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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 sure where to start. Can you please help?

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    $\begingroup$ For (2), since the topology of the Sorgenfrey line is strictly finer than the usual topology, the identity map will be continuous. $\endgroup$
    – Aaron
    Commented Jul 5, 2011 at 1:08
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    $\begingroup$ @Aaron: thanks, yeah, I was about to type that. Still stuck with 1). Any idea? $\endgroup$
    – user10
    Commented Jul 5, 2011 at 1:09
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    $\begingroup$ 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. $\endgroup$
    – Aaron
    Commented Jul 5, 2011 at 1:16
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    $\begingroup$ @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. $\endgroup$
    – user10
    Commented Jul 5, 2011 at 1:25
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    $\begingroup$ @user10 Yes, it would work. You have now answered your own question! You can indeed answer your own question if you wish to record the answers that you have found for the future. (You can write the answers in the "Your Answer" box below and click "Post Your Answer" at the bottom of the page.) $\endgroup$ Commented Jul 5, 2011 at 1:44

1 Answer 1

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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.

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