Embedding vs continuous injection (in topological vector spaces)Are continuous self-bijections of connected...
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Embedding vs continuous injection (in topological vector spaces)
Are continuous self-bijections of connected spaces homeomorphisms?If we have an embedding $f:X rightarrow A$, where $A subset Y$, do we have to show $f^{-1}$ is continuous?Why topological embedding continuous?Can every Hausdorff topological space be homeomorphically embedded in a topological vector space?An example of a space $X$ which doesn't embed in $mathbb{R}^n$ for any $n$?Show that a function is an embeddingA continuous bijection between finite topological spaces which is not a homeomorphismAre locally injective maps between topological manifolds topological immersions?Definition of Topological Embeddingtopology of continuous functions
$begingroup$
When working with topological vector spaces (say $X,Y$), the term “embedding” is often used for a continuous injection $f:Xrightarrow Y$.
Now, $f$
is of course a bijection onto its image, but it's not necessarily a homeomorphism (as we would normally require of an embedding in topology).
Can anyone explain to me exactly what we're loosing by not having $X$
and $f(X)$ homeomorphic? And is there something special about linear spaces that makes whatever we're missing (more or less) irrelevant?
I suppose that the subspace topology on $f(X)$
must in some sense be coarser than the topology on $X$
if the inverse of $f$
(defined on the range of $f$
) fails to be continuous
general-topology functional-analysis
edited Mar 19 at 13:48
John Hughes
65.2k24293
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
$endgroup$
add a comment |
$begingroup$
When working with topological vector spaces (say $X,Y$), the term “embedding” is often used for a continuous injection $f:Xrightarrow Y$.
Now, $f$
is of course a bijection onto its image, but it's not necessarily a homeomorphism (as we would normally require of an embedding in topology).
Can anyone explain to me exactly what we're loosing by not having $X$
and $f(X)$ homeomorphic? And is there something special about linear spaces that makes whatever we're missing (more or less) irrelevant?
I suppose that the subspace topology on $f(X)$
must in some sense be coarser than the topology on $X$
if the inverse of $f$
(defined on the range of $f$
) fails to be continuous
general-topology functional-analysis
edited Mar 19 at 13:48
John Hughes
65.2k24293
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
$endgroup$
$begingroup$
in case of topological space if $X$ is compact or $f$ is a proper map or a closed map then continuous injection is an embedding
$endgroup$
– Anubhav Mukherjee
Mar 7 '15 at 12:47
$begingroup$
Where does the algebraic structure becomes involved here? (It seems the question is about merely topological spaces.)
$endgroup$
– C-Star-Puppy
Mar 8 '15 at 20:08
add a comment |
$begingroup$
When working with topological vector spaces (say $X,Y$), the term “embedding” is often used for a continuous injection $f:Xrightarrow Y$.
Now, $f$
is of course a bijection onto its image, but it's not necessarily a homeomorphism (as we would normally require of an embedding in topology).
Can anyone explain to me exactly what we're loosing by not having $X$
and $f(X)$ homeomorphic? And is there something special about linear spaces that makes whatever we're missing (more or less) irrelevant?
I suppose that the subspace topology on $f(X)$
must in some sense be coarser than the topology on $X$
if the inverse of $f$
(defined on the range of $f$
) fails to be continuous
general-topology functional-analysis
edited Mar 19 at 13:48
John Hughes
65.2k24293
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
$endgroup$
When working with topological vector spaces (say $X,Y$), the term “embedding” is often used for a continuous injection $f:Xrightarrow Y$.
Now, $f$
is of course a bijection onto its image, but it's not necessarily a homeomorphism (as we would normally require of an embedding in topology).
Can anyone explain to me exactly what we're loosing by not having $X$
and $f(X)$ homeomorphic? And is there something special about linear spaces that makes whatever we're missing (more or less) irrelevant?
I suppose that the subspace topology on $f(X)$
must in some sense be coarser than the topology on $X$
if the inverse of $f$
(defined on the range of $f$
) fails to be continuous
general-topology functional-analysis
general-topology functional-analysis
edited Mar 19 at 13:48
John Hughes
65.2k24293
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
edited Mar 19 at 13:48
John Hughes
65.2k24293
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
edited Mar 19 at 13:48
John Hughes
65.2k24293
edited Mar 19 at 13:48
John Hughes
65.2k24293
edited Mar 19 at 13:48
John Hughes
65.2k24293
65.2k24293
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
asked Mar 7 '15 at 12:21
ProbStudentProbStudent
1248
1248
$begingroup$
in case of topological space if $X$ is compact or $f$ is a proper map or a closed map then continuous injection is an embedding
$endgroup$
– Anubhav Mukherjee
Mar 7 '15 at 12:47
$begingroup$
Where does the algebraic structure becomes involved here? (It seems the question is about merely topological spaces.)
