# Obtaining the usual metric on $\mathbb R$ by "pulling back" $\arctan$

This is a part from an e-mail that a professor wrote to me. I couldn't understand what he meant by

Consider two different metrics on the real line $\mathbb R$ inducing its usual topology: one is the usual metric, the other is the metric obtained by choosing a homeomorphism $\mathbb R \cong (0,1)$ (use arctan, for example) and “pulling back" the usual metric on $(0,1)$.

How did he define the second metric on $\mathbb R$? I guess it is something like

$$d(x,y) = \frac{2}{\pi} \arctan |x-y|$$ But of course my guess is not true. Can you give a formula for the "pullback" metric?

Thanks.

• The title did not match the question : you are pulling back the usual metric, but the resulting metric is not the usual one.
– Pece
Oct 8 '14 at 19:14

Define $f(x)=\frac{1}{\pi}\arctan x+\frac{1}{2}$.
Then taking $d(x,y)=|f(x)-f(y)|$ gives $d(x,y)=\frac{1}{\pi}\left|\arctan x-\arctan y\right|$.
Well, simple simply choosing $\phi(x)=\frac{2}{\pi}\arctan x$ creates a homeomorphism between $\mathbb R$ and $(-1,1)$. So you need to adjust it somewhat to get a homeomorphism between $\mathbb R$ and $(0,1)$.
Once you have the homeomorphism, $\phi$, you'd use $d(x,y)=\left|\phi(x)-\phi(y)\right|$. You seem to have chosen $\phi(|x-y|)$, which might or might not be a metric.
The homeomorphism between $\mathbb{R}$ and $(0,1)$ is something like $f(x) = \frac{1}{\pi} \arctan(x) + \frac{1}{2}$. Now pulling back means that we define $d(x,y)$ for $x,y \in \mathbb{R}$ as the distance of their images $|f(x) - f(y)| = \frac{1}{\pi}\left|\arctan(x) - \arctan(y)\right|$. Of course we get an equivalent metric when we multiply by $\pi$, so we could use $d(x,y) = \left|\arctan(x) - \arctan(y)\right|$ as well.