Example of cont. map. $\psi: \mathbb{R}_l \to \mathbb{R}_l$ such that $ \phi: \mathbb{R}\to \mathbb{R} $, def. by $ \phi(x)=\psi(x) $, is not cont.

Question

$ \mathbb{R}_l $ denotes the Sorgenfrey line or the Lower limit topology generated by the half-open intervals $ [a,b) $ and $ \mathbb{R} $ denotes the usual euclidean topology in $ \mathbb{R} $. Can someone help me to find a continuous mapping $\psi: \mathbb{R}_l \longrightarrow \mathbb{R}_l$ such that $ \phi: \mathbb{R}\longrightarrow \mathbb{R} $, defined by $ \phi(x)=\psi(x) $, is not continuous? I can't find the same question here on this site, and I have a hard time coming up with any mapping myself.

Answer 1

Let $\phi(x)=1$ if $x\geq 0$ and $\phi(x)=0$ if $x<0.$

For any $S\subset \mathbb R_l$ we have $$\phi^{-1}S\in \{\emptyset, \mathbb R_l, [0,\infty), (-\infty,0)\},$$ so $\phi^{-1}S$ is Sorgenfrey-open, and $\phi$ is Sorgenfey-continuous. But $\phi$ is obviously not $\mathbb R$-continuous.

Remark: It is not necessary to use a different symbol ($\psi$) for the function when considered as a function on $\mathbb R.$ It's the same function as $\phi.$

Remark: This is the same idea as in the answer by Dejan Govc, which seems to have generated a lengthy discussion.

Answer 2

There are a lot of ways to achieve this. For instance, you can notice that $\mathbb R_l$ is disconnected, since $(-\infty,0)\cup[0,\infty)$ is a separation. In particular, there is a surjective continuous map $\mathbb R_l\to\{0,1\}$. Compose this with an injective map $\{0,1\}\to\mathbb R_l$ and you are done.