How to prove that the quotient space of the punctured plane under dilation is homeomorphic to a torus?

Question

Suppose we're considering the map $f:\mathbb R^2\backslash (0,0)\rightarrow \mathbb R^2\backslash (0,0)$ given by $f(x_1,x_2)=(cx_1,cx_2)$ with $c$ a positive real number. How does one show that the quotient space of $\mathbb R^2\backslash (0,0)$ under this map is homeomorphic to the torus $S^1\times S^1$?

Answer 1

The idea of the proof is to guess at a formula for a quotient map $$p : \mathbb{R}^2-\{(0,0)\} \to S^1 \times S^1$$ having the property that $p(x_1,x_2)=p(y_1,y_2)$ if and only if $(x_1,x_2) \sim (y_1,y_2)$ where $\sim$ is the equivalence relation generated by $(x_1,x_2) \sim (c y_1, c y_2)$, as suggested in the comment of Najib. Then, once you have guessed the appropriate formula for $p$, prove that it is indeed a quotient map.

The hard part, I suppose, is the initial guess.

The function $p$ will have coordinate functions $$p(x_1,x_2) = (p_1(x_1,x_2),p_2(x_1,x_2)) $$ As a preliminary, define $r(x_1,x_2) = \sqrt{x_1^2+x_2^2}$ which is just the distance from the origin to $(x_1,x_2)$.

Define $$p_1(x_1,x_2) = \frac{1}{r(x_1,x_2)} (x_1,x_2) $$ This map is just "radial projection" of the point $(x_1,x_2)$ to the circle, meaning that you draw the ray from the origin through $(x_1,x_2)$ and you intersect that ray with $S^1$ to get the point $p_1(x_1,x_2)$.

Next define $$p_2(x_1,x_2) = exp\biggl(\frac{2 \pi i}{\ln(c)} \ln(r(x_1,x_2))\biggr) $$ The idea for this one is that you take any ray from the origin, then you take the point of radius $r$ on this ray, then you map that point to $\ln(r)$ on the real line, and then you wrap the result around the circle in the usual way using the complex exponential map, but scaled so that the fundamental domain is $[0,\ln(c)]$ rather then the usual fundamental domain $[0,2\pi]$.

Now verify that the map $p$ is a quotient map: this is not hard to do directly from the definition of a quotient map; alternatively, you could verify that $p$ is a covering map and use the theorem that every covering map is a quotient map.

Finally, verify using the formula for $p$ that if you are given two points $(x_1,x_2)$ and $(y_1,y_2)$ then $p(x_1,x_2)=p(y_1,y_2)$ if and only if there exists an integer $k$ such that $(y_1,y_2) = (c^k x_1, c^k x_2)$.

Answer 2

Here's an pictorial alternative to Lee Mosher's answer that uses fundamental domains:

As you correctly guessed, a fundamental domain is given by an annulus. Let $X = \mathbb{R}^2 \setminus (0,0)$ for convenience. Then for example you can take $D = \{ x \in X : 1 \le \|x\| < c \}$ as a fundamental domain. Now if you look at the closure of the fundamental domain and what identifications you need to make you get this:

If you cut up this annulus along a closed segment, you then get the usual picture of a torus as a square with opposing sides identified (with the correct orientation).

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Another possibility (basically Lee Mosher's answer in a different form) is to notice that $X \cong S^1 \times (0,+\infty)$ through $\varphi(x) = (\arg(x), \|x\|)$ where $\arg(x)$ is the angle that the vector $x$ represents makes with the horizontal (or simply view $X$ as $\mathbb{C}^*$). Then through the isomorphism, the equivalence relation becomes $$(\theta, u) \sim (\theta, cu).$$ If you go through yet another isomorphism $S^1 \times (0,+\infty) \cong S^1 \times \mathbb{R}$, namely $\psi(\theta, u) = (\theta, \ln u)$, then finally the identification becomes: $$(\theta, r) \sim (\theta, r + \ln c)$$ and now it's easy to see that it's $S^1 \times (\mathbb{R} / (\ln c) \mathbb{Z}) \cong S^1 \times S^1$.