Describing continuous functions on $S^1/\sim$

Question

Let $S^1=\{e^{2\pi it}|t\in\mathbb{R}\}$ be the unit sphere. Define $\sim$ on $S^1$ where two points are identified if $t_1-t_2=\sqrt{2}k$, for some $k\in\mathbb{Z}$. Is there a way to describe continuous functions on $S^1/\sim$? I know that if $p:S^1\to S^1/\sim$ is the quotient map and $g:S^1\to Z$ is a map that is constant on $p^{-1}\{y\}\,\forall y\in S^1/\sim$ then it induces a map $f:S^1/\sim \to Z$ such that $f\circ p=g$ and $f$ is continuous $\iff$ $g$ is continuous. Any help to find such continuous maps will be a great help! Thank you

Hint: Each equivalence class of your relation is dense in the circle. — Andrew D. Hwang, Jul 31 '17 at 18:57
@AndrewD.Hwang Yes I already knew that but I don't know how that helps — Heisenberg, Jul 31 '17 at 19:07
A mapping $f$ factors through an equivalence relation precisely when $f$ is constant on equivalence classes. — Andrew D. Hwang, Jul 31 '17 at 19:30
You have not defined what $t_1$ and $t_2$ are. Do you want to say the two points are $e^{2\pi it_1}$ and $e^{2\pi it_2}$ respectively? — edm, Aug 01 '17 at 00:24

John Griffin · Answer 1 · 2017-08-01T05:11:34.923

3

Note that every pair of nonempty open sets in the quotient space $S^1/\sim$ intersect (see edit below for a proof). This is a very strong property called hyperconnectedness.

Now we can proceed using the following lemma. If $X$ is hyperconnected and $Z$ is Hausdorff, then the only continuous functions from $X$ into $Z$ are constant.

Proof: Clearly constant functions are continuous. Suppose $f:X\to Z$ is a nonconstant continuous function. Then there exists $x,y\in X$ such that $f(x)\ne f(y)$. This yields disjoint open subsets $U$ and $V$ of $Y$ such that $f(x)\in U$ and $f(y) \in V$. But then $f^{-1}(U)$ and $f^{-1}(V)$ are nonempty disjoint open sets. Contradiction. $\square$

Other examples of hyperconnected spaces include the cofinite and cocountable topologies on infinite and uncountable sets, respectively.

Edit: We use the fact that the equivalence class $[1]$ of $1$ is dense in $S^1$ to show that every pair of nonempty open sets in the quotient space $X:=S^1/\sim$ intersect.

Let $U$ be a nonempty open set in $X$ and denote by $\pi$ the quotient map $\pi:S^1\to X$. Then $\pi^{-1}(U)$ is a nonempty open set in $S^1$. By the density of $[1]$, we know that $[1]\cap \pi^{-1}(U)\ne\emptyset$. Hence there is an integer $k$ such that $e^{2\pi i\sqrt{2}k}\in\pi^{-1}(U)$. Then $\pi(e^{2\pi i\sqrt{2}k})=[1]$ implies $[1] \in \pi(\pi^{-1}(U)) \subseteq U$.

We have just shown that every nonempty open set in $X$ contains $[1]$. Therefore any two nonempty open sets must intersect, so $X$ is hyperconnected (and thus, since $|X|>1$, not Hausdorff).

edited Aug 01 '17 at 05:11

answered Aug 01 '17 at 00:16

John Griffin

10,668

Cool. Do you know any necessary conditions for hyperconnectedness? – Heisenberg Aug 01 '17 at 00:45
@Heisenberg Not too many. Check here and here. – John Griffin Aug 01 '17 at 00:58
However, I have a concern. In this case, $S^1/\sim$ is Hausdorff. If $U\cap V\neq \emptyset$ for every $U,V$ open in $S^1/\sim$, $S^1/\sim$ cannot be Hausdorff – Heisenberg Aug 01 '17 at 04:04
@Heisenberg You are correct. However, what leads you to believe that $S^1/\sim$ is Hausdorff? – John Griffin Aug 01 '17 at 04:18
Ah, it is a completely new proof. Do you think it is not Hausdorff? – Heisenberg Aug 01 '17 at 04:19
Check the answer by k.stm. Even he considers the space to be Hausdorff – Heisenberg Aug 01 '17 at 04:22
@Heisenberg You commented earlier that you are aware that every equivalence class of the relation is dense in $S^1$. This implies that the space isn't Hausdorff. I'll write up a proof. – John Griffin Aug 01 '17 at 04:50
Sorry I think I am confused. I proof will be very helpful. Thanks! – Heisenberg Aug 01 '17 at 04:58

k.stm · Answer 2 · 2017-08-01T05:43:11.197

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Let $Z = S^1 / \sim$.

Let $Y$ be a topological space and $f\colon Z → Y$ be a continuous map. Then $f ∘ p \colon S^1 → Y$ is continuous and constant on any fibre of $p$, and in particular on the dense set $T = \{\exp(2\mathrm{πi}\sqrt 2 k);~k ∈ ℤ\}$.

For any Hausdorff space $X$, any topological space $Y$ and any two continuous maps $f\colon X → Y$, $g \colon X → Y$: If $f$ and $g$ coincide on a dense subspace of $X$, then both are equal. (See here.)

In particular, since $f∘p$ (as above) is constant on a dense subspace of the Hausdorff space $X = S^1$, $f∘p$ is itself constant. By the surjectivity of $p$, $f$ must be constant as well.

edited Aug 01 '17 at 05:43

answered Jul 31 '17 at 19:24

k.stm

18,539

I don't see how it is constant on each fibre of $p$ – Heisenberg Jul 31 '17 at 19:28
@Heisenberg By definition. For any $x, y ∈ S^1$, if they belong to the same fibre of $p$, then $p(x) = p(y)$ and so $(f∘p)(x) = (f∘p)(y)$. – k.stm Jul 31 '17 at 19:31
Oh yes, by definition. Sorry – Heisenberg Jul 31 '17 at 19:31
So in this case $T=p^{-1}[0]?$ – Heisenberg Jul 31 '17 at 19:42
@Heisenberg $T = p^{-1}({[1]})$. – k.stm Jul 31 '17 at 19:56
$[1]={e^{2\pi it}|t=1+\sqrt{2}k }$ – Heisenberg Jul 31 '17 at 19:59
@Heisenberg No, as $1 = \mathrm{e}^{2\mathrm{πi}·0}$, we have $p^{-1}({[1]}) = [1] = {\mathrm{e}^{2\mathrm{πι}t};~t ∈ \sqrt 2 ℤ} = T$, i.e. $T$ is the collection of points equivalent to $1 = \mathrm{e}^{2\mathrm{πi}·0}$. – k.stm Jul 31 '17 at 20:12
Why is Z Hausdorff? – Heisenberg Aug 01 '17 at 05:41
@Heisenberg Er, sorry. It doesn’t need to be, we only need $S^1$ to be Hausdorff to conclude that. I’ll change the answer. – k.stm Aug 01 '17 at 05:41

Describing continuous functions on $S^1/\sim$

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