# Quotient group of $\mathbb{R}^2$ by irrational line

In a section about topological groups, exercise 4.10 in I.M. James' General Topology and Homotopy Theory asks

Show that for irrational values of $\alpha$ the factor group of the real plane $\mathbb{R} \times \mathbb{R}$ by the line of gradient $\alpha$ is isomorphic to the real line $\mathbb{R}$, with the indiscrete topology.

There's something about this I'm not understanding, because I get that it's isomorphic to $\mathbb{R}$ with the normal topology.

As I understand it, the factor group $$F = \frac{\mathbb{R} \times \mathbb{R}}{(a,b) \sim (a + x, b + \alpha x)}$$ is given the quotient topology. A preimage of a set in $F$ under the projection map $\pi$ is a collection of lines in $\mathbb{R}^2$ with slope $\alpha$. So we can parameterize them by their $y$ intercept, which yields $\mathbb{R}$ with the usual topology.

More formally, the map $\phi: \mathbb{R} \times \mathbb{R} \to \mathbb{R}: (a,b) \mapsto b - \alpha a$ descends to a bijective map $\tilde{\phi}: F \to \mathbb{R}$. The map $\phi$ is such that $\phi(\mathcal{O})$ is open in $\mathbb{R}$ whenever $\mathcal{O}$ is an open set in $\mathbb{R}^2$ which is saturated with respect to $\pi$. Therefore $\tilde{\phi}$ is an open bijection, so a homeomorphism. It is also plainly a group isomorphism.

Where am I going wrong?

• Looks like nonsense to me too. You do get an indiscrete continuum if you start with the torus instead of the plane. – Kevin Carlson Jun 11 '15 at 9:06