Mixing
Continuous bijection which is not a homeomorphism.
Clash Royale CLAN TAG#URR8PPP
up vote
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Given the function $f:[0,2pi)to S^1$, $varphimapsto (cos(varphi), sin(varphi))^t$. Show that $f$ is continuous and a bijection, but not a homeomorphism.
That $f$ is continuous is clear, since every component is continuous.
Furthermore it is continuous differentiable.
When I want to show, that $f$ is a bijection it is easy to see, that $f$ is injective, since for
$f(x)=f(y)Leftrightarrow (cos(x), sin(x))=(cos(y),sin(y))Leftrightarrow cos(x)=cos(y)wedgesin(x)=sin(y)stackrelx,yin [0,2pi)Leftrightarrow x=y$
But how can I show, that $f$ is a surjection?
To show, that $f$ is not a homeomorphism, I have to verify, that $f^-1$ is not continuous.
Can I use the inverse function theorem?
I get:
$Df(varphi)=beginpmatrix-sin(varphi)&0\0&cos(varphi)endpmatrix$
With determinant $operatornamedetDf(varphi)=-sin(varphi)cos(varphi)$
Where $Df(varphi)$ is not invertible for $varphi=0$.
Thanks in advance for hints and comments.
general-topology
asked Sep 1 at 17:19
Cornman
2,83221228
 |Â
show 2 more comments
up vote
4
down vote
favorite
Given the function $f:[0,2pi)to S^1$, $varphimapsto (cos(varphi), sin(varphi))^t$. Show that $f$ is continuous and a bijection, but not a homeomorphism.
That $f$ is continuous is clear, since every component is continuous.
Furthermore it is continuous differentiable.
When I want to show, that $f$ is a bijection it is easy to see, that $f$ is injective, since for
$f(x)=f(y)Leftrightarrow (cos(x), sin(x))=(cos(y),sin(y))Leftrightarrow cos(x)=cos(y)wedgesin(x)=sin(y)stackrelx,yin [0,2pi)Leftrightarrow x=y$
But how can I show, that $f$ is a surjection?
To show, that $f$ is not a homeomorphism, I have to verify, that $f^-1$ is not continuous.
Can I use the inverse function theorem?
I get:
$Df(varphi)=beginpmatrix-sin(varphi)&0\0&cos(varphi)endpmatrix$
With determinant $operatornamedetDf(varphi)=-sin(varphi)cos(varphi)$
Where $Df(varphi)$ is not invertible for $varphi=0$.
Thanks in advance for hints and comments.
general-topology
asked Sep 1 at 17:19
Cornman
2,83221228
You can show it's not a homeomorphism by showing the two spaces are different. What happens when you remove a point from the interval? What about the circle?
– Steve D
Sep 1 at 17:31
Can you explain what is it $varphimapsto (cos(varphi), sin(varphi))^t$, please? and $t$?
– Piquito
Sep 1 at 17:31
@Piquito $t$ notes the transposed vector. It is for esthetics. Instead of writing $beginpmatrixcos(varphi)\sin(varphi)endpmatrix$ you write simply $(cos(varphi),sin(varphi))^t$.
– Cornman
Sep 1 at 17:34
@SteveD If you remove a point from the interval it is not connected anymore, but the circle would stay connected. I would like the "most elementary" way to solve this. Is my approach wrong? Could it be fixed? How about showing that $f$ is a surjection?
– Cornman
Sep 1 at 17:35
1
I'm not sure how anything could be more elementary than removing a point :)
– Steve D
Sep 1 at 17:44
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given the function $f:[0,2pi)to S^1$, $varphimapsto (cos(varphi), sin(varphi))^t$. Show that $f$ is continuous and a bijection, but not a homeomorphism.
That $f$ is continuous is clear, since every component is continuous.
Furthermore it is continuous differentiable.
When I want to show, that $f$ is a bijection it is easy to see, that $f$ is injective, since for
$f(x)=f(y)Leftrightarrow (cos(x), sin(x))=(cos(y),sin(y))Leftrightarrow cos(x)=cos(y)wedgesin(x)=sin(y)stackrelx,yin [0,2pi)Leftrightarrow x=y$
But how can I show, that $f$ is a surjection?
