A lot of results concerning the direct product 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 sums.

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

## No duplicates

In the construction, this is just the trivial fact that if you know both and , then you know .

Theorem:

For every , if and , then .

Proof:

Let us suppose that we have such that and .

Let us take a singleton topological space, that is that is a singleton set (singleton sets exist!) and we equip it with the only possible topology, which is both the indiscrete and the discrete one.

We define and as the constant mappings, and respectively, that map the only element of to , which is also , and to , which is also , respectively.

We define and as the constant mappings, both , that map the only element of to and to respectively.

Since these four mappings are constant, or, alternatively, since the topology on is discrete, they are all continuous.

Furthermore, for both and we have:

Since is a direct product of and , there can be only one morphism satisfying these two relations. Hence necessarily , which implies .

## Existence of an antecedent

If we have and , does there exist a such that and ? Again, this is trivial in the representation: is the answer. Sticking to the universal definition of the direct product, we have:

For any and , there exists such that and .

Proof:

We take as a singleton topological space as above, and and as the mappings and respectively that map the unique element of to and to respectively. These two mappings again are continuous, since has the discrete topology.

The universal definition of as a direct product implies that there exists such that and . Taking as the image by of the unique element of , we thus have and . .

This result and the preceding one come together as:

For any and , there exists a unique such that and .

## The associated morphisms are surjective

At least, usually.

If or , then is surjective. If or , then is surjective.

This is just a consequence of the general result proved in exercise 7 (corrected): in any category, if is a direct product of and , and if , then is an epimorphism (and similarly, exchanging the roles of and , and and , respectively); plus the fact that in the category of topological spaces, epimorphisms are surjective mappings.

## Characterisation of the topology

That the topology on the the direct product space is the one generated by the collection of all and for open in and open in is a stipulation of the explicit construction of as . It can however be proven without reference to this construction.

The topology of is the topology generated in by the collection of all and for open in and open in . A subset of is open if and only if it is the union of some collection of subsets of the form with open in and open in .

Proof:

Let be the direct product topology on and the topology on generated in by the collection of all and for open in and open in .

Since and are defined as continuous, necessarily contains all the and for open in and open in . Hence is finer than . We wish to show that is also finer than .

The mapping , considered from to , remains continuous, since all the for open in are in . Similarly, is continuous . Hence we have a topological space and two continuous mappings and , from to and respectively. Since is a direct product of and , there exists a unique continuous mapping such that and .

This implies that any element has the same image as by both and . The first theorem above implies that . Thus .

Hence is continous ; which means that is finer than .

Hence , that is the topology of the direct product is the topology of generated by the collection of all and for open in and open in .

The open sets of the topology generated by a collection of sets are the unions of any number of finite intersections of elements of .

In this case, . A finite intersection of elements of will be of the form with the ‘s open subsets of and the ‘s open subsets of . Since intersections are well-behaved respective to inverse images, and since finite intersections of open subsets are open subsets, this can be written with open subset of and open subset of . Hence an open subset of is a union of subsets of of the form with open subset of and open subset of .

## Continuity of a mapping to a product space

If is a topological space and a mapping , then is continuous if and only if both and are continuous.

If is continuous, since both and are continuous, by composition of continuous mappings we conclude that and are continuous.

Conversely, suppose that we know that and are continuous. Let be an eventual open subset of . We wish to show that is an open subset of .

Let be an eventual point of . Then . Since is open in and (see above) “a subset of is openif and only if it is the union of some collection of subsets of the form with open in and open in ”, there exist open in and open in such that . Taking of this expression, we obtain . But . Since we know that both and are continuous, this last expression is an open subset of , that contains and is included in . Thus is a neighbourhood of . This being the case for any , is an open subset of .