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In our topology class we learn that in $\mathbb{R}^2,$ circles and rectangles are homeomorphic to each others.
I can understand the underline idea intuitively.
But can we find an explicit homeomorphic between them?
If so how?

Also our professor said that, "we can describe any point in the rectangle $[0,1]\times[0,1]$ using a single coordinate."
I wonder how such thing is possible. As I think, for this we need a bijection between $[0,1]\times[0,1]$ and some (closed?) interval in $\mathbb{R}.$
Can some one explain this phenomena?

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  • $\begingroup$ It is a fact that $[0,1]\times[0,1]$ and $[0,1]$ have same cardinality (I'm sure you can find reference). The most natural homeomorphisms between a circle and a rectangle are defined by parts, so it can become a little cumbersome $\endgroup$ – Luiz Cordeiro Sep 15 '16 at 19:48
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One very visual way of seeing such a homeomorphism is to place one inside the other, choose a point that's inside both (say point $P$), and let the image of a point $Q$ in the circle be where the line emanating from $P$ and through $Q$ intersects the rectangle.

EDIT: Added an example image:

enter image description here

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  • $\begingroup$ I hope the image helps explain the homeomorphism. I'm not sure I understand your question. $\endgroup$ – Fimpellizieri Sep 15 '16 at 19:53
  • $\begingroup$ Wow. yes I got it. $\endgroup$ – Bumblebee Sep 15 '16 at 19:55
  • $\begingroup$ Also, what do you think aout the second question? $\endgroup$ – Bumblebee Sep 15 '16 at 20:53
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consider $S^1$ parametrized by $(\cos(t),\sin(t))$. Extend this vector from the origin to its first intersection with the square. You can explicate the first quadrant and then just argue by symmetry. For $0 \leq t \leq \pi/4$, you're going to intersect with the $x=1$ line segment, and for $\pi/4 \leq t \leq \pi/2$, the vector will intersect $y=1$ at $\cot(t)$.

However, the real idea is given in the answer by Fimpellizieri.

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