Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

which of the following statements are true?
1.every homeomorphism of the $2$-sphere to itself has a fixed point.
2.the intervals [$0,1$] and ($0,1$) are homeomorphic.
3.there exists a continuous surjective function from $S^1$ onto $\mathbb{R}$.
4.there exists a continuous surjective function from complex plane onto the non-zero reals.

my effort:

1.true as $2$-sphere is a compact set.
2.true.
3.true as the function $f(α)=re^{|iα|}$ exist.
4. no idea.

can anyone help me please.....

share|improve this question
    
For $2$ through $4$, the following two facts should come in handy at least once: (A) The continuous image of a compact set is compact. (B) The continuous image of a connected set is connected. –  Cameron Buie Dec 10 '12 at 1:42
    
For number 2, what happens if you delete a point from $(0,1)$? What about $[0,1]$? –  andybenji Dec 10 '12 at 2:11
add comment

2 Answers

up vote 4 down vote accepted
  1. $X$ compact does not imply that every homeomorphism of $X$ with itself has a fixed point. Consider $S^1$ where the homeomorphism is rotation by $\pi/2$. You need a better argument here.

  2. This is incorrect. Hint: think about counting special types of points in $[0,1]$ and $(0,1)$.

  3. Also incorrect. Think about compactness here.

  4. What is the "complane"? If the question is "Is there a surjective continuous mapping from the complex plane to the non-zero reals?", then you should draw pictures of both of these sets and look at them- what property does one have which the other does not? Can you use this?

share|improve this answer
    
can u explain question 4 please. –  poton Dec 10 '12 at 3:19
add comment
  1. False. The antipodal map $ f: \mathbb{S}^{2} \rightarrow \mathbb{S}^{2} $ defined by $ f(x) = -x $ does not have a fixed point.

  2. False. The interval $ [0,1] $ is compact, but the interval $ (0,1) $ is not. Note Compactness is a topological invariant, i.e., it is preserved exactly by homeomorphisms.

  3. False. The unit circle $ \mathbb{S}^{1} $ is compact, but $ \mathbb{R} $ is not compact. Note The image of a compact space under a continuous map is also a compact space.

  4. False. The complex plane $ \mathbb{C} $ is connected, but the set $ \mathbb{R} \setminus \{ 0 \} $ of non-zero real numbers is not connected. Note The image of a connected space under a continuous map is also a connected space.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.