A lot of results concerning the direct sum of two topological spaces and are trivial if you consider the usual particular construction of these entities. I find it more satisfying to derive them directly from the universal definitions.

See also: the equivalent page on direct products.

In the following, we will always have and topological spaces and a direct sum thereof.

## Non-overlapping images

and are disjoint.

Let and be eventual elements respectively of and .

Let and be any two distinct objects (such exist), and , equipped with any topology (the discrete one will do). Then the constant mappings and are continuous.

Hence there exists a continuous mapping such that and .

This implies in particular that and , that is, and . Since , it follows that .

Thus and cannot have a common point, which would both be the image by of a point of and by of a point of .

## The images cover the direct sum

Let with the topology induced on by that of .

Let be the mapping and the mapping .

Since an open set on is the intersection of with an open set on and since certainly contains no points of , its preimage by is equal to the preimage of by and hence is open in . Thus is continuous. Similarly, is continuous.

We can thus apply the universal property of the direct sum , which implies that there exists a continuous mapping such that and .

Let be the mapping . The preimage by of an open set of is again the same as its preimage by , hence is an open set of . Hence is continuous.

For any , ; hence . Similarly, .

But we also have and , with continuous . The uniqueness requirement of the universal property of the direct sum thus implies that .

But the image of is that of , and is included in . Thus the image of , which is , is included in . It follows that , that is, that .

## Topology on the direct sum

The topology on the direct sum is the topology generated by the collection of the and for all open in and all open in .

//TODO

## 2 thoughts on “Theorems about the direct sum of two topological spaces”