$[0,1]$ quotient space of $]0,1[$, but not the other way around.

Question

I want to show that $[0,1]$ is a quotient space of $]0,1[$, but not the other way around. I found so far that $]0,1[$ is no quotient space of $[0,1]$.

Assume it was, then there would be a surjective function $f: [0,1] \longrightarrow]0,1[$, such that $\forall A \subset \, ]0,1[: A \, is \, open \, in \, ]0,1[ \Longleftrightarrow f^{-1}(A) \, is \, open \, in \, [0,1]$. Then $f$ is continuous and since $[0,1]$ is compact, $f([0,1])$ is compact too, so $]0,1[$ is compact, which is wrong. Hence there is no such function, therefore $]0,1[$ is not a quotient space of $[0,1]$.

Now to show that $[0,1]$ is a quotient space of $]0,1[$, I tried this. Let be:

$$f:\,]0,1[ \longrightarrow [0,1]: \, x \longmapsto \lvert cos(\frac{1}{x}) \rvert $$

Then f is surjective. Then all I need to show is that $\forall A \subset \, [0,1]: A \,is\,open \Longleftrightarrow f^{-1}(A) \,is\,open$.

One way is obvious. Since $f$ is continuous, if $A$ is open $\Longrightarrow$ $f^{-1}(A)$ is open.

Now I need to show that if for some $A \subset \, [0,1]$, $f^{-1}(A)$ is open, then $A$ is open too. Any hints? Or is there maybe another function that makes this way easier. Or should I try a different approach?

Answer 1

For the quotient map, you can try the much simpler (geometrically) map $$f(x)=\begin{cases}0&x\leq \frac{1}{3}\\3x-1&\frac13<x<\frac 23\\1&x\geq \frac 23\end{cases}$$

This makes describing $f^{-1}(A)$ much easier.

Answer 2

Hint: a slightly less wiggly example for the quotient mapping is $$ f(x) = \frac{1 + \sin(2\pi x)}{2} $$ But the proof is much the same for either your example or the above: show that for each $x \in ]), 1[$, and for all sufficiently small $\delta$, $f(]x - \delta, x + \delta[)$ is an open subset of $[0, 1]$ of one of the three forms $$[0, \epsilon[\\ ]y - \epsilon,y + \epsilon[\\ ]1 - \epsilon, 1] $$ for some $\epsilon > 0$ and some $y \in ]\epsilon, 1 - \epsilon[$ (The three cases correspond to the cases $f(x) = 0$, $f(x) \in ]0, 1[$ and $f(x) = 1$, respectively.) Then if $B$ is any open subset of $]0, 1[$, write $B$ as the union of sets $]x - \delta, x + \delta[$ where $x$ in $B$ and where $\delta$ is sufficiently small for one of the above three cases to hold and for $]x - \delta, x + \delta[$ to be contained in $B$. Hence write $f(B)$ as a union of open subsets of $[0, 1]$. (This shows that $f$ is an open mapping.) Taking $B = f^{-1}(A)$ gives that $A = f(B)$ is open if $f^{-1}(A)$ is open.