Theorems about the direct product of two topological spaces
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
.