Proposition. Let $I$ be an open interval and $f:I\to\mathbb{R}$ be a continuous map. Then $f$ is a local homeomorphism if and only if $f$ is a homeomorphism onto image.

Proof. ,,$\Leftarrow$ '' If $f$ is a homeomorphism onto image, then (in particular) $f$ is monotonic and continuous, thus $f(I)$ is open in $\mathbb{R}$ (please, see this entry for more details). It is easy to see that therefore $f$ is a local homeomorphism.

,,$\Rightarrow$ '' Assume that $f$ is not a homeomorphism onto image. It is well known, that this implies that $f$ is not injective (please, see this entry for more details). Let $x,y\in I$ be such that $x<y$ and $f(x)=f(y)$ . Then there exists $c\in I$ such that $x<c<y$ and $c$ is a local maximum of $f$ . Thus (since $f$ is a Darboux function) for any $\varepsilon>0$ there are points $x_{\varepsilon}, y_{\varepsilon}\in(c-\varepsilon,c+\varepsilon)$ such that $f(x_{\varepsilon})=f(y_{\varepsilon})$ . This obviously implies that $f$ cannot be locally inverted around $c$ . $\square$