$endgroup$
– C-Star-Puppy
Mar 8 '15 at 20:08
add a comment |
$begingroup$
in case of topological space if $X$ is compact or $f$ is a proper map or a closed map then continuous injection is an embedding
$endgroup$
– Anubhav Mukherjee
Mar 7 '15 at 12:47
$begingroup$
Where does the algebraic structure becomes involved here? (It seems the question is about merely topological spaces.)
$endgroup$
– C-Star-Puppy
Mar 8 '15 at 20:08
$begingroup$
in case of topological space if $X$ is compact or $f$ is a proper map or a closed map then continuous injection is an embedding
$endgroup$
– Anubhav Mukherjee
Mar 7 '15 at 12:47
$begingroup$
in case of topological space if $X$ is compact or $f$ is a proper map or a closed map then continuous injection is an embedding
$endgroup$
– Anubhav Mukherjee
Mar 7 '15 at 12:47
$begingroup$
Where does the algebraic structure becomes involved here? (It seems the question is about merely topological spaces.)
$endgroup$
– C-Star-Puppy
Mar 8 '15 at 20:08
$begingroup$
Where does the algebraic structure becomes involved here? (It seems the question is about merely topological spaces.)
$endgroup$
– C-Star-Puppy
Mar 8 '15 at 20:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Think of the torus as the quotient of $mathbb R^2$ in the usual way: $(x, y) mapsto (x bmod 1, y bmod 1)$.
Now look at the line $y = pi x$ in the plane. Under this quotient, it becomes a curve in the torus, and the map is 1-1 and continuous, but locally, in the image, it doesn't look the way you expect, i.e., like a line in $mathbb R^2$, where near the line there is "no other stuff".
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
$endgroup$
add a comment |
$begingroup$
As $X$ and $f(X)$ are not homeomorphic, if we identify $X$ with $f(X)$, the topology on $X$ is not the same as the subspace topology, so that one has to deal with the topology on $X$ seperately. This is important e.g. in the difference between immersed and embedded submanifolds.
$endgroup$
add a comment |
$begingroup$
Both refer to induced structures...
Subspace* and embedding:
$$iota:ShookrightarrowOmega:quadmathcal{T}(S)={Usubseteq S:Uiniota^{-1}mathcal{T}(Omega)}$$
(The embedding being a homeomorphism onto its image.)
Projection and quotient space:
$$pi:OmegatwoheadrightarrowtildeOmega:quadpi^{-1}mathcal{T}(tildeOmega)={tilde{U}subseteqtildeOmega:pi^{-1}tilde{U}inmathcal{T}(Omega)}$$
(The projection being an open continuous map.)
...There are many more considerable structures.
*Subspace in the general sense: $SnsubseteqOmega$
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Think of the torus as the quotient of $mathbb R^2$ in the usual way: $(x, y) mapsto (x bmod 1, y bmod 1)$.
Now look at the line $y = pi x$ in the plane. Under this quotient, it becomes a curve in the torus, and the map is 1-1 and continuous, but locally, in the image, it doesn't look the way you expect, i.e., like a line in $mathbb R^2$, where near the line there is "no other stuff".
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
$endgroup$
add a comment |
$begingroup$
Think of the torus as the quotient of $mathbb R^2$ in the usual way: $(x, y) mapsto (x bmod 1, y bmod 1)$.
Now look at the line $y = pi x$ in the plane. Under this quotient, it becomes a curve in the torus, and the map is 1-1 and continuous, but locally, in the image, it doesn't look the way you expect, i.e., like a line in $mathbb R^2$, where near the line there is "no other stuff".
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
$endgroup$
add a comment |
$begingroup$
Think of the torus as the quotient of $mathbb R^2$ in the usual way: $(x, y) mapsto (x bmod 1, y bmod 1)$.
Now look at the line $y = pi x$ in the plane. Under this quotient, it becomes a curve in the torus, and the map is 1-1 and continuous, but locally, in the image, it doesn't look the way you expect, i.e., like a line in $mathbb R^2$, where near the line there is "no other stuff".
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
$endgroup$
Think of the torus as the quotient of $mathbb R^2$ in the usual way: $(x, y) mapsto (x bmod 1, y bmod 1)$.
Now look at the line $y = pi x$ in the plane. Under this quotient, it becomes a curve in the torus, and the map is 1-1 and continuous, but locally, in the image, it doesn't look the way you expect, i.e., like a line in $mathbb R^2$, where near the line there is "no other stuff".
