In my previous post, I showed that compact objects in the category of topological spaces have to be finite. Today we improve this to a full characterisation.
Lemma. Let be a topological space. Then
is a compact object in
if and only if
is finite discrete.
This result dates back to Gabriel and Ulmer [GU71, 6.4], as was pointed out to me by Jiří Rosický in reply to my MO question and answer of this account (of which this post is essentially a retelling). Our proof is different from the one given in [GU71], instead using a variant of an argument given in the n-Lab.
Before giving the proof, we construct an auxiliary space against which we will be testing compactness. It is essentially the colimit constructed in the n-Lab, except that we swapped the roles of and
(the reason for this will become clear in the proof).
Definition. For all , let
be the topological space
, where the nonempty open sets are given by
for
. They form a topology since
Define the map by
This is continuous since equals
if
and
if
. Let
be the colimit of this diagram.
Since the elements map to the same element in
, we conclude that
is the two-point space
, where the map
is the second coordinate projection. Moreover, the colimit topology on
is the indiscrete topology. Indeed, neither
nor
are open.
Proof of Lemma. If is compact, then my previous post shows that
is finite. Let
be any subset, and let
be the indicator function
. It is continuous because
has the indiscrete topology. Since
is a compact object,
has to factor through some
. Let
be the first coordinate projection, i.e.
Let be a number such that
for all
; this exists because
is finite. Then
, which shows that
is open. Since
was arbitrary, we conclude that
is discrete.
Conversely, every finite discrete space is a compact object. Indeed, any map out of
is continuous, and finite sets are compact in
.
[GU71] Gabriel, Peter and Ulmer, Friedrich, Lokal präsentierbare Kategorien. Lecture Notes in Mathematics 221. Springer-Verlag, Berlin-New York, 1971. DOI: 10.1007/BFb0059396.