local homeomorphisms between real numbers

Proposition. Let $I$ be an open interval and $f:I\to ℝ$ 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 $ℝ$ (please, see this entry (http://planetmath.org/InjectiveMapBetweenRealNumbersIsAHomeomorphism) 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 (http://planetmath.org/InjectiveMapBetweenRealNumbersIsAHomeomorphism) for more details). Let $x,y\in I$ be such that and $f(x)=f(y)$. Then there exists $c\in I$ such that and $c$ is a local maximum of $f$. Thus (since $f$ is a Darboux function) for any $\epsilon >0$ there are points $x_{\epsilon },y_{\epsilon }\in (c-\epsilon ,c+\epsilon )$ such that $f(x_{\epsilon })=f(y_{\epsilon })$. This obviously implies that $f$ cannot be locally inverted around $c$. $\mathrm{\square }$