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Problem: Define the equivalence relation $R$ on $\mathbb R$ by $xRy$ iff $x-y\in \mathbb Z$ and define the equivalence relation $Q$ on $[0,1]$ by $xQy$ iff $x=y$ or $\{x,y\}\subset \mathbb Q\cap [0,1]$, i.e. both are rational. (So $\mathbb R$ essentially gets compressed down to $[0,1)$ (?) and all the rational numbers in $[0,1]$ get glued together.)

(i) Show that there exists a continuous surjection $f\colon \mathbb R/R\to [0,1]/Q$.

(ii) Show that every continuous map $g\colon [0,1]/Q\to \mathbb R/R$ is constant.

So far: For (i) I'm pretty much completely stuck. I was thinking of using some universal property of the coinduced topology/final topology and/or quotient maps. However this would give me the existence of a continuous such $f$, but not surjectivity. For (ii) I think the central idea lies in the fact that all rationals get clumped together under $Q$. If I assume $g([x]_Q)\neq g([x']_Q)$ for some $[x]_Q,[x']_Q\in [0,1]/Q$, then I should (?) be able to reach some form of contradiction.

categoricallystupid
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  • If this is a quotient topology, then $\mathbb R/R$ is not "compressd down to" $[0:1).$ That set is a proper set of representatives, but te topology is different than the usual topology on $[0,1).$ – Thomas Andrews Sep 30 '24 at 18:50
  • Yes, I know the topology on the quotient is coinduced by the quotient map, that's just how I visualize the space. Why do you say those elements are equivalent under $Q$? – categoricallystupid Sep 30 '24 at 18:58
  • Sorry, misread the second condition. – Thomas Andrews Sep 30 '24 at 19:08

1 Answers1

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So $\mathbb{R}/R$ is just circle $S^1$ (see here), so lets use it for simplicity. But it doesn't really matter that much. The reasoning will actually work for a wide range of spaces, at least any normal space that contains a homeomorphic image of $[0,1]$.

Then there is an obvious surjective map $p:S^1\to [0,1]$, namely the absolute projection $p(x,y)=|x|$. Here we treat $S^1$ as the unit circle in $\mathbb{R}^2$. Such map will also exist for the mentioned above wide range of spaces, due to Tietze extension theorem (applied over the embedded interval). This gives us a straightforward composition of surjective continuous maps:

$$\mathbb{R}/R\to S^1\to [0,1]\to [0,1]/Q$$

where the last arrow is the quotient map $\pi:[0,1]\to[0,1]/Q$.

As for the second point, we need to have a closer look at the topology of $Z=[0,1]/Q$. Denote by $\omega$ the special point in $Z$, the equivalence class of $\mathbb{Q}\cap [0,1]$. The topology of $Z$ can be described as follows: $U\subseteq Z$ is open if and only if $\pi^{-1}(U)$ is open in $[0,1]$ and $\mathbb{Q}\cap[0,1]\subseteq\pi^{-1}(U)$. In particular every open subset in $Z$ contains $\omega$.

Now consider a continuous function $f:Z\to H$, where $H$ is any Hausdorff space with at least two points. Assume that this function is not constant. In particular there are $\alpha, \beta\in Z$ such that $f(\alpha)\neq f(\beta)$. Now pick two disjoint open neighbourhoods $U,V\subseteq H$ of $f(\alpha)$ and $f(\beta)$ respectively. Then $f^{-1}(U)$ and $f^{-1}(V)$ have to be disjoint as well. But that cannot happen since every open subset in $Z$ contains $\omega$. Contradiction. And so $f$ is constant.

freakish
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