Simplest way to prove that $e^{ix}$ is an open mapping into $S^1$ Let $S^1$ be the unit circle in $\mathbb R^2$ and give it the subspace topology. What's the simplest way to prove that $f:R\to S^1$, $f(x)=e^{i x}$ is an open mapping, that is $f(U)$ is open when $U$ is open.
 A: You can use the fact that the map $q:[0,1]\to S^1,t\mapsto e^{2\pi i t}$, is a quotient map, being a continuous map from a compact space to a Hausdorff space. The map $f:\Bbb R\to S^1,\ x\mapsto e^{2\pi i x}$, is open because it maps an open interval $(a,b)$ to the image of the open saturated set
$$[0,1]\cap((a-\lfloor a\rfloor,b-\lfloor a\rfloor)\cup(a-\lfloor a\rfloor-1,b-\lfloor a\rfloor-1))$$ under $q$.
A: It follows from $f$ being locally a homeomorphism:
For $U \subset R$ an open set, if $y \in f(U)$, then take $x \in U$ s.t. $f(x) = y$.  Now restrict to a sufficiently small neighborhood $V \subset U$ of $x$, then $f$ is a homeomorphism on this neighborhood (you can give the inverse explicitly with some branch of $\log$.)  Thus, $f(V) \subset f(U)$ is open.
So, for each point $y \in f(U)$, there is an open nbhd $y \in W \subset f(U)$, so $f(U)$ is open.
A: I'm not sure what you mean by "simplest." Perhaps the simplest (in the sense of easiest to state succinctly) is to note that $z\mapsto e^z$ is a holomorphic map, all (nonconstant) holomorphic maps are open maps,[*] and so the restriction to the real axis is an open map onto its image in the subspace topology.
This is, however, certainly not elementary.
[*] This is the open mapping theorem.
A: For every open interval $I$ in $\mathbb{R}$, we can easily check that  $f(I)$ is open in $S^1$.
And every open set of $\mathbb{R}$ is union of some open intervals.
Let $U$ be an open subset of $\mathbb{R}$.
Then $U = \bigcup_{j \in J} I_j$ for some open intervals $I_j$ with an index set $J$.
Then $f(U) = f(\bigcup_{j \in J} I_j) = \bigcup_{j \in J} f(I_j)$,
and since it is the union of open sets, $f(U)$ is open.
Therefore $f$ is an open map.
