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\begin{document}

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\section*{Understanding limits in Set}

[[!redirects Understanding Limits in Set]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{terminal_object}{Terminal Object}\dotfill \pageref*{terminal_object} \linebreak
\noindent\hyperlink{product}{Product}\dotfill \pageref*{product} \linebreak
\noindent\hyperlink{equalizer}{Equalizer}\dotfill \pageref*{equalizer} \linebreak
\noindent\hyperlink{pullbacks}{Pullbacks}\dotfill \pageref*{pullbacks} \linebreak
\noindent\hyperlink{fibred_products}{Fibred Products}\dotfill \pageref*{fibred_products} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

On this page, we work work through several of the key examples of \textbf{[[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 $T$.

\hypertarget{terminal_object}{}\subsection*{{Terminal Object}}\label{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 $Set$.

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

\begin{displaymath}
f:C\to\bullet.
\end{displaymath}
Singletons fill the bill because for any element $c\in C$ we have the unique function defined by

\begin{displaymath}
f(c) = *.
\end{displaymath}
Therefore any singleton is a terminal object in $Set$.

\hypertarget{product}{}\subsection*{{Product}}\label{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 $Set$.

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

\begin{displaymath}
\f_i:T\to A_i
\end{displaymath}
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

\begin{displaymath}
f:T\to\prod_i A_i.
\end{displaymath}
Cartesian product fills the bill because for any $t\in T$, we have the unique function defined by

\begin{displaymath}
f(t) = \prod_i f_i(t).
\end{displaymath}
This function is unique because with any other function

\begin{displaymath}
g:T\to\prod_i A_i
\end{displaymath}
with $\pi_i\circ g = f_i$, then for any element $t\in T$

\begin{displaymath}
g(t) = \prod_i f_i(t) = \left(\prod_i f_i\right)(t) = f(t).
\end{displaymath}
Therefore $f = g$.

\hypertarget{equalizer}{}\subsection*{{Equalizer}}\label{equalizer}

An [[equalizer]] is the universal cone over a parallel diagram

\begin{displaymath}
\bullet\rightrightarrows\bullet.
\end{displaymath}
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 morphism|parallel functions]] $f,g:X\to Y$. A cone over the parallel diagram consists of a set $T$ and two components

\begin{displaymath}
T_X:T\to X\quad\text{and}\quad T_Y:T\to Y.
\end{displaymath}
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

\begin{displaymath}
f\circ T_X = T_Y\quad\text{and}\quad g\circ T_X = T_Y.
\end{displaymath}
Therefore, we are looking for a universal cone $Eq$ such that for any other cone $T$, there is a unique function

\begin{displaymath}
f:T\to Eq.
\end{displaymath}
Let's first define the set

\begin{displaymath}
Eq = \left\{x\in X|f(x)=g(x)\right\}
\end{displaymath}
with two functions $Eq_X:Eq\to X$ and $Eq_Y:Eq\to Y$ defined by

\begin{displaymath}
Eq_X(x) = x\quad\text{and}\quad Eq_Y(x) = f(x)
\end{displaymath}
and verify that it is indeed a cone. The only thing we need to check is that

\begin{displaymath}
g\circ Eq_X = Eq_Y
\end{displaymath}
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

\begin{displaymath}
T_X(t) = Eq_X\circ\phi(t) = Eq_X\circ\psi(t)\implies \phi(t)=\psi(t)
\end{displaymath}
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.

\hypertarget{pullbacks}{}\subsection*{{Pullbacks}}\label{pullbacks}

A [[pullback]] is a universal cone over a [[cospan]]

\begin{displaymath}
\itexarray{
     && \bullet &&&& \bullet
      \\
      & 
      && \searrow
       &
       & \swarrow
      && 
     \\
     
     &&&&
     \bullet
     &&&&
  }.
\end{displaymath}
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

\begin{displaymath}
T_X:T\to X,\quad T_Y:T\to Y,\quad\text{and}\quad T_Z:T\to Z
\end{displaymath}
satisfying

\begin{displaymath}
f\circ T_X = T_Z\quad\text{and}\quad g\circ T_Y = T_Z,
\end{displaymath}
which implies, of course,

\begin{displaymath}
f\circ T_X = g\circ T_Y.
\end{displaymath}
Therefore, we are looking for a universal cone $Pb$ such that for any other cone $T$, there is a unique function

\begin{displaymath}
f:T\to Pb.
\end{displaymath}
Let's first define the set

\begin{displaymath}
Pb = \left\{(x,y)|f(x)=g(y)\right\}
\end{displaymath}
with three functions $\pi_X:Pb\to X$, $\pi_Y:Pb\to Y$, and $T_X:Pb\to Z$ defined by

\begin{displaymath}
\pi_X(x,y) = x,\quad\text{and}\quad \pi_Y(x,y) = y,\quad\text{and}\quad T_Z(x,y) = f(x)
\end{displaymath}
and verify that it is indeed a cone. The only thing we need to check is that

\begin{displaymath}
f\circ\pi_X = g\circ \pi_Y = T_Z
\end{displaymath}
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

\begin{displaymath}
\phi(t) = (\phi_X(t),\phi_Y(t)\quad\text{and}\quad\psi(t) = (\psi_X(t),\psi_Y(t)
\end{displaymath}
for functions

\begin{displaymath}
\phi_X,\psi_X:T\to X\quad\text{and}\quad\phi_Y,\psi_Y:T\to Y.
\end{displaymath}
For any element $t\in T$, we have

\begin{displaymath}
T_X(t) = \pi_X\circ\phi(t) = \pi_X\circ\psi(t)\implies \phi_X(t)=\psi_X(t).
\end{displaymath}
Similarly

\begin{displaymath}
T_Y(t) = \pi_Y\circ\phi(t) = \pi_Y\circ\psi(t)\implies \phi_Y(t)=\psi_Y(t)
\end{displaymath}
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.

\hypertarget{fibred_products}{}\subsection*{{Fibred Products}}\label{fibred_products}

\textbf{Under Construction} [[fibred products]]



\end{document}