Under what conditions two spaces that are homeomorphic with a point removed are homeomorphic Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Homeomorphism vs. Homotopy (Equivalence)Constructablilty of regular polygons on a sphereShowing some spaces are locally compact and show their 1-point compactification is homeomorphicComparing different topological spaces regarding homeomorphisms and fundamental groups.Is there a general way to tell whether two topological spaces are homeomorphic?Construct a homeomorphis between two spacesDoes the “Number Doughnut” make sense?Two manifolds $X$ and $Y$ with two bijective continuous functions from $X$ to $Y$ and $Y$ to $X$, but not homeomorphicOpen unit disc is homeomorphic to sphere minus a pointThe set $zin mathbb C: $ with point at infinity is homeomorphic to closed unit disk.
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Under what conditions two spaces that are homeomorphic with a point removed are homeomorphic
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Homeomorphism vs. Homotopy (Equivalence)Constructablilty of regular polygons on a sphereShowing some spaces are locally compact and show their 1-point compactification is homeomorphicComparing different topological spaces regarding homeomorphisms and fundamental groups.Is there a general way to tell whether two topological spaces are homeomorphic?Construct a homeomorphis between two spacesDoes the “Number Doughnut” make sense?Two manifolds $X$ and $Y$ with two bijective continuous functions from $X$ to $Y$ and $Y$ to $X$, but not homeomorphicOpen unit disc is homeomorphic to sphere minus a pointThe set $z$ with point at infinity is homeomorphic to closed unit disk.
$begingroup$
I'm trying to state crystal clear that the extended complex plane $widehatmathbb C$ is homeomorphic to the sphere $S^2$ through the stereographic projection. Of course in this case it easy to see that the two spaces $S^2-(0,0,1)$ and $mathbb C$ are homeomorphic. But what is the theorem that states that then $S^2$ and $widehatmathbb C$ are homeomorphic. I know it's a well known theorem but I'd like to have the complete details.
general-topology algebraic-topology
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
$endgroup$
|
show 8 more comments
$begingroup$
I'm trying to state crystal clear that the extended complex plane $widehatmathbb C$ is homeomorphic to the sphere $S^2$ through the stereographic projection. Of course in this case it easy to see that the two spaces $S^2-(0,0,1)$ and $mathbb C$ are homeomorphic. But what is the theorem that states that then $S^2$ and $widehatmathbb C$ are homeomorphic. I know it's a well known theorem but I'd like to have the complete details.
general-topology algebraic-topology
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
$endgroup$
1
$begingroup$
Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension
$endgroup$
– Baran Zadeoglu
Apr 2 at 9:28
$begingroup$
Have you not just written down an explicit homeomorphism using the stereographic projection maps?
$endgroup$
– Tyrone
Apr 2 at 9:37
$begingroup$
@Tyrone yes I did, I don't have problem there. My point is that the homeomorphism is between the two spaces removing one point (namely the north pole in $s^2$ and infinity in the complex extended palne. I want to know how can I prove that adding these two point the space are necessarely homeomorphic
$endgroup$
– Dac0
Apr 2 at 9:40
1
$begingroup$
I mean write down an explicit homeomorphism between the two spaces without removing a point. It seems like you have correctly defined the maps on all but one point. Can you extend your defintion over the final point?
$endgroup$
– Tyrone
Apr 2 at 9:42
$begingroup$
@Tyrone You are right, but then I have to be sure of the continuity of the map in that point... so if there was already a theorem it would be already done
$endgroup$
– Dac0
Apr 2 at 9:49
|
show 8 more comments
$begingroup$
I'm trying to state crystal clear that the extended complex plane $widehatmathbb C$ is homeomorphic to the sphere $S^2$ through the stereographic projection. Of course in this case it easy to see that the two spaces $S^2-(0,0,1)$ and $mathbb C$ are homeomorphic. But what is the theorem that states that then $S^2$ and $widehatmathbb C$ are homeomorphic. I know it's a well known theorem but I'd like to have the complete details.