To show, that $f$ is not a homeomorphism, I have to verify, that $f^-1$ is not continuous.
Can I use the inverse function theorem?
I get:
$Df(varphi)=beginpmatrix-sin(varphi)&0\0&cos(varphi)endpmatrix$
With determinant $operatornamedetDf(varphi)=-sin(varphi)cos(varphi)$
Where $Df(varphi)$ is not invertible for $varphi=0$.
Thanks in advance for hints and comments.
general-topology
asked Sep 1 at 17:19
Cornman
2,83221228
Given the function $f:[0,2pi)to S^1$, $varphimapsto (cos(varphi), sin(varphi))^t$. Show that $f$ is continuous and a bijection, but not a homeomorphism.
That $f$ is continuous is clear, since every component is continuous.
Furthermore it is continuous differentiable.
When I want to show, that $f$ is a bijection it is easy to see, that $f$ is injective, since for
$f(x)=f(y)Leftrightarrow (cos(x), sin(x))=(cos(y),sin(y))Leftrightarrow cos(x)=cos(y)wedgesin(x)=sin(y)stackrelx,yin [0,2pi)Leftrightarrow x=y$
But how can I show, that $f$ is a surjection?
To show, that $f$ is not a homeomorphism, I have to verify, that $f^-1$ is not continuous.
Can I use the inverse function theorem?
I get:
$Df(varphi)=beginpmatrix-sin(varphi)&0\0&cos(varphi)endpmatrix$
With determinant $operatornamedetDf(varphi)=-sin(varphi)cos(varphi)$
Where $Df(varphi)$ is not invertible for $varphi=0$.
Thanks in advance for hints and comments.
general-topology
asked Sep 1 at 17:19
Cornman
2,83221228
asked Sep 1 at 17:19
Cornman
2,83221228
asked Sep 1 at 17:19
Cornman
2,83221228
asked Sep 1 at 17:19
Cornman
2,83221228
2,83221228
You can show it's not a homeomorphism by showing the two spaces are different. What happens when you remove a point from the interval? What about the circle?
– Steve D
Sep 1 at 17:31
Can you explain what is it $varphimapsto (cos(varphi), sin(varphi))^t$, please? and $t$?
– Piquito
Sep 1 at 17:31
@Piquito $t$ notes the transposed vector. It is for esthetics. Instead of writing $beginpmatrixcos(varphi)\sin(varphi)endpmatrix$ you write simply $(cos(varphi),sin(varphi))^t$.
– Cornman
Sep 1 at 17:34
@SteveD If you remove a point from the interval it is not connected anymore, but the circle would stay connected. I would like the "most elementary" way to solve this. Is my approach wrong? Could it be fixed? How about showing that $f$ is a surjection?
– Cornman
Sep 1 at 17:35
1
I'm not sure how anything could be more elementary than removing a point :)
– Steve D
Sep 1 at 17:44
 |Â
show 2 more comments
You can show it's not a homeomorphism by showing the two spaces are different. What happens when you remove a point from the interval? What about the circle?
– Steve D
Sep 1 at 17:31
Can you explain what is it $varphimapsto (cos(varphi), sin(varphi))^t$, please? and $t$?
– Piquito
Sep 1 at 17:31
@Piquito $t$ notes the transposed vector. It is for esthetics. Instead of writing $beginpmatrixcos(varphi)\sin(varphi)endpmatrix$ you write simply $(cos(varphi),sin(varphi))^t$.
– Cornman
Sep 1 at 17:34
@SteveD If you remove a point from the interval it is not connected anymore, but the circle would stay connected. I would like the "most elementary" way to solve this. Is my approach wrong? Could it be fixed? How about showing that $f$ is a surjection?
– Cornman
Sep 1 at 17:35
1
I'm not sure how anything could be more elementary than removing a point :)
– Steve D
Sep 1 at 17:44
You can show it's not a homeomorphism by showing the two spaces are different. What happens when you remove a point from the interval? What about the circle?
– Steve D
Sep 1 at 17:31
You can show it's not a homeomorphism by showing the two spaces are different. What happens when you remove a point from the interval? What about the circle?
– Steve D
Sep 1 at 17:31
Can you explain what is it $varphimapsto (cos(varphi), sin(varphi))^t$, please? and $t$?
