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Show that f is a homeomorphism?

Homeomorphism questionOn level set of concave functionHomeomorphism from $Usubset mathbbR^2 rightarrow U$ which permutes points.Explicit Homeomorphism from $mathbbR^2$ to open diskConflict between homeomorphism definition and continuity of inverse theoremChange of variables for multiple integrals for $f(x,y) = -f(y,x)$How to show a homeomorphism exists?Continuous bijection which is not a homeomorphism.Real analysis / Topology - Homeomorphism of functions $f(x) = x^3$ and $f(x) = x+ sin x$Can a Homeomorphism exist between two discontinuous spaces.

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3

0

$begingroup$

I'm trying to show that $vecf: D to R, vecf(r, varphi)=
beginpmatrixrcos(varphi)\ rsin(varphi)endpmatrix
$ is a homeomorphism, with $D = rin[1,2], varphi in [o, fracpi2]$ and on the other hand with $R = ygeq 0,xgeq 0, 1 leq x^2+y^2 leq 4$. Now, I'm not quite sure how I should go about this. Do I just find any function that assigns values to $x$ and $y$ in terms of $r$ and $varphi$ and then show that continuity, bijection and continuity of the inverse are given? Any help is greatly appreciated since I'm fairly new to this.
Best regards.

real-analysis general-topology continuity

share|cite|improve this question

edited Jul 4 at 7:26

psyph

asked Jul 4 at 7:06

psyphpsyph

7477 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: A continuous bijection on a compact set is a homeomorphism.
  $endgroup$
  – EpsilonDelta
  Jul 4 at 7:33
  

  
  
  
  

add a comment | 

3

0

$begingroup$

I'm trying to show that $vecf: D to R, vecf(r, varphi)=
beginpmatrixrcos(varphi)\ rsin(varphi)endpmatrix
$ is a homeomorphism, with $D = rin[1,2], varphi in [o, fracpi2]$ and on the other hand with $R = ygeq 0,xgeq 0, 1 leq x^2+y^2 leq 4$. Now, I'm not quite sure how I should go about this. Do I just find any function that assigns values to $x$ and $y$ in terms of $r$ and $varphi$ and then show that continuity, bijection and continuity of the inverse are given? Any help is greatly appreciated since I'm fairly new to this.
Best regards.

real-analysis general-topology continuity

share|cite|improve this question

edited Jul 4 at 7:26

psyph

asked Jul 4 at 7:06

psyphpsyph

7477 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: A continuous bijection on a compact set is a homeomorphism.
  $endgroup$
  – EpsilonDelta
  Jul 4 at 7:33
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to show that $vecf: D to R, vecf(r, varphi)=
beginpmatrixrcos(varphi)\ rsin(varphi)endpmatrix
$ is a homeomorphism, with $D = rin[1,2], varphi in [o, fracpi2]$ and on the other hand with $R = ygeq 0,xgeq 0, 1 leq x^2+y^2 leq 4$. Now, I'm not quite sure how I should go about this. Do I just find any function that assigns values to $x$ and $y$ in terms of $r$ and $varphi$ and then show that continuity, bijection and continuity of the inverse are given? Any help is greatly appreciated since I'm fairly new to this.
Best regards.

real-analysis general-topology continuity

share|cite|improve this question

edited Jul 4 at 7:26

psyph

asked Jul 4 at 7:06

psyphpsyph

7477 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 4 at 7:26

psyph

asked Jul 4 at 7:06

psyphpsyph

7477 bronze badges

share|cite|improve this question

edited Jul 4 at 7:26

psyph

asked Jul 4 at 7:06

psyphpsyph

7477 bronze badges

asked Jul 4 at 7:06

psyphpsyph

7477 bronze badges

asked Jul 4 at 7:06

psyphpsyph

7477 bronze badges

2 Answers
2

active

oldest

votes

2

$begingroup$

This really depends on which results about homeomorphisms you assume to be true, but keeping it quite simple, in the roum of what you know, I would attack the problem this way:

• Prove $vecf$ is continuous (simple as $f$ is a product of elementary functions)


• Find an inverse for $hatf$ (it won't even look ugly!)


