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 $T$.

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 $Set$.

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

$f:C\to\bullet.$

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

$f(c) = *.$

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

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 $Set$.

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

$\f_i:T\to A_i$

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

$f:T\to\prod_i A_i.$

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

$f(t) = \prod_i f_i(t).$

This function is unique because with any other function

$g:T\to\prod_i A_i$

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

$g(t) = \prod_i f_i(t) = \left(\prod_i f_i\right)(t) = f(t).$

Therefore $f = g$.

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 $F:J\to Set$ 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

$T_X:T\to X\quad\text{and}\quad T_Y:T\to Y.$

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

$f\circ T_X = T_Y\quad\text{and}\quad g\circ T_X = T_Y.$

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

$f:T\to Eq.$

Let’s first define the set

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

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

$Eq_X(x) = x\quad\text{and}\quad Eq_Y(x) = f(x)$

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

$g\circ Eq_X = Eq_Y$

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

$T_X(t) = Eq_X\circ\phi(t) = Eq_X\circ\psi(t)\implies \phi(t)=\psi(t)$

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.

A pullback is a universal cone over a cospan

$\array{
&& \bullet &&&& \bullet
\\
&
&& \searrow
&
& \swarrow
&&
\\
&&&&
\bullet
&&&&
}.$

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

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

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

$T_X:T\to X,\quad T_Y:T\to Y,\quad\text{and}\quad T_Z:T\to Z$

satisfying

$f\circ T_X = T_Z\quad\text{and}\quad g\circ T_Y = T_Z,$

which implies, of course,

$f\circ T_X = g\circ T_Y.$

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

$f:T\to Pb.$

Let’s first define the set

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

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

$\pi_X(x,y) = x,\quad\text{and}\quad \pi_Y(x,y) = y,\quad\text{and}\quad T_Z(x,y) = f(x)$

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

$f\circ\pi_X = g\circ \pi_Y = T_Z$

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

$\phi(t) = (\phi_X(t),\phi_Y(t)\quad\text{and}\quad\psi(t) = (\psi_X(t),\psi_Y(t)$

for functions

$\phi_X,\psi_X:T\to X\quad\text{and}\quad\phi_Y,\psi_Y:T\to Y.$

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

$T_X(t) = \pi_X\circ\phi(t) = \pi_X\circ\psi(t)\implies \phi_X(t)=\psi_X(t).$

Similarly

$T_Y(t) = \pi_Y\circ\phi(t) = \pi_Y\circ\psi(t)\implies \phi_Y(t)=\psi_Y(t)$

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.

**Under Construction** fibred products

Last revised on October 9, 2010 at 14:55:06. See the history of this page for a list of all contributions to it.