# gluing together continuous functions

\PMlinkescapephrase

closed sets^{}
\PMlinkescapephraseopen set
\PMlinkescapephraseopen sets
\PMlinkescapephraseopen subset

## Introduction

Suppose we have a collection^{} $\mathcal{S}$ of subsets of a topological space^{} $X$,
and for each $A\in \mathcal{S}$ we have a continuous function^{} ${f}_{A}:A\to Y$,
where $Y$ is another topological space.
If the functions ${f}_{A}$ agree wherever their domains intersect,
then we can glue them together in the obvious way to form a new function
$f:\cup \mathcal{S}\to Y$.
The theorems in this entry
give some sufficient conditions for $f$ to be continuous.

## Theorems

###### Theorem 1.

Let $X$ and $Y$ be topological spaces,
let $\mathrm{S}$ be a locally finite collection of closed subsets (http://planetmath.org/ClosedSet) of $X$,
and let $f\mathrm{:}\mathrm{\cup}\mathrm{S}\mathrm{\to}Y$ be a function
such that the restriction^{} (http://planetmath.org/RestrictionOfAFunction) ${f\mathrm{|}}_{A}$
is continuous for all $A\mathrm{\in}\mathrm{S}$.
Then $f$ is continuous.

###### Theorem 2.

Let $X$ and $Y$ be topological spaces, let $\mathrm{S}$ be a collection of open subsets (http://planetmath.org/OpenSet) of $X$, and let $f\mathrm{:}\mathrm{\cup}\mathrm{S}\mathrm{\to}Y$ be a function such that the restriction ${f\mathrm{|}}_{A}$ is continuous for all $A\mathrm{\in}\mathrm{S}$. Then $f$ is continuous.

## Notes

Note that the theorem for closed subsets requires the collection to be locally finite. To see that this is condition cannot be omitted, notice that any function $f:\mathbb{R}\to \mathbb{R}$ restricts to a continuous function on each singleton, yet need not be continuous itself.

## Proofs

The two theorems are proved in essentially in the same way, but for the first theorem we need to make use of the fact that the union of a locally finite collection of closed sets is closed.

Proof of Theorem 1.
Let $C$ be a closed subset of $Y$.
Then ${f}^{-1}(C)={\bigcup}_{A\in \mathcal{S}}(A\cap {f}^{-1}(C))={\bigcup}_{A\in \mathcal{S}}{({f|}_{A})}^{-1}(C)$.
By continuity, each ${({f|}_{A})}^{-1}(C)$ is closed in $A$.
But by assumption^{} each $A$ is closed in $X$,
so it follows that each ${({f|}_{A})}^{-1}(C)$ is closed in $X$.
Thus ${f}^{-1}(C)$ is the union of a locally finite collection of closed sets,
and is therefore closed in $X$, and so closed in $\cup \mathcal{S}$.
So $f$ is continuous.
∎

Proof of Theorem 2. Let $U$ be an open subset of $Y$. Then ${f}^{-1}(U)={\bigcup}_{A\in \mathcal{S}}(A\cap {f}^{-1}(U))={\bigcup}_{A\in \mathcal{S}}{({f|}_{A})}^{-1}(U)$. By continuity, each ${({f|}_{A})}^{-1}(U)$ is open in $A$. But by assumption each $A$ is open in $X$, so it follows that each ${({f|}_{A})}^{-1}(U)$ is open in $X$. Thus ${f}^{-1}(U)$ is the union of a collection of open sets, and is therefore open in $X$, and so open in $\cup \mathcal{S}$. So $f$ is continuous. ∎

Title | gluing together continuous functions |
---|---|

Canonical name | GluingTogetherContinuousFunctions |

Date of creation | 2013-03-22 15:17:20 |

Last modified on | 2013-03-22 15:17:20 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 16 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 54C05 |