• Prove that inverse is also continuous (also straight-forward)

share|cite|improve this answer

answered Jul 4 at 7:18

DavidDavid

8211111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't continuity of $vecf$ obvious from just looking at it?
  $endgroup$
  – psyph
  Jul 4 at 7:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, once again, it depends on what you have to prove and what you can assume as true. "It's obvious from looking at it" does not constitute a valid mathematical argument, but can be accepted as such at some courses that want to focus on more complex stuff
  $endgroup$
  – David
  Jul 4 at 7:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First of all, thanks for your great tips and sorry for my sloppy wording. At this point in my course, we can assume that $vecf$ is continuous since it is a composition of trigonometric functions. But could you help me with getting the inverse? I'm a little stuck here.
  $endgroup$
  – psyph
  Jul 4 at 8:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  psyph If you pay attention to $vecf$, you'll realize that it's just a polar-to-cartesian coortinates transformation. Check if a cartesian-to-polar does the job! Can you get "$r$ and $theta$" knowing "$x$ and $y$"?
  $endgroup$
  – David
  Jul 4 at 8:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $vecf$ has an inverse (by the way, don't forget to prove the function you just described is both a right and left-inverse!), then it is always bijective
  $endgroup$
  – David
  Jul 4 at 8:56
  

  
  
  
  

 | 
show 1 more comment

4

$begingroup$

Clearly $f$ is continuous.

Take $(x,y) in R$ and assume that $f(r,varphi)= (x,y)$ for some $(r, varphi) in D$. We get
$$(rcosvarphi, rsinvarphi) = f(r, varphi) = (x,y)$$
so it follows $r = sqrtx^2+y^2$ and $cosvarphi = fracxr$ so $$varphi= arccosfracxsqrtx^2+y^2$$

After checking that indeed $(r, varphi) in D$ and $f(r,varphi) = (x,y)$, we conclude that $f circ g = operatornameid_R$ where $g : R to D$ is given by $$g(x,y) = left(sqrtx^2+y^2, arccosfracxsqrtx^2+y^2right)$$
which is a continuous function $R to D$. Now also check that $g circ f = operatornameid_D$ to conclude that $f$ is bijective with inverse $f^-1 =g$.

Hence $f$ is a homeomorphism.

share|cite|improve this answer

answered Jul 4 at 8:58

mechanodroidmechanodroid

30.8k66 gold badges2828 silver badges4949 bronze badges

$endgroup$

add a comment | 

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2

$begingroup$

This really depends on which results about homeomorphisms you assume to be true, but keeping it quite simple, in the roum of what you know, I would attack the problem this way:

• Prove $vecf$ is continuous (simple as $f$ is a product of elementary functions)


• Find an inverse for $hatf$ (it won't even look ugly!)


• Prove that inverse is also continuous (also straight-forward)

share|cite|improve this answer

answered Jul 4 at 7:18

DavidDavid

8211111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't continuity of $vecf$ obvious from just looking at it?
  $endgroup$
  – psyph
  Jul 4 at 7:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, once again, it depends on what you have to prove and what you can assume as true. "It's obvious from looking at it" does not constitute a valid mathematical argument, but can be accepted as such at some courses that want to focus on more complex stuff
  $endgroup$
  – David
  Jul 4 at 7:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First of all, thanks for your great tips and sorry for my sloppy wording. At this point in my course, we can assume that $vecf$ is continuous since it is a composition of trigonometric functions. But could you help me with getting the inverse? I'm a little stuck here.
  $endgroup$
  – psyph
  Jul 4 at 8:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  psyph If you pay attention to $vecf$, you'll realize that it's just a polar-to-cartesian coortinates transformation. Check if a cartesian-to-polar does the job! Can you get "$r$ and $theta$" knowing "$x$ and $y$"?
  $endgroup$
  – David
  Jul 4 at 8:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $vecf$ has an inverse (by the way, don't forget to prove the function you just described is both a right and left-inverse!), then it is always bijective
  $endgroup$
  – David
  Jul 4 at 8:56
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

This really depends on which results about homeomorphisms you assume to be true, but keeping it quite simple, in the roum of what you know, I would attack the problem this way:

• Prove $vecf$ is continuous (simple as $f$ is a product of elementary functions)


• Find an inverse for $hatf$ (it won't even look ugly!)


• Prove that inverse is also continuous (also straight-forward)

share|cite|improve this answer

answered Jul 4 at 7:18

DavidDavid

8211111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't continuity of $vecf$ obvious from just looking at it?
  $endgroup$
  – psyph
  Jul 4 at 7:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, once again, it depends on what you have to prove and what you can assume as true. "It's obvious from looking at it" does not constitute a valid mathematical argument, but can be accepted as such at some courses that want to focus on more complex stuff
  $endgroup$
  – David
  Jul 4 at 7:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First of all, thanks for your great tips and sorry for my sloppy wording. At this point in my course, we can assume that $vecf$ is continuous since it is a composition of trigonometric functions. But could you help me with getting the inverse? I'm a little stuck here.
  $endgroup$
  – psyph
  Jul 4 at 8:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  psyph If you pay attention to $vecf$, you'll realize that it's just a polar-to-cartesian coortinates transformation. Check if a cartesian-to-polar does the job! Can you get "$r$ and $theta$" knowing "$x$ and $y$"?
  $endgroup$
  – David
  Jul 4 at 8:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $vecf$ has an inverse (by the way, don't forget to prove the function you just described is both a right and left-inverse!), then it is always bijective
  $endgroup$
  – David
  Jul 4 at 8:56
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

This really depends on which results about homeomorphisms you assume to be true, but keeping it quite simple, in the roum of what you know, I would attack the problem this way:

• Prove $vecf$ is continuous (simple as $f$ is a product of elementary functions)


• Find an inverse for $hatf$ (it won't even look ugly!)


