# homeomorphisms preserve connected components

Let $X,Y$ be topological spaces^{} and $X=\bigcup {X}_{i}$, $Y=\bigcup {Y}_{j}$ be decompositions into connected components^{}.

Proposition^{}. Assume that $f:X\to Y$ is a homeomorphism^{}. Then for any $i$ there exists $j$ such that $f({X}_{i})={Y}_{j}$.

Proof. Take any $i$. Because $f$ is continuous^{} $f({X}_{i})$ is connected, then there exists $j$ such that $f({X}_{i})\subseteq {Y}_{j}$ (because ${Y}_{j}$ is a connected component). Now $f$ is a homeomorphism, ${f}^{-1}({Y}_{j})\cap {X}_{i}\ne \mathrm{\varnothing}$, ${Y}_{j}$ is connected and ${X}_{i}$ is a connected component, so ${f}^{-1}({Y}_{j})\subseteq {X}_{i}$. Thus ${Y}_{j}\subseteq f({X}_{i})$, which completes^{} the proof. $\mathrm{\square}$

Title | homeomorphisms preserve connected components |
---|---|

Canonical name | HomeomorphismsPreserveConnectedComponents |

Date of creation | 2013-03-22 18:45:32 |

Last modified on | 2013-03-22 18:45:32 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Derivation^{} |

Classification | msc 54D05 |