# gluing together continuous functions

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## Introduction

Suppose we have a collection $\cal S$ of subsets of a topological space $X$, and for each $A\in\cal S$ we have a continuous function $f_{A}\colon A\to Y$, 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 $f\colon\cup{\cal S}\to Y$. 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 $\cal S$ be a locally finite collection of closed subsets (http://planetmath.org/ClosedSet) of $X$, and let $f\colon\cup{\cal S}\to Y$ be a function such that the restriction (http://planetmath.org/RestrictionOfAFunction) $f|_{A}$ is continuous for all $A\in\cal S$. Then $f$ is continuous.

###### Theorem 2.

Let $X$ and $Y$ be topological spaces, let $\cal S$ be a collection of open subsets (http://planetmath.org/OpenSet) of $X$, and let $f\colon\cup{\cal S}\to Y$ be a function such that the restriction $f|_{A}$ is continuous for all $A\in\cal S$. 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 $f\colon\mathbb{R}\to\mathbb{R}$ 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 $f^{-1}(C)=\bigcup_{A\in\cal S}(A\cap f^{-1}(C))=\bigcup_{A\in\cal S}(f|_{A})^{% -1}(C)$. 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 $\cup\cal S$. So $f$ is continuous. ∎

Proof of Theorem 2. Let $U$ be an open subset of $Y$. Then $f^{-1}(U)=\bigcup_{A\in\cal S}(A\cap f^{-1}(U))=\bigcup_{A\in\cal S}(f|_{A})^{% -1}(U)$. 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 $\cup\cal S$. So $f$ is continuous. ∎

Title gluing together continuous functions GluingTogetherContinuousFunctions 2013-03-22 15:17:20 2013-03-22 15:17:20 yark (2760) yark (2760) 16 yark (2760) Theorem msc 54C05