Finding fundamental groups.
0
$begingroup$
What are the fundamental groups of
$Bbb R^3 setminus {rm (two parallel lines)}$ and
$Bbb R^3 setminus {rm (two intersecting lines)}$.
How do I compute those groups?
Thank you very much.
algebraic-topology fundamental-groups
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
$endgroup$
|
show 2 more comments
0
$begingroup$
What are the fundamental groups of
$Bbb R^3 setminus {rm (two parallel lines)}$ and
$Bbb R^3 setminus {rm (two intersecting lines)}$.
How do I compute those groups?
Thank you very much.
algebraic-topology fundamental-groups
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
$endgroup$
$begingroup$
Do you know van Kampen?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:50
1
$begingroup$
Alternatively, see if you can show that the second space is homotopy equivalent to a sphere minus four points.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:51
$begingroup$
Can you show that $Bbb R^3setminus{text{the origin}}$ is homotopy equivalent to a sphere? One way is to take $f:Bbb R^3setminus{text{the origin}}to S^2$ to be projection from the origin onto the sphere and $g:S^2toBbb R^3setminus{text{the origin}}$ to be inclusion. Can you check that this works?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:03
$begingroup$
How do I find maps from one space such that their compositions is homotopic to identity maps? It's quite difficult to me. Can you please help me in this regard? For applying Van Kampen theorem I have to break the space into two spaces which are open, path-connected with simply connected intersection. If we call the intersecting lines to be $C_1$ and $C_2$ then definitely $(2)$ becomes the $(Bbb R^3 setminus C_1) cup (Bbb R^3 setminus C_2)$ which are clearly open,path-connected. Is their intersection simply-connected?
$endgroup$
– Dbchatto67
Nov 28 '18 at 7:06
$begingroup$
$(Bbb R^3setminus C_1)cup(Bbb R^3setminus C_3)$ equals $Bbb R^3setminus(C_1cap C_2)$, not $Bbb R^3setminus(C_1cup C_2)$. I encourage you to draw pictures.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:08
|
show 2 more comments
0
0
0
$begingroup$
What are the fundamental groups of
$Bbb R^3 setminus {rm (two parallel lines)}$ and
$Bbb R^3 setminus {rm (two intersecting lines)}$.
How do I compute those groups?
Thank you very much.
algebraic-topology fundamental-groups
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
$endgroup$
What are the fundamental groups of
$Bbb R^3 setminus {rm (two parallel lines)}$ and
$Bbb R^3 setminus {rm (two intersecting lines)}$.
How do I compute those groups?
Thank you very much.
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
edited Nov 28 '18 at 7:01
C.F.G
1,4251821
1,4251821
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
asked Nov 28 '18 at 6:43
Dbchatto67Dbchatto67
598116
598116
$begingroup$
Do you know van Kampen?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:50
1
$begingroup$
Alternatively, see if you can show that the second space is homotopy equivalent to a sphere minus four points.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:51
$begingroup$
Can you show that $Bbb R^3setminus{text{the origin}}$ is homotopy equivalent to a sphere? One way is to take $f:Bbb R^3setminus{text{the origin}}to S^2$ to be projection from the origin onto the sphere and $g:S^2toBbb R^3setminus{text{the origin}}$ to be inclusion. Can you check that this works?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:03
$begingroup$
How do I find maps from one space such that their compositions is homotopic to identity maps? It's quite difficult to me. Can you please help me in this regard? For applying Van Kampen theorem I have to break the space into two spaces which are open, path-connected with simply connected intersection. If we call the intersecting lines to be $C_1$ and $C_2$ then definitely $(2)$ becomes the $(Bbb R^3 setminus C_1) cup (Bbb R^3 setminus C_2)$ which are clearly open,path-connected. Is their intersection simply-connected?
$endgroup$
– Dbchatto67
Nov 28 '18 at 7:06
$begingroup$
$(Bbb R^3setminus C_1)cup(Bbb R^3setminus C_3)$ equals $Bbb R^3setminus(C_1cap C_2)$, not $Bbb R^3setminus(C_1cup C_2)$. I encourage you to draw pictures.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:08
|
show 2 more comments
$begingroup$
Do you know van Kampen?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:50
1
$begingroup$
Alternatively, see if you can show that the second space is homotopy equivalent to a sphere minus four points.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:51
$begingroup$
Can you show that $Bbb R^3setminus{text{the origin}}$ is homotopy equivalent to a sphere? One way is to take $f:Bbb R^3setminus{text{the origin}}to S^2$ to be projection from the origin onto the sphere and $g:S^2toBbb R^3setminus{text{the origin}}$ to be inclusion. Can you check that this works?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:03
$begingroup$
How do I find maps from one space such that their compositions is homotopic to identity maps? It's quite difficult to me. Can you please help me in this regard? For applying Van Kampen theorem I have to break the space into two spaces which are open, path-connected with simply connected intersection. If we call the intersecting lines to be $C_1$ and $C_2$ then definitely $(2)$ becomes the $(Bbb R^3 setminus C_1) cup (Bbb R^3 setminus C_2)$ which are clearly open,path-connected. Is their intersection simply-connected?
