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Proving a condition at which a function is not a homomorphism

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1

$begingroup$

Let $A subset R^k$ be an open set and $f:A to R^n$
be a continuous injection $(k < n)$.

The task is to prove that:

$f:A to f(A)$ is not
a homeomorphism if and only if there exists a sequence $ x_k subset A$ that converges to some point on the boundary of $A$ or to $infty$ so that the $ a = limlimits_k to infty f(x_k)$ exists and belongs to $f(A)$.

First it can be noticed that $f: A to f(A)$ is a bijection. The second thing is that $f$ is continuous, since it was given. So' the only missing thing here for $f$ to be a homomorphism is that $f^-1$ needs to be continuous.

In the first direction, I assume that such a sequence exists. If $x_k$ converges to a point on the boundary or to $infty$ then I can take the sequence $ f(x_k) $ which converges to $a$, however $limlimits_k to infty f(x_k) = a$ and $f^-1(a) ne limlimits_k to infty f^-1(f(x_k))$, since it is given that $a$ belong to $f(A)$ so there is some $y in A$ that satisfies $f(y) = a$.

So I thing I managed to prove here the first direction.

However, I don't really know how to start in the second direction.

Does anyone here has any suggestion?

Help would be appreciated.

real-analysis calculus continuous-homomorphisms

share|cite|improve this question

asked Mar 10 at 21:01

Gabi GGabi G

442110

$endgroup$

add a comment | 

1

$begingroup$

Let $A subset R^k$ be an open set and $f:A to R^n$
be a continuous injection $(k < n)$.

The task is to prove that:

$f:A to f(A)$ is not
a homeomorphism if and only if there exists a sequence $ x_k subset A$ that converges to some point on the boundary of $A$ or to $infty$ so that the $ a = limlimits_k to infty f(x_k)$ exists and belongs to $f(A)$.

First it can be noticed that $f: A to f(A)$ is a bijection. The second thing is that $f$ is continuous, since it was given. So' the only missing thing here for $f$ to be a homomorphism is that $f^-1$ needs to be continuous.

In the first direction, I assume that such a sequence exists. If $x_k$ converges to a point on the boundary or to $infty$ then I can take the sequence $ f(x_k) $ which converges to $a$, however $limlimits_k to infty f(x_k) = a$ and $f^-1(a) ne limlimits_k to infty f^-1(f(x_k))$, since it is given that $a$ belong to $f(A)$ so there is some $y in A$ that satisfies $f(y) = a$.

So I thing I managed to prove here the first direction.

However, I don't really know how to start in the second direction.

Does anyone here has any suggestion?

Help would be appreciated.

real-analysis calculus continuous-homomorphisms

share|cite|improve this question

asked Mar 10 at 21:01

Gabi GGabi G

442110

$endgroup$

add a comment | 

$begingroup$

Let $A subset R^k$ be an open set and $f:A to R^n$
be a continuous injection $(k < n)$.

The task is to prove that:

$f:A to f(A)$ is not
a homeomorphism if and only if there exists a sequence $ x_k subset A$ that converges to some point on the boundary of $A$ or to $infty$ so that the $ a = limlimits_k to infty f(x_k)$ exists and belongs to $f(A)$.

First it can be noticed that $f: A to f(A)$ is a bijection. The second thing is that $f$ is continuous, since it was given. So' the only missing thing here for $f$ to be a homomorphism is that $f^-1$ needs to be continuous.

In the first direction, I assume that such a sequence exists. If $x_k$ converges to a point on the boundary or to $infty$ then I can take the sequence $ f(x_k) $ which converges to $a$, however $limlimits_k to infty f(x_k) = a$ and $f^-1(a) ne limlimits_k to infty f^-1(f(x_k))$, since it is given that $a$ belong to $f(A)$ so there is some $y in A$ that satisfies $f(y) = a$.

So I thing I managed to prove here the first direction.

However, I don't really know how to start in the second direction.

Does anyone here has any suggestion?

Help would be appreciated.

real-analysis calculus continuous-homomorphisms

share|cite|improve this question

asked Mar 10 at 21:01

Gabi GGabi G

442110

$endgroup$

share|cite|improve this question

asked Mar 10 at 21:01

Gabi GGabi G

442110

share|cite|improve this question

asked Mar 10 at 21:01

Gabi GGabi G

442110

asked Mar 10 at 21:01

Gabi GGabi G

442110

asked Mar 10 at 21:01

Gabi GGabi G

442110

1 Answer
1

active

oldest

votes

1

$begingroup$

Well, if $f$ is not a homeomorphism, then $f^-1$ is not continuous, thus there exists a convergent sequence $y_n=f(x_n)$ of elements of $f(A)$ such that $f^-1(y_n)=x_n$ does not converge to an element of $A$.

share|cite|improve this answer

answered Mar 10 at 21:25

MindlackMindlack

4,920211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure it's just that simple?
  $endgroup$
  – Gabi G
  Mar 10 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nevermind, I think that I understand now
  $endgroup$
  – Gabi G
  Mar 10 at 23:07
  

  
  
  
  

add a comment | 

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1

$begingroup$

Well, if $f$ is not a homeomorphism, then $f^-1$ is not continuous, thus there exists a convergent sequence $y_n=f(x_n)$ of elements of $f(A)$ such that $f^-1(y_n)=x_n$ does not converge to an element of $A$.

share|cite|improve this answer

answered Mar 10 at 21:25

MindlackMindlack

4,920211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure it's just that simple?
  $endgroup$
  – Gabi G
  Mar 10 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nevermind, I think that I understand now
  $endgroup$
  – Gabi G
  Mar 10 at 23:07
  

  
  
  
  

add a comment | 

1

$begingroup$

Well, if $f$ is not a homeomorphism, then $f^-1$ is not continuous, thus there exists a convergent sequence $y_n=f(x_n)$ of elements of $f(A)$ such that $f^-1(y_n)=x_n$ does not converge to an element of $A$.

share|cite|improve this answer

answered Mar 10 at 21:25

MindlackMindlack

4,920211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure it's just that simple?
  $endgroup$
  – Gabi G
  Mar 10 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nevermind, I think that I understand now
  $endgroup$
  – Gabi G
  Mar 10 at 23:07
  

  
  
  
  

add a comment | 

$begingroup$

Well, if $f$ is not a homeomorphism, then $f^-1$ is not continuous, thus there exists a convergent sequence $y_n=f(x_n)$ of elements of $f(A)$ such that $f^-1(y_n)=x_n$ does not converge to an element of $A$.

share|cite|improve this answer

answered Mar 10 at 21:25

MindlackMindlack

4,920211

$endgroup$

share|cite|improve this answer

answered Mar 10 at 21:25

MindlackMindlack

4,920211

answered Mar 10 at 21:25

MindlackMindlack

4,920211

answered Mar 10 at 21:25

MindlackMindlack

4,920211