Describing continuous functions on $S^1/\sim$ 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 
 A: 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).
A: 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.
