The 2nd chapter of Topology by Munkres discussed the identity function $$ f: \mathbb{R} \rightarrow \mathbb{R}_l $$ from $\mathbb{R}$ (with standard topology) to $\mathbb{R}_l$ (with lower limit topology).
The book reads: the inverse image of the open set $[a,b)$ of $\mathbb{R}_l$ equals itself, which is not open in $\mathbb{R}$.
My questions:
(1) How the identity function can map from $(a,b)$ to $[a,b)$? since $a\notin (a,b)$ but $a\in [a,b)$.
(2) Why the inverse image of the open set $[a,b)$ of $\mathbb{R}_l$ equals itself? I think the inverse image should be $(a,b)$, which is the "opposite" situation that $f$ maps from $[a,b)$ to $(a,b)$.
It's the identity function he discusses, i.e. $f(x)=x$, which is a well-defined function from $\Bbb R$ to $\Bbb R_l$ because both spaces have the same elements (all real numbers). And of course $a \in \Bbb R$. Maybe the arbitrary $a$ and $b$ confuse you? To show $f$ is not continuous it is enough to find one open set in $\Bbb R_l$ such that $f^{-1}[O]$ is not open in $\Bbb R$. I'll take $O=[0,1)$ for definiteness, which is open in $\Bbb R_l$ by definition.
Recall that $f^{-1}[[0,1)]= \{x \in \Bbb R \mid f(x) \in [0,1)\} = \{x \in \Bbb R \mid x \in [0,1)\} = [0,1)$ and $[0,1)$ is not open in $\Bbb R$ as no open interval that contains $0$ sits inside $[0,1)$ so that $0$ is not an interior point of $[0,1)$ in $\Bbb R$.
So we are done showing $f$ non-continuous. Note that inverse images of the identity function are very simple: $f^{-1}[O]=O$ for all $O$. And also $f[O]=O$.