# 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 $B\otimes_A C$ 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 $A\downarrow \mathcal C$ whose objects are morphisms $A\to B$ (as B runs through objects of $\mathcal C$), and morphisms are just morphisms $B\to B'$ in $\mathcal C$ making the diagram commute. If we unwind the definition, coproduct in the coslice category means the following.

Definition.

Let $\beta: A\to B$, $\beta':A \to B'$ be morphisms in the category $\mathcal C$. The pushout of $\beta$ and $\beta'$ is a triplet:

$(C, \epsilon : B \to C, \epsilon' : B'\to C)$,

where $C := B \coprod_A B'\in \mathcal C$ is an object, $\epsilon$, $\epsilon'$ are morphisms in $\mathcal C$ satisfying $\epsilon\circ \beta = \epsilon' \circ \beta'$ such that for any triplet:

$(D, \alpha : B \to D, \alpha' : B'\to D)$

of object $D\in \mathcal C$ and morphisms $\alpha$, $\alpha'$ satisfying $\alpha \circ \beta = \alpha' \circ \beta'$, there is a unique morphism $f : C\to D$ such that $f\circ \epsilon = \alpha$ and $f\circ \epsilon' = \alpha'$.

Pictorially, the pushout gives a correspondence as follows.

The idea is that $B\coprod_A B'$ 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 $A\to B$ and $A\to B'$ is $B\otimes_A B'$.

2. Take $\beta : S\to T$ and $\beta' : S\to T$ in the category $\mathbf{Set}$. Then:

$T \coprod_S T' = \{T \coprod T' \} / (\beta(s) \sim \beta'(s), s\in S)$

where $T\coprod T'$ is the disjoint union of T and T’.

3. Let $\beta : H\to G$ and $\beta' : H\to G'$ be injective group homomorphisms in $\mathbf{Gp}$. 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:

• $\alpha : X\to Y$ and $\alpha' :X \to Y'$ in the category $\mathbf{Top}$ of topological spaces;
• $\alpha : M\to N$ and $\alpha' : M\to N'$ in the category $A\text{-}\mathbf{Mod}$.

# Pullbacks (Fibre Products)

Dually, we can define the following.

Definition.

Let $\beta :B\to A$, $\beta' :B' \to A$ be morphisms in $\mathcal C$. The pullback (or fibre product) of $\beta$ and $\beta'$ corresponds to the pushforward of $\beta^{\text{op}}$ and $\beta'^{\text{op}}$ in the opposite category $\mathcal C^{\text{op}}$.

The underlying object of the pullback is denoted by $B\times_A B'$.

Easy exercise

Write out the details of the definition (see below diagram).

## Examples

1. Let $\gamma : T \to S$ and $\gamma' : T' \to S$ be functions in $\mathbf{Set}$. The pullback is given by

$T\times_S T' = \{(t, t') \in T\times T' : \gamma(t) = \gamma'(t') \in S\}$

with projection maps $T\times_S T' \to T$, $T\times_S T' \to T'$ taking $(t,t')$ to $t, t'$ respectively.

2. As a special case suppose $T\subseteq S$ and $\gamma : T\to S$ is the inclusion map. Then $T\times_S T' = \gamma'^{-1}(T)$, i.e. the fibre space of $\gamma' : T' \to S$ over the subset T.

3. As a special case of special case, suppose $T, T'\subseteq S$ and $\gamma, \gamma'$ are both inclusions. Then $T\times_S T' = T\cap T'$.

Exercise B

Do the same constructions work for $\mathbf{Gp}$, $\mathbf{Ring}$, $\mathbf{Top}$, $A\text{-}\mathbf{Mod}$, $A\text{-}\mathbf{Alg}$?

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 $i\in I$ 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 $J$ be an index category; a diagram of type $J$ in the category $\mathcal C$ is a covariant functor $D: J \to \mathcal C$.

Thus to each vertex $i \in J$ we assign an object $A_i \in \mathcal C$ and to each arrow $e : i \to j$ in J, we assign a morphism $\beta_{e} : A_i \to A_j$.

