# local homeomorphisms between real numbers

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 (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 $c$. $\mathrm{\square}$

Title | local homeomorphisms between real numbers |
---|---|

Canonical name | LocalHomeomorphismsBetweenRealNumbers |

Date of creation | 2013-03-22 18:53:50 |

Last modified on | 2013-03-22 18:53:50 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 54C05 |