# restriction of a continuous mapping is continuous

Theorem Suppose $X$ and $Y$ are topological spaces^{}, and suppose
$f:X\to Y$ is a continuous function^{}. For a subset $A\subset X$,
the restriction^{} (http://planetmath.org/RestrictionOfAFunction)
of $f$ to $A$ (that is ${f|}_{A}$) is a continuous
mapping ${f|}_{A}:A\to Y$, where $A$ is given the subspace topology
from $X$.

*Proof.* We need to show that for any open set $V\subset Y$, we
can write ${({f|}_{A})}^{-1}(V)=A\cap U$ for some set $U$ that is open in $X$.
However, by the properties of the inverse image (see
this page (http://planetmath.org/InverseImage)), we have for any open set $V\subset Y$,

$${({f|}_{A})}^{-1}(V)=A\cap {f}^{-1}(V).$$ |

Since $f:X\to Y$ is continuous, ${f}^{-1}(V)$ is open in $X$, and our claim follows. $\mathrm{\square}$

Title | restriction of a continuous mapping is continuous |
---|---|

Canonical name | RestrictionOfAContinuousMappingIsContinuous |

Date of creation | 2013-03-22 13:55:53 |

Last modified on | 2013-03-22 13:55:53 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 6 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 54C05 |

Classification | msc 26A15 |