general-topology algebraic-topology
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
$endgroup$
I'm trying to state crystal clear that the extended complex plane $widehatmathbb C$ is homeomorphic to the sphere $S^2$ through the stereographic projection. Of course in this case it easy to see that the two spaces $S^2-(0,0,1)$ and $mathbb C$ are homeomorphic. But what is the theorem that states that then $S^2$ and $widehatmathbb C$ are homeomorphic. I know it's a well known theorem but I'd like to have the complete details.
general-topology algebraic-topology
general-topology algebraic-topology
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
edited Apr 2 at 11:17
Dac0
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
asked Apr 2 at 9:25
Dac0Dac0
6,0921937
6,0921937
1
$begingroup$
Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension
$endgroup$
– Baran Zadeoglu
Apr 2 at 9:28
$begingroup$
Have you not just written down an explicit homeomorphism using the stereographic projection maps?
$endgroup$
– Tyrone
Apr 2 at 9:37
$begingroup$
@Tyrone yes I did, I don't have problem there. My point is that the homeomorphism is between the two spaces removing one point (namely the north pole in $s^2$ and infinity in the complex extended palne. I want to know how can I prove that adding these two point the space are necessarely homeomorphic
$endgroup$
– Dac0
Apr 2 at 9:40
1
$begingroup$
I mean write down an explicit homeomorphism between the two spaces without removing a point. It seems like you have correctly defined the maps on all but one point. Can you extend your defintion over the final point?
$endgroup$
– Tyrone
Apr 2 at 9:42
$begingroup$
@Tyrone You are right, but then I have to be sure of the continuity of the map in that point... so if there was already a theorem it would be already done
$endgroup$
– Dac0
Apr 2 at 9:49
|
show 8 more comments
1
$begingroup$
Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension
$endgroup$
– Baran Zadeoglu
Apr 2 at 9:28
$begingroup$
Have you not just written down an explicit homeomorphism using the stereographic projection maps?
$endgroup$
– Tyrone
Apr 2 at 9:37
$begingroup$
@Tyrone yes I did, I don't have problem there. My point is that the homeomorphism is between the two spaces removing one point (namely the north pole in $s^2$ and infinity in the complex extended palne. I want to know how can I prove that adding these two point the space are necessarely homeomorphic
$endgroup$
– Dac0
Apr 2 at 9:40
1
$begingroup$
I mean write down an explicit homeomorphism between the two spaces without removing a point. It seems like you have correctly defined the maps on all but one point. Can you extend your defintion over the final point?
$endgroup$
– Tyrone
Apr 2 at 9:42
$begingroup$
@Tyrone You are right, but then I have to be sure of the continuity of the map in that point... so if there was already a theorem it would be already done
$endgroup$
– Dac0
Apr 2 at 9:49
1
1
$begingroup$
Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension
$endgroup$
– Baran Zadeoglu
Apr 2 at 9:28
$begingroup$
Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension
$endgroup$
– Baran Zadeoglu
Apr 2 at 9:28
$begingroup$
Have you not just written down an explicit homeomorphism using the stereographic projection maps?
$endgroup$
– Tyrone
Apr 2 at 9:37
$begingroup$
Have you not just written down an explicit homeomorphism using the stereographic projection maps?
$endgroup$
– Tyrone
Apr 2 at 9:37
$begingroup$
@Tyrone yes I did, I don't have problem there. My point is that the homeomorphism is between the two spaces removing one point (namely the north pole in $s^2$ and infinity in the complex extended palne. I want to know how can I prove that adding these two point the space are necessarely homeomorphic
$endgroup$
– Dac0
Apr 2 at 9:40
$begingroup$
@Tyrone yes I did, I don't have problem there. My point is that the homeomorphism is between the two spaces removing one point (namely the north pole in $s^2$ and infinity in the complex extended palne. I want to know how can I prove that adding these two point the space are necessarely homeomorphic
$endgroup$
– Dac0
Apr 2 at 9:40
1
1
$begingroup$
I mean write down an explicit homeomorphism between the two spaces without removing a point. It seems like you have correctly defined the maps on all but one point. Can you extend your defintion over the final point?
