# Topology/Universal property of topological sum in slightly sharper form

A slightly sharper result than the universal property of topological sum is introduced, which is just an excercise given in ^{[1]}^{:68, Exercise 3.2.5}.

## Introduction

editA **topological sum** of two topological spaces , is a triple of a topological space and two continuous functions , that satisfies the following universal property.

- For every topological space and continuous functions , , there is a unique continuous function such that and , i.e. the following diagram commutes.

A topological sum of two spaces always exists, and is unique up to homeomorphism. So we can denote it by . Since the inclusion functions , are both embeddings we can denote , simply by , . With this notation the topological sum is as a set the disjoint union of and , and a subset is open if and only if is open in and is open in .

The following theorem characterizes when a topological space can be seen as a topological sum of two given subspaces.

Theorem 1. Let be a topological space and two subsets. Let , , , be inclusion functions. Then the following two conditions are equivalent.

- (a) . That is, the unique function defined by , is a homeomorphism.
- (b) , , and both , are open in

Proof. (a) ⇒ (b). is open in , since is open in and is open in . is open for a similar reason. (b) ⇒ (a). Suppose is open in and is open in . Then, because , are both open in , , is open in and so is open in .

## Main result

editAccording to the universal property of topological sum, a function from the topological sum of topological spaces , to another topological space is continuous, if both and are continuous. This is a special case of the following theorem.

Theorem 2. Let be a topological space and two subsets. Suppose that , and that . Let be a topological space. Then a function is continuous if both and are continuous.

## Proof of the main result

editLemma 1. Let , be topological spaces, and . Let be a neighbourhood of in . If is continuous at , then is continuous at .

Proof. Let be any neighbourhood of in . Then is a neighbourhood of in . So there is an such that . This and imply that , and so .

Proof of Theorem 2. We show that is continuous at every point . If or , then by Lemma 1 is continuous at . If and , then it is easily verified that . Suppose otherwise that . Then is open in and so there is a such that . Then and this is impossible since .

Now let be any neighbourhood of . Then there are such that and . Then we have and , from which is continuous at .

## References

edit- ↑ Brown, Ronald (2006).
*Topology and groupoids. A geometric account of general topology, homotopy types and the fundamental groupoid*(in en) (3 ed.). ISBN 1-4196-2722-8.