# a few questions on homeomorphism and continuous surjective mapping

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.

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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

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?

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can u explain question 4 please. – poton Dec 10 '12 at 3:19
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.

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