$endgroup$
– Tyrone
Apr 2 at 9:42
$begingroup$
I mean write down an explicit homeomorphism between the two spaces without removing a point. It seems like you have correctly defined the maps on all but one point. Can you extend your defintion over the final point?
$endgroup$
– Tyrone
Apr 2 at 9:42
$begingroup$
@Tyrone You are right, but then I have to be sure of the continuity of the map in that point... so if there was already a theorem it would be already done
$endgroup$
– Dac0
Apr 2 at 9:49
$begingroup$
@Tyrone You are right, but then I have to be sure of the continuity of the map in that point... so if there was already a theorem it would be already done
$endgroup$
– Dac0
Apr 2 at 9:49
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The extended complex plane $widehatmathbb C$ is defined by adjoining to $mathbb C $ an additional point at infinity, that is $widehatmathbb C = mathbb C cup infty $. Algebraic operations are defined in the obvious way. However, this does not automatically provide a topology on $widehatmathbb C$.
The standard topological model of the extended complex plane $widehatmathbb C$ is the Riemann sphere $S^2 subset mathbbR^3$. In fact, many authors use the stereographic projection $p : S^2 setminus (0,0,1) to mathbb C$ to introduce $widehatmathbb C$. This map is a homeomorphism, and $i = p^-1$ embeds $mathbb C$ as an open subset into $S^2$. Clearly $i$ extends to a bijection $h : widehatmathbb C to S^2$ by defining $h(infty) = (0,0,1)$. Then $h$ induces a unique topology on $widehatmathbb C$ making $h$ a homeomorphism. With this topology $widehatmathbb C$ is a compact metrizable space and the subspace $mathbb C$ receives its original topology.
If you use this construction as the definition of $widehatmathbb C$ as a topological space, then nothing remains to be shown.
On the other hand, it suggests itself to define the space $widehatmathbb C$ as the Alexandroff compactification of $mathbb C$. Open neighborhoods of $infty$ are the complements of compact subsets of $mathbb C$.
Here are some well-known facts.
(1) The Alexandroff compactification of a space $X$ is a compact Hausdorff space if and only if $X$ is a locally compact Hausdorff space.
(2) For any two embeddings $i_1: X to C_1, i_2: X to C_2$ of a locally compact Hausdorff space $X$ into compact Hausdorff spaces $C_k$ such that $C_k setminus i_k(X)$ is a one-point set, there exists a unique homeomorphism $g : C_1 to C_2$ such that $gi_1 = i_2$.
If we apply this to $i : mathbb C to S^2$ and $mathbb C hookrightarrow widehatmathbb C$, we get the desired homeomorphism.
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
$endgroup$
add a comment |
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$begingroup$
The extended complex plane $widehatmathbb C$ is defined by adjoining to $mathbb C $ an additional point at infinity, that is $widehatmathbb C = mathbb C cup infty $. Algebraic operations are defined in the obvious way. However, this does not automatically provide a topology on $widehatmathbb C$.
The standard topological model of the extended complex plane $widehatmathbb C$ is the Riemann sphere $S^2 subset mathbbR^3$. In fact, many authors use the stereographic projection $p : S^2 setminus (0,0,1) to mathbb C$ to introduce $widehatmathbb C$. This map is a homeomorphism, and $i = p^-1$ embeds $mathbb C$ as an open subset into $S^2$. Clearly $i$ extends to a bijection $h : widehatmathbb C to S^2$ by defining $h(infty) = (0,0,1)$. Then $h$ induces a unique topology on $widehatmathbb C$ making $h$ a homeomorphism. With this topology $widehatmathbb C$ is a compact metrizable space and the subspace $mathbb C$ receives its original topology.
If you use this construction as the definition of $widehatmathbb C$ as a topological space, then nothing remains to be shown.