[ Note: when representing a diagram in pictorial form, we often remove the arrows which are implied, e.g. the arrow corresponding to $\beta_{35}\circ \beta_{13}$ 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 $(A_i)$ indexed by $i\in I$ (with the identity arrows mapped to the identity morphisms).

Definition.

Let $D : J\to \mathcal C$ be a diagram, written as

$\left((A_i)_{i\in J}, (\beta_{e}: A_i \to A_j)_{(e : i\to j)}\right)$.

The colimit of this diagram comprises of the tuple

$(A, (\epsilon_i : A_i\to A)_{i\in J})$

where $A = \mathrm{colim }_{i \in J} A_i \in \mathcal C$ is an object, $\epsilon_i$ is a morphism for each $i\in J$, such that for any arrow $e: i\to j$, we have $\epsilon_j \circ \beta_e = \epsilon_i$ as morphisms $A \to A_j$.

We require the following universal property to hold: for any tuple

$(B, (\alpha_i : A_i \to B)_{i\in I})$

where $B\in \mathcal C$ is an object, $\alpha_i$ is a morphism for each $i\in J$, such that for any arrow $e:i\to j$ in $J$ we have $\alpha_j \circ\beta_e = \alpha_i$, there is a unique morphism $f:A\to B$ in $\mathcal C$ such that

$f\circ \epsilon_i = \alpha_i \text{ for each } i\in J$.

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 $\coprod_{i\in I} A_i$.

### Example 2: Pushout

If J is the following index category, the resulting colimit is the pushout.

### Example 3: Coequalizers

Definition.

The coequalizer of $\beta_1, \beta_2 :A\to B$ in a category $\mathcal C$ is the colimit of the following diagram.

This is a pair $(C, \epsilon : B\to C)$ where $C\in \mathcal C$ is an object such that $\epsilon \circ \beta_1 = \epsilon\circ \beta_2$ and, for any pair $(D, \alpha : B\to D)$ satisfying $\alpha\circ \beta_1 = \alpha\circ \beta_2$, there exists a unique $f : C \to D$ such that $f\circ \epsilon = \alpha$.

Exercise C

1. Describe the colimits for each of the following two diagrams.

2. Compute the coequalizer in the categories $\mathbf{Set}$, $\mathbf{Top}$ and $\mathbf{Gp}$.

3. Let $S\subseteq A$ be a multiplicative subset and M an A-module. Prove that we have an isomorphism of A-modules:

$\mathrm{colim}_{f\in S} M_f \cong M_S$,

with morphisms $M_f \to M_g$ defined as follows. If g = af we take the canonical map $M_f \to M_g$ via $\frac m f \mapsto \frac{am}g$. Otherwisee there is no map between $M_f$ and $M_g$. Prove also that we get a ring isomorphism $\mathrm{colim}_{f\in S} A_f \cong A_S$.

In particular for any prime $\mathfrak p \subset A$ we have

$\mathrm{colim}_{f\in A-\mathfrak p} M_f \cong M_{\mathfrak p}, \quad \mathrm{colim}_{f\in A-\mathfrak p} A_f \stackrel{\text{rings}}\cong A_S$.

## 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.

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### 4 Responses to Commutative Algebra 48

1. Vanya says:

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 $A \to B \coprod B'$ instead of just “latex B \coprod B’\$?

• limsup says:

The general coproduct of two objects $A, B$ comprises of an object $A\coprod B$ and morphisms $A \to A\coprod B$ and $B \to A\coprod B$.

But when you take specifically the coslice cateogry $A\downarrow \mathcal C$, the coproduct between $A\to B$ and $A\to B'$ has three things:
(i) an object in the coslice category, i.e. $A \to B\coprod_A B'$,
(ii) morphisms in the coslice cateogry $B\to B\coprod_A B'$ and $B' \to B\coprod_A B'$ which “make the diagram commute”.

However, if the commutativity holds, you can recover $A\to B\coprod_A B'$ in (i) from the two morphisms in (ii).

2. Vanya says:

Did you mean $T \times \gamma'^{-1}(T)$ as the fibre product in Example 2 of fibre products?

• limsup says:

No it’s not an exact product like that.