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
edited Mar 22 '18 at 14:13
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
answered Mar 7 '15 at 12:51
John HughesJohn Hughes
65.2k24293
65.2k24293
add a comment |
add a comment |
$begingroup$
As $X$ and $f(X)$ are not homeomorphic, if we identify $X$ with $f(X)$, the topology on $X$ is not the same as the subspace topology, so that one has to deal with the topology on $X$ seperately. This is important e.g. in the difference between immersed and embedded submanifolds.
$endgroup$
add a comment |
$begingroup$
As $X$ and $f(X)$ are not homeomorphic, if we identify $X$ with $f(X)$, the topology on $X$ is not the same as the subspace topology, so that one has to deal with the topology on $X$ seperately. This is important e.g. in the difference between immersed and embedded submanifolds.
$endgroup$
add a comment |
$begingroup$
As $X$ and $f(X)$ are not homeomorphic, if we identify $X$ with $f(X)$, the topology on $X$ is not the same as the subspace topology, so that one has to deal with the topology on $X$ seperately. This is important e.g. in the difference between immersed and embedded submanifolds.
$endgroup$
As $X$ and $f(X)$ are not homeomorphic, if we identify $X$ with $f(X)$, the topology on $X$ is not the same as the subspace topology, so that one has to deal with the topology on $X$ seperately. This is important e.g. in the difference between immersed and embedded submanifolds.
answered Mar 7 '15 at 12:34
user220467
add a comment |
add a comment |
$begingroup$
Both refer to induced structures...
Subspace* and embedding:
$$iota:ShookrightarrowOmega:quadmathcal{T}(S)={Usubseteq S:Uiniota^{-1}mathcal{T}(Omega)}$$
(The embedding being a homeomorphism onto its image.)
Projection and quotient space:
$$pi:OmegatwoheadrightarrowtildeOmega:quadpi^{-1}mathcal{T}(tildeOmega)={tilde{U}subseteqtildeOmega:pi^{-1}tilde{U}inmathcal{T}(Omega)}$$
(The projection being an open continuous map.)
...There are many more considerable structures.
*Subspace in the general sense: $SnsubseteqOmega$
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
$endgroup$
add a comment |
$begingroup$
Both refer to induced structures...
Subspace* and embedding:
$$iota:ShookrightarrowOmega:quadmathcal{T}(S)={Usubseteq S:Uiniota^{-1}mathcal{T}(Omega)}$$
(The embedding being a homeomorphism onto its image.)
Projection and quotient space:
$$pi:OmegatwoheadrightarrowtildeOmega:quadpi^{-1}mathcal{T}(tildeOmega)={tilde{U}subseteqtildeOmega:pi^{-1}tilde{U}inmathcal{T}(Omega)}$$
(The projection being an open continuous map.)
...There are many more considerable structures.
*Subspace in the general sense: $SnsubseteqOmega$
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
$endgroup$
add a comment |
$begingroup$
Both refer to induced structures...
Subspace* and embedding:
$$iota:ShookrightarrowOmega:quadmathcal{T}(S)={Usubseteq S:Uiniota^{-1}mathcal{T}(Omega)}$$
(The embedding being a homeomorphism onto its image.)
Projection and quotient space:
$$pi:OmegatwoheadrightarrowtildeOmega:quadpi^{-1}mathcal{T}(tildeOmega)={tilde{U}subseteqtildeOmega:pi^{-1}tilde{U}inmathcal{T}(Omega)}$$
(The projection being an open continuous map.)
...There are many more considerable structures.
*Subspace in the general sense: $SnsubseteqOmega$
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
$endgroup$
Both refer to induced structures...
Subspace* and embedding:
$$iota:ShookrightarrowOmega:quadmathcal{T}(S)={Usubseteq S:Uiniota^{-1}mathcal{T}(Omega)}$$
(The embedding being a homeomorphism onto its image.)
Projection and quotient space:
$$pi:OmegatwoheadrightarrowtildeOmega:quadpi^{-1}mathcal{T}(tildeOmega)={tilde{U}subseteqtildeOmega:pi^{-1}tilde{U}inmathcal{T}(Omega)}$$
(The projection being an open continuous map.)
...There are many more considerable structures.
*Subspace in the general sense: $SnsubseteqOmega$
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
edited Mar 8 '15 at 21:09
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
answered Mar 8 '15 at 20:47
C-Star-PuppyC-Star-Puppy
8,35642165
8,35642165
add a comment |
add a comment |
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$begingroup$
in case of topological space if $X$ is compact or $f$ is a proper map or a closed map then continuous injection is an embedding
$endgroup$
– Anubhav Mukherjee
Mar 7 '15 at 12:47
$begingroup$
Where does the algebraic structure becomes involved here? (It seems the question is about merely topological spaces.)
$endgroup$
– C-Star-Puppy
Mar 8 '15 at 20:08