On the other hand, it suggests itself to define the space $widehatmathbb C$ as the Alexandroff compactification of $mathbb C$. Open neighborhoods of $infty$ are the complements of compact subsets of $mathbb C$.
Here are some well-known facts.
(1) The Alexandroff compactification of a space $X$ is a compact Hausdorff space if and only if $X$ is a locally compact Hausdorff space.
(2) For any two embeddings $i_1: X to C_1, i_2: X to C_2$ of a locally compact Hausdorff space $X$ into compact Hausdorff spaces $C_k$ such that $C_k setminus i_k(X)$ is a one-point set, there exists a unique homeomorphism $g : C_1 to C_2$ such that $gi_1 = i_2$.
If we apply this to $i : mathbb C to S^2$ and $mathbb C hookrightarrow widehatmathbb C$, we get the desired homeomorphism.
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
$endgroup$
add a comment |
$begingroup$
The extended complex plane $widehatmathbb C$ is defined by adjoining to $mathbb C $ an additional point at infinity, that is $widehatmathbb C = mathbb C cup infty $. Algebraic operations are defined in the obvious way. However, this does not automatically provide a topology on $widehatmathbb C$.
The standard topological model of the extended complex plane $widehatmathbb C$ is the Riemann sphere $S^2 subset mathbbR^3$. In fact, many authors use the stereographic projection $p : S^2 setminus (0,0,1) to mathbb C$ to introduce $widehatmathbb C$. This map is a homeomorphism, and $i = p^-1$ embeds $mathbb C$ as an open subset into $S^2$. Clearly $i$ extends to a bijection $h : widehatmathbb C to S^2$ by defining $h(infty) = (0,0,1)$. Then $h$ induces a unique topology on $widehatmathbb C$ making $h$ a homeomorphism. With this topology $widehatmathbb C$ is a compact metrizable space and the subspace $mathbb C$ receives its original topology.
If you use this construction as the definition of $widehatmathbb C$ as a topological space, then nothing remains to be shown.
On the other hand, it suggests itself to define the space $widehatmathbb C$ as the Alexandroff compactification of $mathbb C$. Open neighborhoods of $infty$ are the complements of compact subsets of $mathbb C$.
Here are some well-known facts.
(1) The Alexandroff compactification of a space $X$ is a compact Hausdorff space if and only if $X$ is a locally compact Hausdorff space.
(2) For any two embeddings $i_1: X to C_1, i_2: X to C_2$ of a locally compact Hausdorff space $X$ into compact Hausdorff spaces $C_k$ such that $C_k setminus i_k(X)$ is a one-point set, there exists a unique homeomorphism $g : C_1 to C_2$ such that $gi_1 = i_2$.
If we apply this to $i : mathbb C to S^2$ and $mathbb C hookrightarrow widehatmathbb C$, we get the desired homeomorphism.
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
$endgroup$
add a comment |
$begingroup$
The extended complex plane $widehatmathbb C$ is defined by adjoining to $mathbb C $ an additional point at infinity, that is $widehatmathbb C = mathbb C cup infty $. Algebraic operations are defined in the obvious way. However, this does not automatically provide a topology on $widehatmathbb C$.
The standard topological model of the extended complex plane $widehatmathbb C$ is the Riemann sphere $S^2 subset mathbbR^3$. In fact, many authors use the stereographic projection $p : S^2 setminus (0,0,1) to mathbb C$ to introduce $widehatmathbb C$. This map is a homeomorphism, and $i = p^-1$ embeds $mathbb C$ as an open subset into $S^2$. Clearly $i$ extends to a bijection $h : widehatmathbb C to S^2$ by defining $h(infty) = (0,0,1)$. Then $h$ induces a unique topology on $widehatmathbb C$ making $h$ a homeomorphism. With this topology $widehatmathbb C$ is a compact metrizable space and the subspace $mathbb C$ receives its original topology.
If you use this construction as the definition of $widehatmathbb C$ as a topological space, then nothing remains to be shown.
