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Old November 1st, 2007, 04:16 PM
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Default fixed point mapping

Trying to figure out these two questions, got the rest of them but can't remember my linear algebra for the life of me thanks!

a. make a continuous function (0,1] --> (0,1] without fixed points.
note: there is supposed to be some simple map but honestly I can't see it.

b. prove that the unit square [0,1]x[0,1] is homeomorphic to the parellelogram in RČ with vertices (3,2), (6,5), (4,4), and (7,7). Use a suitable map and an appropriate translation.

Thanks so much for the help appreciated.
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Old November 1st, 2007, 06:40 PM
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Trying to figure out these two questions, got the rest of them but can't remember my linear algebra for the life of me thanks!

a. make a continuous function (0,1] --> (0,1] without fixed points.
note: there is supposed to be some simple map but honestly I can't see it.
If it was [0,1] --> [0,1] it would be impossible by the the Brouwer fixed point theorem. However, here it is (0,1] hence take f(x) = x^2.
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Old November 1st, 2007, 07:06 PM
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However, here it is (0,1] hence take f(x) = x^2.
Surely x=1 is a fixed point for that mapping.

However, the following continuous function has no fixed point.
f(x) = \left\{ {\begin{array}{cr}
   {x^2 } & {0 < x \le \frac{1}{2}}  \\
   {\left( {1 - x} \right)^2 } & {\frac{1}{2} < x \le 1}  \\
\end{array}} \right.
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Old November 1st, 2007, 08:09 PM
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thanks for the help, i'm trying to figure out how to do the second part, I forgot how to map a vector <2,1> to the unit vector <0,1>, and the vector <3,3> to <1,0>, thanks
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