Does there exist a continous surjective map between $mathbb R^2to S^1$
$begingroup$
Does there exist a continuous surjective map between $mathbb R^2to S^1$?
I had find such map with domain $mathbb R^2/${$0$}.
But I do not able to find if we insert 0 there?
Any help will be appreciated
real-analysis general-topology continuity examples-counterexamples
edited Jan 6 at 8:24
Henno Brandsma
113k348123
asked Jan 6 at 4:42
MathLoverMathLover
53710
$endgroup$
add a comment |
$begingroup$
Does there exist a continuous surjective map between $mathbb R^2to S^1$?
I had find such map with domain $mathbb R^2/${$0$}.
But I do not able to find if we insert 0 there?
Any help will be appreciated
real-analysis general-topology continuity examples-counterexamples
edited Jan 6 at 8:24
Henno Brandsma
113k348123
asked Jan 6 at 4:42
MathLoverMathLover
53710
$endgroup$
$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49
add a comment |
$begingroup$
Does there exist a continuous surjective map between $mathbb R^2to S^1$?
I had find such map with domain $mathbb R^2/${$0$}.
But I do not able to find if we insert 0 there?
Any help will be appreciated
real-analysis general-topology continuity examples-counterexamples
edited Jan 6 at 8:24
Henno Brandsma
113k348123
asked Jan 6 at 4:42
MathLoverMathLover
53710
$endgroup$
Does there exist a continuous surjective map between $mathbb R^2to S^1$?
I had find such map with domain $mathbb R^2/${$0$}.
But I do not able to find if we insert 0 there?
Any help will be appreciated
real-analysis general-topology continuity examples-counterexamples
real-analysis general-topology continuity examples-counterexamples
edited Jan 6 at 8:24
Henno Brandsma
113k348123
asked Jan 6 at 4:42
MathLoverMathLover
53710
edited Jan 6 at 8:24
Henno Brandsma
113k348123
asked Jan 6 at 4:42
MathLoverMathLover
53710
edited Jan 6 at 8:24
Henno Brandsma
113k348123
edited Jan 6 at 8:24
Henno Brandsma
113k348123
edited Jan 6 at 8:24
Henno Brandsma
113k348123
113k348123
asked Jan 6 at 4:42
MathLoverMathLover
53710
asked Jan 6 at 4:42
MathLoverMathLover
53710
asked Jan 6 at 4:42
MathLoverMathLover
53710
53710
$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49
add a comment |
$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49
$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49
$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
How about $f(x,y)=(cos x,sin x)$?
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
$endgroup$
add a comment |
$begingroup$
R^2 to R projection and then from R to circle the exponential map.
answered Jan 6 at 4:46
NeelNeel
572314
$endgroup$
add a comment |
$begingroup$
Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How about $f(x,y)=(cos x,sin x)$?
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
$endgroup$
add a comment |
$begingroup$
How about $f(x,y)=(cos x,sin x)$?
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
$endgroup$
add a comment |
$begingroup$
How about $f(x,y)=(cos x,sin x)$?
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
$endgroup$
How about $f(x,y)=(cos x,sin x)$?
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
answered Jan 6 at 4:45
SmileyCraftSmileyCraft
3,749519
3,749519
add a comment |
add a comment |
$begingroup$
R^2 to R projection and then from R to circle the exponential map.
answered Jan 6 at 4:46
NeelNeel
572314
$endgroup$
add a comment |
$begingroup$
R^2 to R projection and then from R to circle the exponential map.
answered Jan 6 at 4:46
NeelNeel
572314
$endgroup$
add a comment |
$begingroup$
R^2 to R projection and then from R to circle the exponential map.
answered Jan 6 at 4:46
NeelNeel
572314
$endgroup$
R^2 to R projection and then from R to circle the exponential map.
answered Jan 6 at 4:46
NeelNeel
572314
answered Jan 6 at 4:46
NeelNeel
572314
answered Jan 6 at 4:46
NeelNeel
572314
answered Jan 6 at 4:46
NeelNeel
572314
572314
add a comment |
add a comment |
$begingroup$
Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 6 at 4:45
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
10.5k41741
add a comment |
add a comment |
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$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49