Introduction
For the next few articles we are back to discussing category theory to develop even more concepts. First we will look at limits and colimits, which greatly generalize the concept of products and coproducts and cover loads of interesting cases.
As a starting example, recall that for A-algebras B and C, we have which is the coproduct of B and C in the category of A-algebras. But the category of A-algebras corresponds to the coslice category
whose objects are morphisms
(as B runs through objects of
), and morphisms are just morphisms
in
making the diagram commute. If we unwind the definition, coproduct in the coslice category means the following.
Definition.
Let
,
be morphisms in the category
. The pushout of
and
is a triplet:
,
where
is an object,
,
are morphisms in
satisfying
such that for any triplet:
of object
and morphisms
,
satisfying
, there is a unique morphism
such that
and
.
Pictorially, the pushout gives a correspondence as follows.
The idea is that classifies all morphisms “from” the diagram in red.
The pushout may not exist; if it does, it is unique up to unique isomorphism.
Examples
1. We already saw that in the category of rings, the pushout of and
is
.
2. Take and
in the category
. Then:
where is the disjoint union of T and T’.
3. Let and
be injective group homomorphisms in
. The pushout is called the amalgamation of G and G’ over H and can be concretely described in terms of words in H, coset representatives of G/H and those of G’/H. A description is given in Trees by Jean-Pierre Serre.
Exercise A
1. Prove example 2.
2. Find the pushforward for:
and
in the category
of topological spaces;
and
in the category
.
Pullbacks (Fibre Products)
Dually, we can define the following.
Definition.
Let
,
be morphisms in
. The pullback (or fibre product) of
and
corresponds to the pushforward of
and
in the opposite category
.
The underlying object of the pullback is denoted by
.
Easy exercise
Write out the details of the definition (see below diagram).
Examples
1. Let and
be functions in
. The pullback is given by
with projection maps ,
taking
to
respectively.
2. As a special case suppose and
is the inclusion map. Then
, i.e. the fibre space of
over the subset T.
3. As a special case of special case, suppose and
are both inclusions. Then
.
Exercise B
Do the same constructions work for ,
,
,
,
?
Explain how the three constructions at the end of Chapter 30 are all fibre products in the category of k-schemes (or pushouts, if we take the dual category of finitely generated k-algebras).
Colimits
Now we will generalize the above constructions, by taking limits and colimits over a diagram in a category.
Definition.
An index category J is a category such that its class of objects is a set.
If you are concerned about the difference between classes and sets, please read the note at the end.
In summary, an index category consists of a set of vertices and arrows between them such that we can compose arrows. For convenience, we will employ this terminology for an index category: objects are called vertices and edges are called arrows.
Now an index set I may be regarded as an index category J, where we pick a vertex for each and the only morphisms are the identities. E.g. the index set {1,2,3} has 3 vertices and one identity arrow for each object.
Definition.
Let
be an index category; a diagram of type
in the category
is a covariant functor
.
Thus to each vertex we assign an object
and to each arrow
in J, we assign a morphism
.
[ Note: when representing a diagram in pictorial form, we often remove the arrows which are implied, e.g. the arrow corresponding to is not drawn. In particular, identity maps are not drawn. ]
In particular, if J is obtained from an index set I, a diagram of type J is just a collection of objects indexed by
(with the identity arrows mapped to the identity morphisms).
Definition.
Let
be a diagram, written as
.
The colimit of this diagram comprises of the tuple
where
is an object,
is a morphism for each
, such that for any arrow
, we have
as morphisms
.
We require the following universal property to hold: for any tuple
where
is an object,
is a morphism for each
, such that for any arrow
in
we have
, there is a unique morphism
in
such that
.
Clearly, the colimit is unique up to unique isomorphism if it exists.
Example 1: Coproducts
If J is obtained from an index set I, the resulting colimit is the coproduct .
Example 2: Pushout
If J is the following index category, the resulting colimit is the pushout.
Example 3: Coequalizers
Definition.
The coequalizer of
in a category
is the colimit of the following diagram.
This is a pair where
is an object such that
and, for any pair
satisfying
, there exists a unique
such that
.
Exercise C
1. Describe the colimits for each of the following two diagrams.
2. Compute the coequalizer in the categories ,
and
.
3. Let be a multiplicative subset and M an A-module. Prove that we have an isomorphism of A-modules:
,
with morphisms defined as follows. If g = af we take the canonical map
via
. Otherwisee there is no map between
and
. Prove also that we get a ring isomorphism
.
In particular for any prime we have
.
Note
For readers who are unfamiliar with the set-theoretic notion of class and set, try not to fret too much over this. Roughly, a set is a class which is “not too huge”. If we allow arbitrary collections to be sets, then Russell’s paradox occurs (where one can take the set of all sets not containing themselves). A common way out of this paradox is to restrict the type of collections which can be considered as sets.
As a very rough guide, in Zermelo-Fraenkel set theory, a collection is a set if it can be “built from the set of natural numbers ℕ”. Thus the collection of real numbers ℝ is a set because its cardinality is the same as the power set of ℕ. Similarly, the set of open subsets of ℝ forms a set because it is a subset of the power set of ℝ, so we can do topology on ℝ. On the other hand, the collection of all groups does not form a set.
Can you explain how the definition of pushout is ” If we unwind the definition, coproduct in the coslice category means the following.” ? Shouldn’t the coproduct be an object
instead of just “latex B \coprod B’$?
The general coproduct of two objects
comprises of an object
and morphisms
and
.
But when you take specifically the coslice cateogry
, the coproduct between
and
has three things:
,
and
which “make the diagram commute”.
(i) an object in the coslice category, i.e.
(ii) morphisms in the coslice cateogry
However, if the commutativity holds, you can recover
in (i) from the two morphisms in (ii).
Did you mean
as the fibre product in Example 2 of fibre products?
No it’s not an exact product like that.