Does a bijection $S^1 := \{ (x,y): x^2 + y^2 = 1\} \to \mathbb{R}$ exist? This is a question I encountered on the first Linear Algebra Exercise sheet of a friend, who just started studying maths. I have already completed basic modules like linear algebra and analysis, but was stumped by this question. 

Find a surjective mapping $\mathbb{R} \to S^1$.
  Does there exists a injective mapping $\mathbb{R} \to S^1$?

Easily, $f: \mathbb{R} \to S^1, \ x \mapsto \left(\cos(x), \sin(x) \right)$ is a surjective mapping, but I having trouble finding a function which is also injective and also simple, since a maths beginner should (dis)prove the second question.
I am also interested how such a bijective mapping would look like.
 A: A standard example is$$\begin{array}{ccc}f\colon&\mathbb R&\longrightarrow&S^1\\&x&\mapsto&\displaystyle\left(\frac{1-x^2}{1+x^2},\frac{2x}{1+x^2}\right).\end{array}$$It is injective and it range is $S^1\setminus\bigl\{(-1,0)\bigr\}$.
A: The following answers the question in the title and on the last line of your post.
For topological reasons a bijective map $f:\>S^1\to{\mathbb R}$ can only be obtained using some trickery. Note that $S^1$ is compact, but ${\mathbb R}$ is not. As a consequence there is no continuous map $f:\>S^1\to{\mathbb R}$ which is surjective. It follows that we cannot set up a simple expression defining a map having all desired properties. 
Here is an example which I made up ad hoc; maybe there are simpler constructions: Beginning with $z_0:=(1,0)$ choose an infinite sequence of distinct points $z_k$ along $S^1$ such that after one full turn the $z_k$ converge to $(1,0)$ "from the other side". These $z_k$ partition $S^1$ into infinitely many half-open arcs $[z_{k-1},z_k[^{\rm (arc)}\>$ $(k\geq1)$. Map the first of these arcs bijectively onto $[0,1[\>\subset{\mathbb R}$, the second onto $[{-1},0[\>$, the third onto $[1,2[\>$, the next onto $[{-2},{-1}[\>$, and so on.
A: Since $S^{1}$ and $\mathbb{R}$ have the same cardinal, there exits a bijection between $S^{1}$ and $\mathbb{R}$.