$endgroup$
– Dbchatto67
Nov 28 '18 at 7:06
$begingroup$
$(Bbb R^3setminus C_1)cup(Bbb R^3setminus C_3)$ equals $Bbb R^3setminus(C_1cap C_2)$, not $Bbb R^3setminus(C_1cup C_2)$. I encourage you to draw pictures.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:08
$begingroup$
Do you know van Kampen?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:50
$begingroup$
Do you know van Kampen?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:50
1
1
$begingroup$
Alternatively, see if you can show that the second space is homotopy equivalent to a sphere minus four points.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:51
$begingroup$
Alternatively, see if you can show that the second space is homotopy equivalent to a sphere minus four points.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:51
$begingroup$
Can you show that $Bbb R^3setminus{text{the origin}}$ is homotopy equivalent to a sphere? One way is to take $f:Bbb R^3setminus{text{the origin}}to S^2$ to be projection from the origin onto the sphere and $g:S^2toBbb R^3setminus{text{the origin}}$ to be inclusion. Can you check that this works?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:03
$begingroup$
Can you show that $Bbb R^3setminus{text{the origin}}$ is homotopy equivalent to a sphere? One way is to take $f:Bbb R^3setminus{text{the origin}}to S^2$ to be projection from the origin onto the sphere and $g:S^2toBbb R^3setminus{text{the origin}}$ to be inclusion. Can you check that this works?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:03
$begingroup$
How do I find maps from one space such that their compositions is homotopic to identity maps? It's quite difficult to me. Can you please help me in this regard? For applying Van Kampen theorem I have to break the space into two spaces which are open, path-connected with simply connected intersection. If we call the intersecting lines to be $C_1$ and $C_2$ then definitely $(2)$ becomes the $(Bbb R^3 setminus C_1) cup (Bbb R^3 setminus C_2)$ which are clearly open,path-connected. Is their intersection simply-connected?
$endgroup$
– Dbchatto67
Nov 28 '18 at 7:06
$begingroup$
How do I find maps from one space such that their compositions is homotopic to identity maps? It's quite difficult to me. Can you please help me in this regard? For applying Van Kampen theorem I have to break the space into two spaces which are open, path-connected with simply connected intersection. If we call the intersecting lines to be $C_1$ and $C_2$ then definitely $(2)$ becomes the $(Bbb R^3 setminus C_1) cup (Bbb R^3 setminus C_2)$ which are clearly open,path-connected. Is their intersection simply-connected?
$endgroup$
– Dbchatto67
Nov 28 '18 at 7:06
$begingroup$
$(Bbb R^3setminus C_1)cup(Bbb R^3setminus C_3)$ equals $Bbb R^3setminus(C_1cap C_2)$, not $Bbb R^3setminus(C_1cup C_2)$. I encourage you to draw pictures.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:08
$begingroup$
$(Bbb R^3setminus C_1)cup(Bbb R^3setminus C_3)$ equals $Bbb R^3setminus(C_1cap C_2)$, not $Bbb R^3setminus(C_1cup C_2)$. I encourage you to draw pictures.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:08
|
show 2 more comments
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$begingroup$
Do you know van Kampen?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:50
1
$begingroup$
Alternatively, see if you can show that the second space is homotopy equivalent to a sphere minus four points.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 6:51
$begingroup$
Can you show that $Bbb R^3setminus{text{the origin}}$ is homotopy equivalent to a sphere? One way is to take $f:Bbb R^3setminus{text{the origin}}to S^2$ to be projection from the origin onto the sphere and $g:S^2toBbb R^3setminus{text{the origin}}$ to be inclusion. Can you check that this works?
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:03
$begingroup$
How do I find maps from one space such that their compositions is homotopic to identity maps? It's quite difficult to me. Can you please help me in this regard? For applying Van Kampen theorem I have to break the space into two spaces which are open, path-connected with simply connected intersection. If we call the intersecting lines to be $C_1$ and $C_2$ then definitely $(2)$ becomes the $(Bbb R^3 setminus C_1) cup (Bbb R^3 setminus C_2)$ which are clearly open,path-connected. Is their intersection simply-connected?
$endgroup$
– Dbchatto67
Nov 28 '18 at 7:06
$begingroup$
$(Bbb R^3setminus C_1)cup(Bbb R^3setminus C_3)$ equals $Bbb R^3setminus(C_1cap C_2)$, not $Bbb R^3setminus(C_1cup C_2)$. I encourage you to draw pictures.
$endgroup$
– Akiva Weinberger
Nov 28 '18 at 7:08