An injective continuous self-map of the unit circle is a homeomorphism 
Let $U$ be the set of complex numbers with magnitude $1$.
Let $f: U \to U $ be an injective, continuous map.
Prove that $f$ is a homeomorphism.

Since $U$ is compact, it suffices to prove that $f$ is surjective. I supposed for contradiction's sake that $f$ is not surjective, that is, there is some $u\in  U\setminus f(U)$.
Since $U\setminus f(U)$ is open, there is an open ball $B$ with radius $r$ centered at $u$ such that $B \subset U\setminus f(U)$.
This implies that an arc $A$ is inside $U\setminus f(U)$.
How should this lead to a contradiction ?
Obviously, $f$ is now an injection from $U$ to the smaller set $U\setminus A$.
 A: Suppose that $f$ is not surjective, then $f(U) \subset U \setminus \{u\}$ for some $u$. It's a well known fact that $U \setminus \{u\}$ is homeomorphic to $(0,1)$. $f(U)$ is a connected subset of $(0,1)$ (because $U$ is connected), thus it's an interval $(a,b)$ with $a<b$. Let $c \in (a,b)$ and $v \in U$ such that $f(v) = c$. Then by injectivity $f(U \setminus \{v\}) = (a,c) \cup (c,b)$. But $U \setminus \{v\}$ is connected, while $(a,c) \cup (c,b)$ isn't, a contradiction.
A: Let me add a different argument. 
Denote your set of complex numbers by $\mathbb{S}^2$. Then, you have a continuous map $f$ between a compact and a Hausdorff space which therefore is closed. 
Since, $f$ is injective you have that $\mathbb{S}_2$ is bijective with $f(\mathbb{S}_2)$ and using the above you have a bijective continuous and closed map $f|_{\mathbb{S}^2}$ that has to be homeomorphism. 
Then, one uses Invariance of Domain theorem which basically states that any set homeomorphic to an open set in $\mathbb{R}^n$ needs to be open itself. 
So take a point $x \in \mathbb{S}^2$ and an open neighborhood $U \in \mathcal{V}_x$. Compute its image $f(x) \in \mathbb{S}^2$ and $f(U)$. By Invariance of Domain, you have that $f(U)$ is an open neighborhood of $f(x)$. In this way you show that $f(\mathbb{S}^2)$ has to be open. 
But certainly, $f(\mathbb{S}^2)$ is closed since $f$ and $\mathbb{S}^2$ are closed. By connectedness you have shown that $f(\mathbb{S}^2) = \mathbb{S}^2$, so that $f$ is an homeomorphism. 