– Piquito
Sep 1 at 17:31
Can you explain what is it $varphimapsto (cos(varphi), sin(varphi))^t$, please? and $t$?
– Piquito
Sep 1 at 17:31
@Piquito $t$ notes the transposed vector. It is for esthetics. Instead of writing $beginpmatrixcos(varphi)\sin(varphi)endpmatrix$ you write simply $(cos(varphi),sin(varphi))^t$.
– Cornman
Sep 1 at 17:34
@Piquito $t$ notes the transposed vector. It is for esthetics. Instead of writing $beginpmatrixcos(varphi)\sin(varphi)endpmatrix$ you write simply $(cos(varphi),sin(varphi))^t$.
– Cornman
Sep 1 at 17:34
@SteveD If you remove a point from the interval it is not connected anymore, but the circle would stay connected. I would like the "most elementary" way to solve this. Is my approach wrong? Could it be fixed? How about showing that $f$ is a surjection?
– Cornman
Sep 1 at 17:35
@SteveD If you remove a point from the interval it is not connected anymore, but the circle would stay connected. I would like the "most elementary" way to solve this. Is my approach wrong? Could it be fixed? How about showing that $f$ is a surjection?
– Cornman
Sep 1 at 17:35
1
1
I'm not sure how anything could be more elementary than removing a point :)
– Steve D
Sep 1 at 17:44
I'm not sure how anything could be more elementary than removing a point :)
– Steve D
Sep 1 at 17:44
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
The function $f$ is surjective becasue if $(x,y)in S^1$, there is a $thetain[0,2pi)$ such that $(x,y)=(costheta,sintheta)$; just take $theta=arccos x$ if $ygeqslant0$ and $theta=2pi-arccos x$ otherwise.
And $f^-1$ is discontinuous because $lim_ntoinftyleft(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)=(1,0)=f(0)$, but $lim_ntoinftyf^-1left(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)$ doesn't exist (in $[0,2pi)$).
edited Sep 2 at 16:22
Cornman
2,83221228
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
How do you get $lim_ntoinfty sincos(2pi-tfrac1n))=0$? Should it not be $sin(1)neq 0$?
– Cornman
Sep 2 at 16:17
I think you mean just $sin(2pi-tfrac1n)$ and it is a typo.
– Cornman
Sep 2 at 16:18
add a comment |Â
up vote
3
down vote
To see that the inverse is not continuous note that there exists a sequence of points $(y_n)$ s.t $y_nto (1,0)$ but $f^-1(y_n)to 2pine f^-1(1,0)=0.$ Consider the sequence of points given by
$$
(cos (2pi-n^-1), sin (2pi-n^-1) )
$$
for example.
answered Sep 1 at 17:41
Foobaz John
18.6k41245
How do you know what $f^-1$ is?
– Cornman
Sep 1 at 17:43
2
Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $yin S^1$, $f^-1(y)$ is the unique value of $xin [0,2pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^-1$. In particular $f^-1(y_n)=2pi-n^-1$ where $y_n=(cos (2pi-n^-1), sin (2pi-n^-1) )$.
– Foobaz John
Sep 1 at 17:49
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The function $f$ is surjective becasue if $(x,y)in S^1$, there is a $thetain[0,2pi)$ such that $(x,y)=(costheta,sintheta)$; just take $theta=arccos x$ if $ygeqslant0$ and $theta=2pi-arccos x$ otherwise.
And $f^-1$ is discontinuous because $lim_ntoinftyleft(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)=(1,0)=f(0)$, but $lim_ntoinftyf^-1left(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)$ doesn't exist (in $[0,2pi)$).
edited Sep 2 at 16:22
Cornman
2,83221228
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
How do you get $lim_ntoinfty sincos(2pi-tfrac1n))=0$? Should it not be $sin(1)neq 0$?
– Cornman
Sep 2 at 16:17
I think you mean just $sin(2pi-tfrac1n)$ and it is a typo.
– Cornman
Sep 2 at 16:18
add a comment |Â
up vote
4
down vote
accepted
The function $f$ is surjective becasue if $(x,y)in S^1$, there is a $thetain[0,2pi)$ such that $(x,y)=(costheta,sintheta)$; just take $theta=arccos x$ if $ygeqslant0$ and $theta=2pi-arccos x$ otherwise.