• Prove that inverse is also continuous (also straight-forward)

share|cite|improve this answer

answered Jul 4 at 7:18

DavidDavid

8211111 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 4 at 7:18

DavidDavid

8211111 bronze badges

answered Jul 4 at 7:18

DavidDavid

8211111 bronze badges

answered Jul 4 at 7:18

DavidDavid

8211111 bronze badges

4

$begingroup$

Clearly $f$ is continuous.

Take $(x,y) in R$ and assume that $f(r,varphi)= (x,y)$ for some $(r, varphi) in D$. We get
$$(rcosvarphi, rsinvarphi) = f(r, varphi) = (x,y)$$
so it follows $r = sqrtx^2+y^2$ and $cosvarphi = fracxr$ so $$varphi= arccosfracxsqrtx^2+y^2$$

After checking that indeed $(r, varphi) in D$ and $f(r,varphi) = (x,y)$, we conclude that $f circ g = operatornameid_R$ where $g : R to D$ is given by $$g(x,y) = left(sqrtx^2+y^2, arccosfracxsqrtx^2+y^2right)$$
which is a continuous function $R to D$. Now also check that $g circ f = operatornameid_D$ to conclude that $f$ is bijective with inverse $f^-1 =g$.

Hence $f$ is a homeomorphism.

share|cite|improve this answer

answered Jul 4 at 8:58

mechanodroidmechanodroid

30.8k66 gold badges2828 silver badges4949 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Clearly $f$ is continuous.

Take $(x,y) in R$ and assume that $f(r,varphi)= (x,y)$ for some $(r, varphi) in D$. We get
$$(rcosvarphi, rsinvarphi) = f(r, varphi) = (x,y)$$
so it follows $r = sqrtx^2+y^2$ and $cosvarphi = fracxr$ so $$varphi= arccosfracxsqrtx^2+y^2$$

After checking that indeed $(r, varphi) in D$ and $f(r,varphi) = (x,y)$, we conclude that $f circ g = operatornameid_R$ where $g : R to D$ is given by $$g(x,y) = left(sqrtx^2+y^2, arccosfracxsqrtx^2+y^2right)$$
which is a continuous function $R to D$. Now also check that $g circ f = operatornameid_D$ to conclude that $f$ is bijective with inverse $f^-1 =g$.

Hence $f$ is a homeomorphism.

share|cite|improve this answer

answered Jul 4 at 8:58

mechanodroidmechanodroid

30.8k66 gold badges2828 silver badges4949 bronze badges

$endgroup$

add a comment | 

$begingroup$

Clearly $f$ is continuous.

Take $(x,y) in R$ and assume that $f(r,varphi)= (x,y)$ for some $(r, varphi) in D$. We get
$$(rcosvarphi, rsinvarphi) = f(r, varphi) = (x,y)$$
so it follows $r = sqrtx^2+y^2$ and $cosvarphi = fracxr$ so $$varphi= arccosfracxsqrtx^2+y^2$$

After checking that indeed $(r, varphi) in D$ and $f(r,varphi) = (x,y)$, we conclude that $f circ g = operatornameid_R$ where $g : R to D$ is given by $$g(x,y) = left(sqrtx^2+y^2, arccosfracxsqrtx^2+y^2right)$$
which is a continuous function $R to D$. Now also check that $g circ f = operatornameid_D$ to conclude that $f$ is bijective with inverse $f^-1 =g$.

Hence $f$ is a homeomorphism.

share|cite|improve this answer

answered Jul 4 at 8:58

mechanodroidmechanodroid

30.8k66 gold badges2828 silver badges4949 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 4 at 8:58

mechanodroidmechanodroid

30.8k66 gold badges2828 silver badges4949 bronze badges

answered Jul 4 at 8:58

mechanodroidmechanodroid

30.8k66 gold badges2828 silver badges4949 bronze badges

answered Jul 4 at 8:58

mechanodroidmechanodroid

30.8k66 gold badges2828 silver badges4949 bronze badges