Find a quotient map $f:(0,1) \rightarrow [0,1]$ where the intervals $(0,1)$ and $[0,1]$ are in $\mathbb{R}$ and endowed with the subspace topology.

Question

Find a quotient map $f:(0,1) \rightarrow [0,1]$ where the intervals $(0,1)$ and $[0,1]$ are in $\mathbb{R}$ and endowed with the subspace topology.

I am really not to sure where to start. I know that quotient map must be continuous and surjective. So, it makes sense to show the following

$$\forall\ U \in [0,1], f^{-1}(U) \in (0,1)$$ with $f^{-1}(U)$ open.

Case 1:

$U \in (0,1)$ then $f^{-1}(U) \in (0,1)$ and $f(f^{-1}(U))=U$ and $0,1 \notin (0,1)$

Case 2:

Let $f^{-1}(0)=f^{-1}(1)= \frac{1}{2}$ so that $U_1=[\frac{1}{2}, 1-\epsilon)$ and $U_2=(0+\epsilon,\frac{1}{2}]$ and $f^{-1}(U_1 \cup U_2)$ and then $$f(f^{-1}(U_1 \cup U_2))=(0,1)$$.

Am I on the right track? Also do I need to show what the equivalence class that I created is? If so, how do I show that?

Answer 1

Define $f: (0,1) \longrightarrow [0,1]$ as follows $$f(x)= \left\{\begin{matrix} 0 & \mbox{if } & 0<x < \frac{1}{3} \\ 3x-1 & \mbox{if } &\frac{1}{3} \leq x \le \frac{2}{3} \\ 1 & \mbox{if} & \frac{2}{3} < x < 1 \end{matrix} \right. $$