On the other hand, it suggests itself to define the space $widehatmathbb C$ as the Alexandroff compactification of $mathbb C$. Open neighborhoods of $infty$ are the complements of compact subsets of $mathbb C$.
Here are some well-known facts.
(1) The Alexandroff compactification of a space $X$ is a compact Hausdorff space if and only if $X$ is a locally compact Hausdorff space.
(2) For any two embeddings $i_1: X to C_1, i_2: X to C_2$ of a locally compact Hausdorff space $X$ into compact Hausdorff spaces $C_k$ such that $C_k setminus i_k(X)$ is a one-point set, there exists a unique homeomorphism $g : C_1 to C_2$ such that $gi_1 = i_2$.
If we apply this to $i : mathbb C to S^2$ and $mathbb C hookrightarrow widehatmathbb C$, we get the desired homeomorphism.
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
$endgroup$
The extended complex plane $widehatmathbb C$ is defined by adjoining to $mathbb C $ an additional point at infinity, that is $widehatmathbb C = mathbb C cup infty $. Algebraic operations are defined in the obvious way. However, this does not automatically provide a topology on $widehatmathbb C$.
The standard topological model of the extended complex plane $widehatmathbb C$ is the Riemann sphere $S^2 subset mathbbR^3$. In fact, many authors use the stereographic projection $p : S^2 setminus (0,0,1) to mathbb C$ to introduce $widehatmathbb C$. This map is a homeomorphism, and $i = p^-1$ embeds $mathbb C$ as an open subset into $S^2$. Clearly $i$ extends to a bijection $h : widehatmathbb C to S^2$ by defining $h(infty) = (0,0,1)$. Then $h$ induces a unique topology on $widehatmathbb C$ making $h$ a homeomorphism. With this topology $widehatmathbb C$ is a compact metrizable space and the subspace $mathbb C$ receives its original topology.
If you use this construction as the definition of $widehatmathbb C$ as a topological space, then nothing remains to be shown.
On the other hand, it suggests itself to define the space $widehatmathbb C$ as the Alexandroff compactification of $mathbb C$. Open neighborhoods of $infty$ are the complements of compact subsets of $mathbb C$.
Here are some well-known facts.
(1) The Alexandroff compactification of a space $X$ is a compact Hausdorff space if and only if $X$ is a locally compact Hausdorff space.
(2) For any two embeddings $i_1: X to C_1, i_2: X to C_2$ of a locally compact Hausdorff space $X$ into compact Hausdorff spaces $C_k$ such that $C_k setminus i_k(X)$ is a one-point set, there exists a unique homeomorphism $g : C_1 to C_2$ such that $gi_1 = i_2$.
If we apply this to $i : mathbb C to S^2$ and $mathbb C hookrightarrow widehatmathbb C$, we get the desired homeomorphism.
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
answered Apr 2 at 12:44
Paul FrostPaul Frost
12.9k31035
12.9k31035
add a comment |
add a comment |
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$begingroup$
Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension
$endgroup$
– Baran Zadeoglu
Apr 2 at 9:28
$begingroup$
Have you not just written down an explicit homeomorphism using the stereographic projection maps?
$endgroup$
– Tyrone
Apr 2 at 9:37
$begingroup$
@Tyrone yes I did, I don't have problem there. My point is that the homeomorphism is between the two spaces removing one point (namely the north pole in $s^2$ and infinity in the complex extended palne. I want to know how can I prove that adding these two point the space are necessarely homeomorphic
$endgroup$
– Dac0
Apr 2 at 9:40
1
$begingroup$
I mean write down an explicit homeomorphism between the two spaces without removing a point. It seems like you have correctly defined the maps on all but one point. Can you extend your defintion over the final point?
$endgroup$
– Tyrone
Apr 2 at 9:42
$begingroup$
@Tyrone You are right, but then I have to be sure of the continuity of the map in that point... so if there was already a theorem it would be already done
$endgroup$
– Dac0
Apr 2 at 9:49