homeomorphism between unit square and unit circle?

Question

r:Square -> Circle has been given as r(x,y) = (x/sqrt(x^2+y^2),y/sqrt(x^2+y^2)), the radial projection from square to circle. I've proved the trivial facts that r is a bijection between the square and the circle. How do I prove r and its inverse are continuous to show r is a homeomorphism?

kb · Accepted Answer

I'm assuming that the square has equation |x| + |y| = 1, and the circle has equation x^2 + y^2 = 1.
----------------------
That r is continuous is clear, since we are avoiding the origin (presumably, (0, 0) is not in the square).

Sketch of r being continuous:
x, y are continuous (trivially)
x^2 + y^2 is continuous (polynomial in x, y)
==> sqrt(x^2 + y^2) is continuous (composition of sqrt(t) and t = x^2 + y^2, both are continuous)
==> x/sqrt(x^2 + y^2) and y/sqrt(x^2 + y^2) are continuous (quotients)
==> r(x, y) is continuous (each component is continuous).
--------------------
As for the inverse, define s: Circle ---> Square by 
s(x, y) = (x, y) = (x/(|x| + |y|), y/(|x| + |y|))

(s o r)(x, y) 
= s(r(x, y)) 
= s(x/sqrt(x^2+y^2), y/sqrt(x^2+y^2)) 
= ( [x/sqrt(x^2+y^2)] / (|x/sqrt(x^2+y^2)| + |y/sqrt(x^2+y^2)|), [y/sqrt(x^2+y^2)]/(|x/sqrt(x^2+y^2)| + |y/sqrt(x^2+y^2)|) )
= (x / (|x| + |y|), y/(|x| + |y|))
= (x, y), since |x| + |y| = 1 for this square.

(r o s)(x, y) 
= r(s(x, y)) 
= r((x/(|x| + |y|), y/(|x| + |y|)))
= ([x/(|x| + |y|)]/sqrt[(x/(|x| + |y|)^2 + (y/(|x| + |y|))^2)], [y/(|x| + |y|)]/sqrt[(x/(|x| + |y|)^2 + (y/(|x| + |y|))^2)])
= (x/sqrt(x^2 + y^2), y/sqrt(x^2 + y^2))
= (x, y), since x^2 + y^2 = 1 for this circle.

Hence, r and s are inverse of one another, each of which are continuous.
So, r is a homeomorphism.
---------------------
I hope this helps!

homeomorphism between unit square and unit circle?

1 Answer