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Let $S^1 = \{(x, y) \in \Bbb R^2 : x^2 + y^2 = 1 \}.$ Let $D = \{(x, y) \in\Bbb R^2 : x^2 + y^2 \le 1 \}$ and $E = \{(x, y) \in\Bbb R^2 : 2x^2 + 3y^2 \le 1\}$ be also considered as subspaces of $\Bbb R^2.$ Which of the following statements are true?

a. If $f : D \to S^1$ is a continuous mapping, then there exists $x \in S^1$ such that $f(x) = x$.

b. If $f : S^1 \to S^1$ is a continuous mapping, then there exists $x \in S^1$ such that $f(x) = x$.

c. If $f : E \to E$ is a continuous mapping, then there exists $x \in E$ such that $f(x) = x$.

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please somebody help how to solve this question.i have no idea. –  poton Aug 24 '12 at 1:04
    
Ok, I've made a formatting edit. Please make sure I didn't change the question. Probably these $R$'s should be $\Bbb{R}$. –  user2468 Aug 24 '12 at 1:11

1 Answer 1

up vote 1 down vote accepted

HINTS:

(a) and (c) The Brouwer fixed-point theorem. For (a) don’t forget that $S^1\subseteq D$.

(b) $S^1$ is just a circle; what happens if you rotate it a half turn?

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please explain in detail –  poton Aug 24 '12 at 1:41
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@poton, are you willing to do any work yourself? Have you looked at the link? Have you thought about rotation half a turn? I'm sure Brian and others are happy to help you, but it would help if you tried to follow up on the suggestions here and gave a more focussed request than just "please explain in detail". –  Gerry Myerson Aug 24 '12 at 1:52
    
i saw the link. and now c is clear. and a will be true if D and S1 are homeomorphic. but how can i show this? –  poton Aug 24 '12 at 6:17
    
and could not understand what do you want to say by about rotation half a turn for b. –  poton Aug 24 '12 at 6:19
    
Do you see that $S^1$ is a circle? Think about rotating a circle halfway around as being a function on the points of the circle. Also, $D$ and $S^1$ are certainly not homeomorphic. –  Gerry Myerson Aug 24 '12 at 9:32

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