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

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A 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

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According 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

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Lemma 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

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  1. 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.