# 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 $F:J\to \mathrm{Set}$ is a cone $T$ over $J$ such that, given any cone $T\prime$, there is a unique cone function from $T\prime$ 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:

Any singleton (a one-element set) is a terminal object in $\mathrm{Set}$.

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” $•$ such that for an other set $C$, there is a unique function

$f:C\to •.$f:C\to\bullet.

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

$f\left(c\right)=*.$f(c) = *.

Therefore any singleton is a terminal object in $\mathrm{Set}$.

## Product

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

Any cartesian product is a product in $\mathrm{Set}$.

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

${f}_{i}:T\to {A}_{i}$\f_i:T\to A_i

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

$f:T\to \prod _{i}{A}_{i}.$f:T\to\prod_i A_i.

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

$f\left(t\right)=\prod _{i}{f}_{i}\left(t\right).$f(t) = \prod_i f_i(t).

This function is unique because with any other function

$g:T\to \prod _{i}{A}_{i}$g:T\to\prod_i A_i

with ${\pi }_{i}\circ g={f}_{i}$, then for any element $t\in T$

$g\left(t\right)=\prod _{i}{f}_{i}\left(t\right)=\left(\prod _{i}{f}_{i}\right)\left(t\right)=f\left(t\right).$g(t) = \prod_i f_i(t) = \left(\prod_i f_i\right)(t) = f(t).

Therefore $f=g$.

## Equalizer

An equalizer is the universal cone over a parallel diagram

$•⇉•.$\bullet\rightrightarrows\bullet.

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

The equalizer of two functions is the subset on which both functions coincide.

To demonstrate, first note that a parallel diagram $F:J\to \mathrm{Set}$ 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

${T}_{X}:T\to X\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{T}_{Y}:T\to Y.$T_X:T\to X\quad\text{and}\quad T_Y:T\to Y.

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

$f\circ {T}_{X}={T}_{Y}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\circ {T}_{X}={T}_{Y}.$f\circ T_X = T_Y\quad\text{and}\quad g\circ T_X = T_Y.

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

$f:T\to \mathrm{Eq}.$f:T\to Eq.

Let’s first define the set

$\mathrm{Eq}=\left\{x\in X\mid f\left(x\right)=g\left(x\right)\right\}$Eq = \left\{x\in X|f(x)=g(x)\right\}

with two functions ${\mathrm{Eq}}_{X}:\mathrm{Eq}\to X$ and ${\mathrm{Eq}}_{Y}:\mathrm{Eq}\to Y$ defined by

${\mathrm{Eq}}_{X}\left(x\right)=x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{Eq}}_{Y}\left(x\right)=f\left(x\right)$Eq_X(x) = x\quad\text{and}\quad Eq_Y(x) = f(x)

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

$g\circ {\mathrm{Eq}}_{X}={\mathrm{Eq}}_{Y}$g\circ Eq_X = Eq_Y

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

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

${T}_{X}\left(t\right)={\mathrm{Eq}}_{X}\circ \varphi \left(t\right)={\mathrm{Eq}}_{X}\circ \psi \left(t\right)⇒\varphi \left(t\right)=\psi \left(t\right)$T_X(t) = Eq_X\circ\phi(t) = Eq_X\circ\psi(t)\implies \phi(t)=\psi(t)

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

Since every cone function $\varphi :T\to \mathrm{Eq}$ is unique, it follows that $\mathrm{Eq}$ together with ${\mathrm{Eq}}_{X}:\mathrm{Eq}\to X$ and ${\mathrm{Eq}}_{Y}:\mathrm{Eq}\to Y$ is a universal cone, i.e. $\mathrm{Eq}$ is an equalizer.

## Pullbacks

A pullback is a universal cone over a cospan

$\begin{array}{ccccccc}& & •& & & & •\\ & & & ↘& & ↙& & \\ & & & & •& & & & \end{array}.$\array{ && \bullet &&&& \bullet \\ & && \searrow & & \swarrow && \\ &&&& \bullet &&&& }.

