If $f:X\to Y$ is defined by $f(x)=1,x<0$; $f(x)=2,x\geq 0$, and $X$ and $Y$ have Sorgenfrey topology, is $f$ continuous?

Question

Let $X$ and $Y$ be spaces. If $f:X\to Y$ is defined by $f(x)=1,x<0$; $f(x)=2,x\geq 0$, and $X$ and $Y$ have Sorgenfrey topology, is $f$ continuous?

I am not sure how to think about this. Any ideas?

Answer 1

If $O$ is open in $Y$ and $1,2 \notin O$ then $f^{-1}[O]=\emptyset$ is open in $X$.

If $O$ is contains both $1$ and $2$ then $f^{-1}[O] = X$ is open in $X$.

If $O$ contains $1$ but not $2$, $f^{-1}[O] = (-\infty,0)$ is Sorgenfrey open (even usual-open)

If $1 \notin O$ and $2 \in O$ then $f^{-1}[O] =[0,\infty)$ is Sorgenfrey open (union of $[n,n+1), n \in \Bbb N$, e.g.)

So whatever $Y$ is and $O \subseteq Y$, $f^{-1}[O]$ is open in the Sorgenfrey line, so the definition of continuity tells us that $f$ is continuous.

Answer 2

Let [a,b) be a open base set of Y.
Show f$^{-1}$([a,b)) is an open subset of X.