# Continuous map from Sorgenfrey line

Let $\mathbb{R}_{\ell}$ the Sorgenfrey line and let $\mathbb{Q}$ endowed with usual topology.

I have two questions:

1) Is there a continuous surjective map $f: \mathbb{R}_{\ell} \rightarrow \mathbb{Q}$?

2) Is there a continuous surjective map $f: \mathbb{R}_{\ell} \rightarrow \mathbb{R}$ where $\mathbb{R}$ has the usual topology?

Not off the top of my head. The most obvious reason why there are no continuous surjective maps $\mathbb{R}\to \mathbb{Q}$ is because $\mathbb{Q}$ is not connected. However, neither is $\mathbb{R}_{\ell}$. If I come up with something later, I let you know. – Aaron Jul 5 '11 at 1:16
@Aaron: would this work? take $f: \mathbb{R}_{\ell} \rightarrow \mathbb{Z}$ by the floor function, then $f$ is surjective and continuous. Now $\mathbb{Z}$ and $\mathbb{Q}$ are both countable so there is a bijection $g: \mathbb{Z} \rightarrow \mathbb{Q}$. Since $\mathbb{Z}$ is discrete this map is cts and surjective so $g \circ f$ is the desired map. – user10 Jul 5 '11 at 1:25
Let $f: \mathbb{R}_{\ell} \rightarrow \mathbb{Z}$ be the map given by the floor function, then $f$ is surjective and continuous. Now $\mathbb{Z}$ and $\mathbb{Q}$ are both countable so there is a bijection $g: \mathbb{Z} \rightarrow \mathbb{Q}$. Since $\mathbb{Z}$ is discrete this map is cts and surjective so $g \circ f$ is the desired map.