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

The pullback of two functions of sets is the set of pairs $\left(x,y\right)$ such that $f\left(x\right)=g\left(y\right)$.

To demonstrate, first note that a cospan $F:J\to \mathrm{Set}$ 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

${T}_{X}:T\to X,\phantom{\rule{1em}{0ex}}{T}_{Y}:T\to Y,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{T}_{Z}:T\to Z$T_X:T\to X,\quad T_Y:T\to Y,\quad\text{and}\quad T_Z:T\to Z

satisfying

$f\circ {T}_{X}={T}_{Z}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\circ {T}_{Y}={T}_{Z},$f\circ T_X = T_Z\quad\text{and}\quad g\circ T_Y = T_Z,

which implies, of course,

$f\circ {T}_{X}=g\circ {T}_{Y}.$f\circ T_X = g\circ T_Y.

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

$f:T\to \mathrm{Pb}.$f:T\to Pb.

Let’s first define the set

$\mathrm{Pb}=\left\{\left(x,y\right)\mid f\left(x\right)=g\left(y\right)\right\}$Pb = \left\{(x,y)|f(x)=g(y)\right\}

with three functions ${\pi }_{X}:\mathrm{Pb}\to X$, ${\pi }_{Y}:\mathrm{Pb}\to Y$, and ${T}_{X}:\mathrm{Pb}\to Z$ defined by

${\pi }_{X}\left(x,y\right)=x,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\pi }_{Y}\left(x,y\right)=y,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{T}_{Z}\left(x,y\right)=f\left(x\right)$\pi_X(x,y) = x,\quad\text{and}\quad \pi_Y(x,y) = y,\quad\text{and}\quad T_Z(x,y) = f(x)

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

$f\circ {\pi }_{X}=g\circ {\pi }_{Y}={T}_{Z}$f\circ\pi_X = g\circ \pi_Y = T_Z

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

Now let $\varphi ,\psi :T\to \mathrm{Eq}$ be two cone functions with

$\varphi \left(t\right)=\left({\varphi }_{X}\left(t\right),{\varphi }_{Y}\left(t\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\psi \left(t\right)=\left({\psi }_{X}\left(t\right),{\psi }_{Y}\left(t\right)$\phi(t) = (\phi_X(t),\phi_Y(t)\quad\text{and}\quad\psi(t) = (\psi_X(t),\psi_Y(t)

for functions

${\varphi }_{X},{\psi }_{X}:T\to X\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\varphi }_{Y},{\psi }_{Y}:T\to Y.$\phi_X,\psi_X:T\to X\quad\text{and}\quad\phi_Y,\psi_Y:T\to Y.

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

${T}_{X}\left(t\right)={\pi }_{X}\circ \varphi \left(t\right)={\pi }_{X}\circ \psi \left(t\right)⇒{\varphi }_{X}\left(t\right)={\psi }_{X}\left(t\right).$T_X(t) = \pi_X\circ\phi(t) = \pi_X\circ\psi(t)\implies \phi_X(t)=\psi_X(t).

Similarly

${T}_{Y}\left(t\right)={\pi }_{Y}\circ \varphi \left(t\right)={\pi }_{Y}\circ \psi \left(t\right)⇒{\varphi }_{Y}\left(t\right)={\psi }_{Y}\left(t\right)$T_Y(t) = \pi_Y\circ\phi(t) = \pi_Y\circ\psi(t)\implies \phi_Y(t)=\psi_Y(t)

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

Since every cone function $\varphi :T\to \mathrm{Pb}$ is unique, it follows that $\mathrm{Pb}$ together with ${\pi }_{X}:\mathrm{Pb}\to X$, ${\pi }_{Y}:\mathrm{Pb}\to Y$, and ${T}_{Z}:\mathrm{Pb}\to Z$ is a universal cone, i.e. $\mathrm{Pb}$ is a pullback.

## Fibred Products

Under Construction fibred products

Revised on October 9, 2010 14:55:06 by Eric Forgy (219.78.39.176)