# composition with coercive function

###### Theorem 1.

Suppose $X\mathrm{,}Y\mathrm{,}Z$ are topological spaces^{},
$f\mathrm{:}X\mathrm{\to}Y$ is a bijective^{} proper map, and
$g\mathrm{:}Y\mathrm{\to}Z$ is a coercive map.
Then $g\mathrm{\circ}f\mathrm{:}X\mathrm{\to}Z$ is a coercive map.

###### Proof.

Let $J\subseteq Z$ be a compact set. As $g$ is coercive, there is a compact set $K\subseteq Y$ such that

$$g(Y\setminus K)\subseteq Z\setminus J.$$ |

Let $I={f}^{-1}(K)$, and since $f$ is a proper map $I$ is compact. Thus

$$(g\circ f)(X\setminus I)=g(Y\setminus K)\subseteq Z\setminus J$$ |

and $g\circ f$ is coercive. ∎

Title | composition with coercive function |
---|---|

Canonical name | CompositionWithCoerciveFunction |

Date of creation | 2013-03-22 15:20:16 |

Last modified on | 2013-03-22 15:20:16 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 6 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 54A05 |