How to show that $p(x)=(cos2pi x,sin2pi x)$ is a homeomorphism?
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Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$
How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.
general-topology
add a comment |Â
up vote
0
down vote
favorite
Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$
How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.
general-topology
Do you mean $S^1-(1,0)$?
â Lord Shark the Unknown
Aug 30 at 6:22
It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
â Uncool
Aug 30 at 6:24
No, it isn't. What is $p(1/4)$ according to your definition?
â Lord Shark the Unknown
Aug 30 at 6:25
Oops! Yes I mean (1,0)
â Uncool
Aug 30 at 6:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$
How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.
general-topology
Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$
How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.
general-topology
general-topology
edited Aug 30 at 6:22
Lord Shark the Unknown
88.8k955115
88.8k955115
asked Aug 30 at 6:16
Uncool
7417
7417
Do you mean $S^1-(1,0)$?
â Lord Shark the Unknown
Aug 30 at 6:22
It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
â Uncool
Aug 30 at 6:24
No, it isn't. What is $p(1/4)$ according to your definition?
â Lord Shark the Unknown
Aug 30 at 6:25
Oops! Yes I mean (1,0)
â Uncool
Aug 30 at 6:30
add a comment |Â
Do you mean $S^1-(1,0)$?
â Lord Shark the Unknown
Aug 30 at 6:22
It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
â Uncool
Aug 30 at 6:24
No, it isn't. What is $p(1/4)$ according to your definition?
â Lord Shark the Unknown
Aug 30 at 6:25
Oops! Yes I mean (1,0)
â Uncool
Aug 30 at 6:30
Do you mean $S^1-(1,0)$?
â Lord Shark the Unknown
Aug 30 at 6:22
Do you mean $S^1-(1,0)$?
â Lord Shark the Unknown
Aug 30 at 6:22
It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
â Uncool
Aug 30 at 6:24
It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
â Uncool
Aug 30 at 6:24
No, it isn't. What is $p(1/4)$ according to your definition?
â Lord Shark the Unknown
Aug 30 at 6:25
No, it isn't. What is $p(1/4)$ according to your definition?
â Lord Shark the Unknown
Aug 30 at 6:25
Oops! Yes I mean (1,0)
â Uncool
Aug 30 at 6:30
Oops! Yes I mean (1,0)
â Uncool
Aug 30 at 6:30
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.
Not sure why you give me this. I want the above function
â Uncool
Aug 30 at 6:25
It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
â Amrat A
Aug 30 at 6:27
Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
â Amrat A
Aug 30 at 6:32
Oa.. ok I'll try
â Uncool
Aug 30 at 6:32
add a comment |Â
up vote
1
down vote
We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.
1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
$
2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$
3) Both function $P$ and its inverse function are continuous.
Thus $P$ it is a homeomorphism.
add a comment |Â
up vote
0
down vote
It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.
What about the inverse? How to show that it's continuous and before that how to define it.
â Uncool
Aug 30 at 7:17
In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
â Hugo
Aug 30 at 8:22
Since it is equivalent, we can from left to the right ,and right to the left.
â Hugo
Aug 30 at 8:27
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.
Not sure why you give me this. I want the above function
â Uncool
Aug 30 at 6:25
It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
â Amrat A
Aug 30 at 6:27
Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
â Amrat A
Aug 30 at 6:32
Oa.. ok I'll try
â Uncool
Aug 30 at 6:32
add a comment |Â
up vote
2
down vote
Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.
Not sure why you give me this. I want the above function
â Uncool
Aug 30 at 6:25
It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
â Amrat A
Aug 30 at 6:27
Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
â Amrat A
Aug 30 at 6:32
Oa.. ok I'll try
â Uncool
Aug 30 at 6:32
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.
Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.
answered Aug 30 at 6:21
Amrat A
1115
1115
Not sure why you give me this. I want the above function
â Uncool
Aug 30 at 6:25
It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
â Amrat A
Aug 30 at 6:27
Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
â Amrat A
Aug 30 at 6:32
Oa.. ok I'll try
â Uncool
Aug 30 at 6:32
add a comment |Â
Not sure why you give me this. I want the above function
â Uncool
Aug 30 at 6:25
It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
â Amrat A
Aug 30 at 6:27
Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
â Amrat A
Aug 30 at 6:32
Oa.. ok I'll try
â Uncool
Aug 30 at 6:32
Not sure why you give me this. I want the above function
â Uncool
Aug 30 at 6:25
Not sure why you give me this. I want the above function
â Uncool
Aug 30 at 6:25
It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
â Amrat A
Aug 30 at 6:27
It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
â Amrat A
Aug 30 at 6:27
Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
â Amrat A
Aug 30 at 6:32
Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
â Amrat A
Aug 30 at 6:32
Oa.. ok I'll try
â Uncool
Aug 30 at 6:32
Oa.. ok I'll try
â Uncool
Aug 30 at 6:32
add a comment |Â
up vote
1
down vote
We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.
1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
$
2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$
3) Both function $P$ and its inverse function are continuous.
Thus $P$ it is a homeomorphism.
add a comment |Â
up vote
1
down vote
We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.
1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
$
2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$
3) Both function $P$ and its inverse function are continuous.
Thus $P$ it is a homeomorphism.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.
1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
$
2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$
3) Both function $P$ and its inverse function are continuous.
Thus $P$ it is a homeomorphism.
We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.
1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
$
2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$
3) Both function $P$ and its inverse function are continuous.
Thus $P$ it is a homeomorphism.
answered Aug 30 at 7:29
Mohammad Riazi-Kermani
31k41853
31k41853
add a comment |Â
add a comment |Â
up vote
0
down vote
It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.
What about the inverse? How to show that it's continuous and before that how to define it.
â Uncool
Aug 30 at 7:17
In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
â Hugo
Aug 30 at 8:22
Since it is equivalent, we can from left to the right ,and right to the left.
â Hugo
Aug 30 at 8:27
add a comment |Â
up vote
0
down vote
It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.
What about the inverse? How to show that it's continuous and before that how to define it.
â Uncool
Aug 30 at 7:17
In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
â Hugo
Aug 30 at 8:22
Since it is equivalent, we can from left to the right ,and right to the left.
â Hugo
Aug 30 at 8:27
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.
It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.
answered Aug 30 at 6:40
Hugo
1029
1029
What about the inverse? How to show that it's continuous and before that how to define it.
â Uncool
Aug 30 at 7:17
In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
â Hugo
Aug 30 at 8:22
Since it is equivalent, we can from left to the right ,and right to the left.
â Hugo
Aug 30 at 8:27
add a comment |Â
What about the inverse? How to show that it's continuous and before that how to define it.
â Uncool
Aug 30 at 7:17
In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
â Hugo
Aug 30 at 8:22
Since it is equivalent, we can from left to the right ,and right to the left.
â Hugo
Aug 30 at 8:27
What about the inverse? How to show that it's continuous and before that how to define it.
â Uncool
Aug 30 at 7:17
What about the inverse? How to show that it's continuous and before that how to define it.
â Uncool
Aug 30 at 7:17
In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
â Hugo
Aug 30 at 8:22
In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
â Hugo
Aug 30 at 8:22
Since it is equivalent, we can from left to the right ,and right to the left.
â Hugo
Aug 30 at 8:27
Since it is equivalent, we can from left to the right ,and right to the left.
â Hugo
Aug 30 at 8:27
add a comment |Â
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Do you mean $S^1-(1,0)$?
â Lord Shark the Unknown
Aug 30 at 6:22
It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
â Uncool
Aug 30 at 6:24
No, it isn't. What is $p(1/4)$ according to your definition?
â Lord Shark the Unknown
Aug 30 at 6:25
Oops! Yes I mean (1,0)
â Uncool
Aug 30 at 6:30