# composition of continuous mappings is continuous

###### Theorem 1.

The composition^{} of two continuous mappings (when
defined) is continuous^{}.

###### Proof.

Let $X,Y,Z$ be topological space^{}, and let $f,g$ be
mappings

$f:X$ | $\to $ | $Y,$ | ||

$g:Y$ | $\to $ | $Z.$ |

We wish to prove that $g\circ f$ is continuous. Suppose $B$ is an open set in $Z$. Since $g$ is continuous, ${g}^{-1}(B)$ is an open set in $Y$, and since $f$ is continuous, ${f}^{-1}({g}^{-1}(B))$ is an open set in $X$. Since ${f}^{-1}({g}^{-1}(B))={(g\circ f)}^{-1}(B)$, it follows that ${(g\circ f)}^{-1}(B)$ is open and the composition if continuous. ∎

Title | composition of continuous mappings is continuous |
---|---|

Canonical name | CompositionOfContinuousMappingsIsContinuous |

Date of creation | 2013-03-22 15:16:52 |

Last modified on | 2013-03-22 15:16:52 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 8 |

Author | mathcam (2727) |

Entry type | Theorem |

Classification | msc 26A15 |

Classification | msc 54C05 |