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?


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

  • $\begingroup$ Is there any reason why the function is defined as 1 for x greater than two thirds ? $\endgroup$ – Zeta10 Apr 17 '15 at 7:39
  • 2
    $\begingroup$ @Zeta10 This quotient map (you can see that it is surjective and that it is continuous) identifies $(0, \frac{1}{3}]$ with $\{ \frac{1}{3} \}$ and identifies $[\frac{2}{3}, 1)$ with $\{ \frac{2}{3} \}$. So, $(0,1)$ is somehow shrinked to $[\frac{1}{3}, \frac{2}{3}]$, which is homeomorphic to $[0,1]$. $\endgroup$ – Crostul Apr 17 '15 at 8:08
  • $\begingroup$ @Crostul How do I show that the continuous onto function $f(x)$ is indeed a quotient map? $\endgroup$ – mr eyeglasses Sep 22 '15 at 19:06
  • $\begingroup$ @morphic It's very easy to see this: take any set of $[0,1]$ whose preimage is open: show that this must be open. $\endgroup$ – Crostul Sep 22 '15 at 19:20

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