Function that wraps unit circle twice around itself I'm looking for an example of a function $f: S^1 \to S^1$ (from unit circle to unit circle), that is continuous, open and surjective but not injective. I have intuitively thought of example, that would be a function that "wraps" unit circle twice around itself. More precisely this is what I had in mind. We can uniquely write points on unit circle as $(\cos t, \sin t)$, where $t \in [0, 2 \pi)$. And the function would be $f: S^1 \to S^1$ with $(\cos t, \sin t) \to (\cos 2t, \sin 2t)$.  Now, I'm not even sure if this would be a good thing to look at, and if it was, I don't know how to show that $f$ is continuous and open.
Keep in mind that this is for course of real analysis and I'm not allowed to use facts and parametrizations from complex analysis.
Any comments or suggestions are very welcomed.
 A: Based on trigonometric equations, your suggested function can be expressed as
$$(x,y)\mapsto (x^2-y^2,\,2xy)$$
which is clearly continuous.
For the part being open, you can either directly argue that an open arc of $S^1$ will be mapped either to an open arc (of double arc length) or to the whole $S^1$.
Or, alternatively, you can show that it's piecewise locally invertible.
A: As your space is metrisable it follows that continuity is equivalent to sequential continuity. It is easy to see that your proposed function is sequentially continuous (although it may be easier to identify $S^1$ with the quotient space of $[0,2\pi]$ with end points identified, and your function $f$ as just the map $x\mapsto 2x$.)
To see that your $f$ is open notice that the collection of all open arcs form a basis for your topology, and see that $f$ must map open arcs to open arcs, or all of $S^1$.
A: To formally define the map you describe consider the map $t\mapsto 2t\mapsto (\cos(2t),\sin(2t))$. Clearly, this is a continuous map $\phi\colon [0,2\pi]\to S^1$ as it is a composition of continuous maps. It is,

*

*Open, since it is a composition of open maps.

*Surjective, since it is a composition of surjective maps.

*Not injective, as it is two-to-one.

The map $\phi$ descends to the quotient of $[0,2\pi]$ identifying the endpoints giving a map $\widetilde{\phi}\colon S^1\to S^1$. This is certainly surjective, and the properties of being open and continuous also decends to the quotient by properties of the quotient topology (look at the properties characterizing the quotient topology and chase around the commutative diagram one obtains).
