Explicit or analytic formula for a homeomorphism

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SUMMARY

This discussion focuses on finding an explicit or analytic formula for a homeomorphism between a circle and a square, as well as between a disc and the Cartesian product [0,1] x [0,1]. A proposed mapping is presented using the function f(x,y) = (x,y)/max{|x|,|y|}, which transforms points from the unit circle to the unit square. While this mapping is continuous, it is noted that it is not differentiable, highlighting a key aspect of the topology involved.

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  • Understanding of homeomorphisms in topology
  • Familiarity with Euclidean spaces and convex sets
  • Knowledge of continuous and differentiable functions
  • Basic proficiency in trigonometric functions
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What would an explicit or analytic formula for a homeomorphism between a circle and a square be?

Or a disc and [0,1] x [0,1]?
 
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I don't know why you would want this; the whole point of topology is to get beyond the analytic strait-jacket, but If you must why not do it in the first quadrant, mapping the quarter circle x^2 + y^2 = 1 to the quarter square made by the line segments x = 1, 0 \leq y \leq 1 and y = 1, 0 \leq x \leq 1? Trig functions should do the job. Think about it.
 
ex-xian said:
What would an explicit or analytic formula for a homeomorphism between a circle and a square be?

Or a disc and [0,1] x [0,1]?
I would try converting the euclidean ball to the infinity norm ball.

That is, take a point in the circle, i.e., a vector (x,y) such that x^2+y^2 = 1. Now take the map:

f(x,y)\; := \; \frac{(x,y)}{\mathrm{max}\{\vert x\vert,\,\vert y\vert\}}

where f(0,0) is undefined.

This map takes any convex set containing (0,0) into the unit square, hence it takes the unit circle in the unit square. It is easy to show that f is continuous, but not differentiable.
 
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