Theorems about the direct sum of two topological spaces
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