1
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.