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I am asked to prove that a square is homeomorphic to a circle.

Now we can construct the homeomorphism explicitly by first having a bijection $\gamma$ that takes an arbitrary square in $\mathbb{R}^2$ to the unit square. The bijection $\gamma$ may include a rotation, scaling and translation.

Afterwards, we can construct another explicit bijection $\beta$ that takes the unit square to the unit circle, and then finally a bijection $\alpha$ that takes the unit circle to any arbitrary circle in $\mathbb{R^2}$ using perhaps a scaling and a translation.

So the explicit homeomorphsim that takes any square to any circle in $\mathbb{R}^2$ is $$\alpha \circ \beta \circ \gamma $$

But wpuld this sufficient as a 'proof' of the square being homeomorphic to a circle?

This 'method of proof' appears to break down if I changed the square to a triangle, where the bijection from an arbitrary triangle to the equilateral triangle is not so obvious.

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Trogdor
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2 Answers2

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The short answer is: yes. It's always the case that if you demonstrate an explicit homeomorphism between two spaces, then the spaces are homeomorphic.

As for the part about triangles, this should work just as easily as the case for the square. Note that homeomorphisms don't require the lengths to be proportional to one another. If you can construct explicit homeomorphisms say $\gamma_1, \gamma_2, \gamma_3$, from each side of an arbitrary triangle to the sides of some equilateral triangle (with origin in the interior), such that all maps agree on their respective endpoints, then we can construct a map $\gamma$ that is pointwise defined by $\gamma_1, \gamma_2, \gamma_3$. Then the composition $\alpha \circ \beta \circ \gamma$ would work the same for the triangle as for the square.

Mnifldz
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Topology is great at (abstractly) determining when two spaces are not homeomorphic, but if you want to show that two spaces are homeomorphic, your best bet is to write a homeomorphism down.

Rick Sanchez
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