And $f^-1$ is discontinuous because $lim_ntoinftyleft(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)=(1,0)=f(0)$, but $lim_ntoinftyf^-1left(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)$ doesn't exist (in $[0,2pi)$).
edited Sep 2 at 16:22
Cornman
2,83221228
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
How do you get $lim_ntoinfty sincos(2pi-tfrac1n))=0$? Should it not be $sin(1)neq 0$?
– Cornman
Sep 2 at 16:17
I think you mean just $sin(2pi-tfrac1n)$ and it is a typo.
– Cornman
Sep 2 at 16:18
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The function $f$ is surjective becasue if $(x,y)in S^1$, there is a $thetain[0,2pi)$ such that $(x,y)=(costheta,sintheta)$; just take $theta=arccos x$ if $ygeqslant0$ and $theta=2pi-arccos x$ otherwise.
And $f^-1$ is discontinuous because $lim_ntoinftyleft(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)=(1,0)=f(0)$, but $lim_ntoinftyf^-1left(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)$ doesn't exist (in $[0,2pi)$).
edited Sep 2 at 16:22
Cornman
2,83221228
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
The function $f$ is surjective becasue if $(x,y)in S^1$, there is a $thetain[0,2pi)$ such that $(x,y)=(costheta,sintheta)$; just take $theta=arccos x$ if $ygeqslant0$ and $theta=2pi-arccos x$ otherwise.
And $f^-1$ is discontinuous because $lim_ntoinftyleft(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)=(1,0)=f(0)$, but $lim_ntoinftyf^-1left(cosleft(2pi-frac1nright),sinleft(2pi-frac1nright)right)$ doesn't exist (in $[0,2pi)$).
edited Sep 2 at 16:22
Cornman
2,83221228
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
edited Sep 2 at 16:22
Cornman
2,83221228
edited Sep 2 at 16:22
Cornman
2,83221228
edited Sep 2 at 16:22
Cornman
2,83221228
2,83221228
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
answered Sep 1 at 17:43
José Carlos Santos
120k16101182
120k16101182
How do you get $lim_ntoinfty sincos(2pi-tfrac1n))=0$? Should it not be $sin(1)neq 0$?
– Cornman
Sep 2 at 16:17
I think you mean just $sin(2pi-tfrac1n)$ and it is a typo.
– Cornman
Sep 2 at 16:18
add a comment |Â
How do you get $lim_ntoinfty sincos(2pi-tfrac1n))=0$? Should it not be $sin(1)neq 0$?
– Cornman
Sep 2 at 16:17
I think you mean just $sin(2pi-tfrac1n)$ and it is a typo.
– Cornman
Sep 2 at 16:18
How do you get $lim_ntoinfty sincos(2pi-tfrac1n))=0$? Should it not be $sin(1)neq 0$?
– Cornman
Sep 2 at 16:17
How do you get $lim_ntoinfty sincos(2pi-tfrac1n))=0$? Should it not be $sin(1)neq 0$?
– Cornman
Sep 2 at 16:17
I think you mean just $sin(2pi-tfrac1n)$ and it is a typo.
– Cornman
Sep 2 at 16:18
I think you mean just $sin(2pi-tfrac1n)$ and it is a typo.
– Cornman
Sep 2 at 16:18
add a comment |Â
up vote
3
down vote
To see that the inverse is not continuous note that there exists a sequence of points $(y_n)$ s.t $y_nto (1,0)$ but $f^-1(y_n)to 2pine f^-1(1,0)=0.$ Consider the sequence of points given by
$$
(cos (2pi-n^-1), sin (2pi-n^-1) )
$$
for example.
answered Sep 1 at 17:41
Foobaz John
18.6k41245
How do you know what $f^-1$ is?
– Cornman
Sep 1 at 17:43
2
Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $yin S^1$, $f^-1(y)$ is the unique value of $xin [0,2pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^-1$. In particular $f^-1(y_n)=2pi-n^-1$ where $y_n=(cos (2pi-n^-1), sin (2pi-n^-1) )$.
