Continuity of Left Inverse

Question

Let $f:X\rightarrow Y$ be an injective continuous function between metric spaces and $f(X)$ be dense in $Y$. Clearly $f$ has a left-inverse but does it have a continuous left-inverse?

Answer 1

Take for example $f : [0,1) \to S^1, f(t) = e^{2\pi i t}$. This is a continuous bijection but certainly $f^{-1}$ is not continuous. Thus, even if $f(X) = Y$, the answer is in general "no".

A neccesary conditon for the existence of a continuous left inverse $g : Y \to X$ is $f(X) = Y$, i.e. that $f$ is a bijection. Too see this, let $y \in Y$. There exists a sequence $(y_n)$ in $f(X)$ such that $y_n \to y$. Letz $x_n \in X$ be the unique element such that $f(x_n) = y_n$. Then $x_n = g(f(x_n)) = g(y_n) \to g(y)$ by continuity of $g$ and $y_n = f(x_n) \to f(g(y))$ by continuity of $f$. By the uniqueness of limits in metric spaces we conclude $y = f(g(y)) \in f(X)$.

A sufficient criterion for the existence of a continuous left inverse is that $f$ is a bijection and $X$ is compact.