In category theory a concept is called absolute if it is preserved
by all functors. Identity arrows and composition are absolute by the definition of
functor. Less trivially, isomorphisms are absolute. In general, anything that
is described by a diagram commuting is absolute as that diagram will be preserved
by any functor. This is generally the case, but if I tell you something is an
absolute epimorphism, it’s not clear what diagram is causing that; the notion
of epimorphism itself doesn’t reduce to the commutativity of a particular diagram.
Below I’ll be focused primarily on absolute colimits
as those are the most commonly used examples. They are an important part of the theory of
monadicity. The trick to
many questions about absolute colimits and related concepts is to see what it
means for functors to preserve them.
Non-Examples
To start, we can show that certain colimits cannot be absolute, at least for -enriched
category theory. In particular, initial objects and coproducts are never absolute. Using the
trick above, this is easily proven.
Absolute Epimorphisms
What do absolute epimorphisms look like? We know that there are absolute epimorphisms because
a split epimorphism is defined by a certain diagram commuting. Are there other absolute epimorphisms?
To find out, we apply our trick.
Let be an arbitrary absolute epimorphism. Absoluteness implies the surjection
but this means that for every arrow , there’s an arrow such
that . As you can no doubt guess, we want to choose , and we then have
that is a split epimorphism. Therefore split epimorphisms are the only
examples of absolute epimorphisms.
Split Coequalizers
Now let’s consider the coequalizer case. Let and be their coequalizer which
we’ll assume is absolute. Before we pull out our trick, we can immediately use the previous result to
show that has a section, i.e. an arrow such that . Moving on,
we use the trick to get the diagram:
Next, we use the explicit construction of the coequalizer in which
is supposed to be canonically isomorphic to. That is, the
coequalizer of and is quotiented by the
equivalence relation generated by the relation, , which identifies
when . Let
represent the equivalence class of . That is,
corresponds to the function , where
is the reflexive, symmetric, transitive closure of ,
witnessing the set-theoretic coequalizer. The claim that is (with
) a coequalizer of the above arrows implies that we have a
canonical isomorphism such
that and thus . Of
course, our next move is to choose giving .
However, so we get
because is invertible.
Expanding the definition of , we have .
Therefore . Now let’s make a simplifying assumption and
assume further that , i.e. that is directly related
to by . By definition of this means there is a
such that and . A given triple of , ,
and such that and equipped with a and
satisfying the previous two equations along with is called a
split coequalizer. This data is
specified diagrammatically and so is preserved by all functors, thus split
coequalizers are absolute. All that we need to show is that this data is
enough, on its own, to show that is a coequalizer.
Given any such that , we need to show that
there exists a unique arrow which factors through via , i.e.
for some (unique) . The obvious candidate is leading to us needing to verify
that . We calculate as follows:
Uniqueness then quickly follows since if then
.
There’s actually an even simpler example where which
corresponds to the trivial case where .
Absolute Coequalizers
Split coequalizers show that (non-trivial) absolute coequalizers can exist, but
they don’t exhaust all the possibilities. The obvious cause of this is the
simplifying assumption we made where we said rather than
. In the general case, the equivalence will be
witnessed by a sequence of arrows such that we have either
or , then
or respectively, and so on until we
get to or . As a diagram, this is a
fan of diamonds of the form or
with a diamond with side on one end of the fan and
a triangle with on the other. All this data is diagrammatic so it is
preserved by all functors making the coequalizer absolute. That it is a
coequalizer uses the same proof as for split coequalizers except that we have a
series of equalities to show that , namely all those
pasted together diamonds. There is no conceptual difficulty here; it’s just
awkward to notate.
Absolute Colimits
The absolute coequalizer case captures the spirit of the general case, but you
can see an explicit description of the general case
here.
I’m not going to work through it, but you could as an exercise. Less tediously,
you could work through absolute pushouts. If is the pushout of
, then the functors to consider are and
. For each, the pushout in can be turned into a
coequalizer of a coproduct. For the first functor, as before, this gives us an
inverse image of which will either be an arrow or an arrow
which will play the role of . The other functor produces a
coequalizer of . The generating
relation of the equivalence relation states that there exists either an arrow
or an arrow such that the appropriate equation holds. The
story plays out much as before except that we have a sequence of arrows from
two different hom-sets.