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 $U\setminus A$.
Best Answer
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.