Homeomorphism between unit square and unit disc

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1. The problem statement, all variables and given/known data
I want to find a bijective function from [0,1] x [0,1] -> D, where D is the closed unit disc.


2. Relevant equations



3. The attempt at a solution
I have been able to find two continuous surjective functions, but neither is injective. they are [tex]f_1(s,t)=\left((1-s)\cos{(2\pi t)}+s,(1-s)\sin{(2\pi t)}\right)[/tex] and [tex]f_2(s,t)=\left((1-s)\cos{(2\pi t)},(1-s)\sin{(2\pi t)}\right)[/tex]. I can't think of one that is injective, but there should be one because these two spaces are homeomorphic, right?

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I even have another: [tex]f_3(s,t)=\left(\sin{(2\pi s)}\cos{(2\pi t)},\sin{(2\pi s)}\sin{(2\pi t)}\right)[/tex]. However, this is also not injective since all points with s=0 are mapped to the origin.

Office_Shredder

Imagine drawing the unit circle inside of the unit disc. For a given angle t from the x-axis, by what factor do you need to multiply the unit vector in that direction to get it to reach the unit square?



Once you map the boundary to the boundary, map the interior to the interior by just stretching a point in the circle by the same factor that boundary point was stretched. You've probably heard of topology as 'rubber sheet geometry' and this is the most basic case of that; you literally want to stretch the circle to cover the square

EDIT: Whoops.. I realized my square is too big here. For some reason I was thinking the unit square would have size two. The basic idea is the same though, center the two at the origin and ask what you need to multiply the boundary of the circle by to make it map to the boundary of the square

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Thanks for the help!! I got it now.