Contents

Idea

On this page, we work work through several of the key examples of limits in the category Set. This is part of a bigger project: Understanding Constructions in Categories.

Recall that a limit, or universal cone, over a diagram $F:J\to Set$ is a cone $T$ over $J$ such that, given any cone $T'$, there is a unique cone function? from $T'$ to $T$.

Terminal Object

A terminal object is a universal cone over the empty diagram. In this section, we demonstrate how this leads us to the statement:

To demonstrate, first note that a cone over an empty diagram is just a set and a corresponding cone function is just a function. Therefore, we are looking for a “universal set” $\bullet$ such that for an other set $C$, there is a unique function

Singletons fill the bill because for any element $c\in C$ we have the unique function defined by

Therefore any singleton is a terminal object in $Set$.

Product

A product is a universal cone over a discrete diagram. In this section, we demonstrate how this leads us to the statement:

To demonstrate, first note that a discrete diagram $F:J\to Set$ produces a family of sets $(A_i)$ with no functions between them. A cone over the discrete diagram consists of a set $T$ and a single component

for each set $A_i$. Therefore, we are looking for a universal cone $\prod_i A_i$ such that for any other cone $T$, there is a unique cone function

Cartesian product fills the bill because for any $t\in T$, we have the unique function defined by

This function is unique because with any other function

with $\pi_i\circ g = f_i$, then for any element $t\in T$

Therefore $f = g$.

Equalizer

An equalizer is the universal cone over a parallel diagram

In this section, we demonstrate how this leads us to the statement:

To demonstrate, first note that a parallel diagram $F:J\to Set$ produces two sets $X,Y$ with two parallel functions $f,g:X\to Y$. A cone over the parallel diagram consists of a set $T$ and two components

This is the first example we encounter where the diagram contains morphisms so recall that with a cone we also want the the component diagrams to commute, i.e. we want

Therefore, we are looking for a universal cone $Eq$ such that for any other cone $T$, there is a unique function

Let’s first define the set

with two functions $Eq_X:Eq\to X$ and $Eq_Y:Eq\to Y$ defined by

and verify that it is indeed a cone. The only thing we need to check is that

which amounts to showing that $f(x) = g(x)$ for all $x\in Eq$, but that is the definition of $Eq$, so we do have a cone.

Now let $\phi,\psi: T\to Eq$ be two cone functions. For any element $t\in T$, we have

so that $\phi = \psi$ and the cone function is unique.

Since every cone function $\phi:T\to Eq$ is unique, it follows that $Eq$ together with $Eq_X:Eq\to X$ and $Eq_Y:Eq\to Y$ is a universal cone, i.e. $Eq$ is an equalizer.

Pullbacks

A pullback is a universal cone over a cospan

In this section, we demonstrate how this leads us to the statement:

To demonstrate, first note that a cospan $F:J\to Set$ produces three sets $X,Y,Z$ with two functions $f:X\to Z$ and $g:Y\to Z$. A cone over the cospan consists of a set $T$ and three components

satisfying

which implies, of course,

Therefore, we are looking for a universal cone $Pb$ such that for any other cone $T$, there is a unique function

Let’s first define the set

with three functions $\pi_X:Pb\to X$, $\pi_Y:Pb\to Y$, and $T_X:Pb\to Z$ defined by

and verify that it is indeed a cone. The only thing we need to check is that

which amounts to showing that $f(x) = g(y)$ for all $(x,y)\in Pb$, but that is the definition of $Pb$, so we do have a cone.

Now let $\phi,\psi: T\to Eq$ be two cone functions with

for functions

For any element $t\in T$, we have

Similarly

so that $\phi = \psi$ and the cone function is unique.

Since every cone function $\phi:T\to Pb$ is unique, it follows that $Pb$ together with $\pi_X:Pb\to X$, $\pi_Y:Pb\to Y$, and $T_Z:Pb\to Z$ is a universal cone, i.e. $Pb$ is a pullback.

Fibred Products

Under Construction fibred products