– Foobaz John
Sep 1 at 17:49
add a comment |Â
up vote
3
down vote
To see that the inverse is not continuous note that there exists a sequence of points $(y_n)$ s.t $y_nto (1,0)$ but $f^-1(y_n)to 2pine f^-1(1,0)=0.$ Consider the sequence of points given by
$$
(cos (2pi-n^-1), sin (2pi-n^-1) )
$$
for example.
answered Sep 1 at 17:41
Foobaz John
18.6k41245
How do you know what $f^-1$ is?
– Cornman
Sep 1 at 17:43
2
Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $yin S^1$, $f^-1(y)$ is the unique value of $xin [0,2pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^-1$. In particular $f^-1(y_n)=2pi-n^-1$ where $y_n=(cos (2pi-n^-1), sin (2pi-n^-1) )$.
– Foobaz John
Sep 1 at 17:49
add a comment |Â
up vote
3
down vote
up vote
3
down vote
To see that the inverse is not continuous note that there exists a sequence of points $(y_n)$ s.t $y_nto (1,0)$ but $f^-1(y_n)to 2pine f^-1(1,0)=0.$ Consider the sequence of points given by
$$
(cos (2pi-n^-1), sin (2pi-n^-1) )
$$
for example.
answered Sep 1 at 17:41
Foobaz John
18.6k41245
To see that the inverse is not continuous note that there exists a sequence of points $(y_n)$ s.t $y_nto (1,0)$ but $f^-1(y_n)to 2pine f^-1(1,0)=0.$ Consider the sequence of points given by
$$
(cos (2pi-n^-1), sin (2pi-n^-1) )
$$
for example.
answered Sep 1 at 17:41
Foobaz John
18.6k41245
answered Sep 1 at 17:41
Foobaz John
18.6k41245
answered Sep 1 at 17:41
Foobaz John
18.6k41245
answered Sep 1 at 17:41
Foobaz John
18.6k41245
18.6k41245
How do you know what $f^-1$ is?
– Cornman
Sep 1 at 17:43
2
Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $yin S^1$, $f^-1(y)$ is the unique value of $xin [0,2pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^-1$. In particular $f^-1(y_n)=2pi-n^-1$ where $y_n=(cos (2pi-n^-1), sin (2pi-n^-1) )$.
– Foobaz John
Sep 1 at 17:49
add a comment |Â
How do you know what $f^-1$ is?
– Cornman
Sep 1 at 17:43
2
Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $yin S^1$, $f^-1(y)$ is the unique value of $xin [0,2pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^-1$. In particular $f^-1(y_n)=2pi-n^-1$ where $y_n=(cos (2pi-n^-1), sin (2pi-n^-1) )$.
– Foobaz John
Sep 1 at 17:49
How do you know what $f^-1$ is?
– Cornman
Sep 1 at 17:43
How do you know what $f^-1$ is?
– Cornman
Sep 1 at 17:43
2
2
Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $yin S^1$, $f^-1(y)$ is the unique value of $xin [0,2pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^-1$. In particular $f^-1(y_n)=2pi-n^-1$ where $y_n=(cos (2pi-n^-1), sin (2pi-n^-1) )$.
– Foobaz John
Sep 1 at 17:49
Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $yin S^1$, $f^-1(y)$ is the unique value of $xin [0,2pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^-1$. In particular $f^-1(y_n)=2pi-n^-1$ where $y_n=(cos (2pi-n^-1), sin (2pi-n^-1) )$.
– Foobaz John
Sep 1 at 17:49
add a comment |Â
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You can show it's not a homeomorphism by showing the two spaces are different. What happens when you remove a point from the interval? What about the circle?
– Steve D
Sep 1 at 17:31
Can you explain what is it $varphimapsto (cos(varphi), sin(varphi))^t$, please? and $t$?
– Piquito
Sep 1 at 17:31
@Piquito $t$ notes the transposed vector. It is for esthetics. Instead of writing $beginpmatrixcos(varphi)\sin(varphi)endpmatrix$ you write simply $(cos(varphi),sin(varphi))^t$.
– Cornman
Sep 1 at 17:34
@SteveD If you remove a point from the interval it is not connected anymore, but the circle would stay connected. I would like the "most elementary" way to solve this. Is my approach wrong? Could it be fixed? How about showing that $f$ is a surjection?
– Cornman
Sep 1 at 17:35
1
I'm not sure how anything could be more elementary than removing a point :)
– Steve D
Sep 1 at 17:44