
– mfl
Jan 21 at 20:05
$begingroup$
Ben,Thanks for your answer. I'm interested in your first Edit. How do you prove that? I know that the inclusion is continuous. But how do you prove that the inverse is also continuous?
$endgroup$
– An old man in the sea.
Jan 21 at 21:12
$begingroup$
@Anoldmaninthesea. It's pretty much by definition.
$endgroup$
– Ben W
Jan 21 at 21:16
|
show 4 more comments
$begingroup$
One needs that the subset has the subspace topology.
$endgroup$
– mfl
Jan 21 at 20:03
$begingroup$
@mfl You mean for an inclusion map? Actually, you can have an inclusion between sets with no topological structure at all.
$endgroup$
– Ben W
Jan 21 at 20:04
$begingroup$
Yes. Your definition of inclusion map is between sets.
$endgroup$
– mfl
Jan 21 at 20:05
$begingroup$
Ben,Thanks for your answer. I'm interested in your first Edit. How do you prove that? I know that the inclusion is continuous. But how do you prove that the inverse is also continuous?
$endgroup$
– An old man in the sea.
Jan 21 at 21:12
$begingroup$
@Anoldmaninthesea. It's pretty much by definition.
$endgroup$
– Ben W
Jan 21 at 21:16
$begingroup$
One needs that the subset has the subspace topology.
$endgroup$
– mfl
Jan 21 at 20:03
$begingroup$
One needs that the subset has the subspace topology.
$endgroup$
– mfl
Jan 21 at 20:03
$begingroup$
@mfl You mean for an inclusion map? Actually, you can have an inclusion between sets with no topological structure at all.
$endgroup$
– Ben W
Jan 21 at 20:04
$begingroup$
@mfl You mean for an inclusion map? Actually, you can have an inclusion between sets with no topological structure at all.
$endgroup$
– Ben W
Jan 21 at 20:04
$begingroup$
Yes. Your definition of inclusion map is between sets.
$endgroup$
– mfl
Jan 21 at 20:05
$begingroup$
Yes. Your definition of inclusion map is between sets.
$endgroup$
– mfl
Jan 21 at 20:05
$begingroup$
Ben,Thanks for your answer. I'm interested in your first Edit. How do you prove that? I know that the inclusion is continuous. But how do you prove that the inverse is also continuous?
$endgroup$
– An old man in the sea.
Jan 21 at 21:12
$begingroup$
Ben,Thanks for your answer. I'm interested in your first Edit. How do you prove that? I know that the inclusion is continuous. But how do you prove that the inverse is also continuous?
$endgroup$
– An old man in the sea.
Jan 21 at 21:12
$begingroup$
@Anoldmaninthesea. It's pretty much by definition.
$endgroup$
– Ben W
Jan 21 at 21:16
$begingroup$
@Anoldmaninthesea. It's pretty much by definition.
$endgroup$
– Ben W
Jan 21 at 21:16
|
show 4 more comments
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1
$begingroup$
Do you mean homeomorphism onto its image?
$endgroup$
– Randall
Jan 21 at 19:58