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.

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.
