Explicit homeomorphism between quotient square and torus

general-topology

Let $X = [0,1] \times [0,1]$, and let consider the quotient topology $X^* := X / ((x,0) \sim (x,1), (0,y) \sim (1,y))$. Given $r_0 > h > 0$, we define explicitly the torus as:
$$
Y_{h,r_0} = \left\{(x,y,z) : z^2 = h^2 – \left(r_0 – \sqrt{x^2 + y^2}\right)^2\right\}
$$
I want to construct an explicit homeomorphism $f : X^* \to Y_{h,r_0}$.

I believe I've managed to come up with the homeomorphism. We express points in $Y_{h,r_0}$ in terms of cylindrical coordinates, and define $f$ as follows:
$$
f(x,y) = (r_0 – h\cos(2\pi y), 2\pi x, h\sin(2\pi y))
$$
One can check that this is well-defined up to quotient topology, and is clearly continuous as it is continuous coordinate-wise. I've also constructed the inverse map $g : Y_{h,r_0} \to X^*$:
$$
g(r,\theta,z) = \left(\frac{\theta}{2\pi}, \frac{1}{2\pi}\arg(r_0 – r + iz)\right)
$$
One can also check that this well defined up to quotient topology, and it is both the left and the right inverse of $f$. However, I'm struggling to prove that it is continuous (I'm not entirely sure if it's even continuous). I believe the second argument is indeed continuous, but it seems complicated. I'm also not sure if it's possible to define $g$ differently such that no complex numbers are involved, and the function will still not be too complicated.

Any help is appreciated.

Best Answer

your construction might work but verifying its validity takes time, so I would recommend the following process. I'll provide two methods in total because the second one aims to help you prove the continuity. :)

Note: the second one is almost the same as your process but I'll change something.

Let $T$ denote the torus.

  1. $T$ can be imbedded in 4-D space by $$T=\{(\cos(2\pi x),\sin(2\pi x),\cos(2\pi y),\sin(2\pi y)):x,y\in[0,1)\}\subset\Bbb{R}^4$$ which implies that any point in $T$ is on two circles simultaneously.

Def $f:X^*\to T$ explicitly by $$f(x,y)=((\cos(2\pi x),\sin(2\pi x)),(\cos(2\pi y),\sin(2\pi y)))$$ This is a continuous bijection from a compact space to a hausdorff space because each coordinate function is continuous and $x,y\in[0,1]$. And also well-defined since $f(0,y)=(1,0,\cos(2\pi y),\sin(2\pi y))=f(1,y)$ & similarly $f(x,0)=f(x,1)$

Now, take a closed set $A\subset X^*$, then $A$ is compact. And, $f(A)\subset T$ is also compact because $f$ is continuous, and is closed because it's a compact subspace of a Hausdorff space. Thus, its inverse is continuous $\implies$ homeomorphism. (does not require complex numbers)

  1. Let $R$ denote the distance from the center of $T$ to the center of the tube (hope this is clear, if not pls let me know) and $r$ denote the radius of the tube.

Define $f:X^*\to T$ using the parametrical definition of a torus. $$(x,y)\mapsto (x',y',z')$$ $$f(x,y)=((R+r\cos (2\pi x))\cos (2\pi y),(R+r\cos (2\pi x))\sin (2\pi y),r\sin (2\pi x))=(x',y',z')$$ which is continuous and bijective cuz you already know that.

In this case $f^{-1}$ should be the following one, not $g$ in your post. $$(x',y',z')\mapsto (x,y)$$ $$f^{-1}(x',y',z')=\bigg(\dfrac{\tan^{-1}(\frac{z'}{\sqrt{x'^2+y'^2}-R})}{2\pi},\dfrac{\tan^{-1}(\frac{y'}{x'})}{2\pi}\bigg)=(x,y)$$ You can substitute the expression for $x',y',z'$ to verify it.

Give a sequence $(x'_n)\to0$, $\tan(y'/x'_n)\to \pi/2$ (the limit exists because our the domain is restricted) $\implies $ the $y$-coordinate function is indeed continuous at that point, similarly for $x$-coordinate function. Thus $f^{-1}$ is continuous, and we're done. Note that I didn't use complex numbers and the argument isn't so long.

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