Construction 2.1.4.1.2. [Construction of Pullbacks of Sets]
The pullback of $A$ and $B$ over $C$ along $f$ and $g$ is the pair $\webleft (A\times _{C}B,\webleft\{ \text{pr}_{1},\text{pr}_{2}\webright\} \webright )$ consisting of:

  1. The Limit. The set $A\times _{C}B$ defined by
    \[ A\times _{C}B\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft\{ \webleft (a,b\webright )\in A\times B\ \middle |\ f\webleft (a\webright )=g\webleft (b\webright )\webright\} . \]
  2. The Cone. The maps1
    \begin{align*} \text{pr}_{1} & \colon A\times _{C}B\to A,\\ \text{pr}_{2} & \colon A\times _{C}B\to B \end{align*}

    defined by

    \begin{align*} \text{pr}_{1}\webleft (a,b\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}a,\\ \text{pr}_{2}\webleft (a,b\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}b \end{align*}

    for each $\webleft (a,b\webright )\in A\times _{C}B$.


1Further Notation: Also written $\text{pr}^{A\times _{C}B}_{1}$ and $\text{pr}^{A\times _{C}B}_{2}$.

We claim that $A\times _{C}B$ is the categorical pullback of $A$ and $B$ over $C$ with respect to $\webleft (f,g\webright )$ in $\mathsf{Sets}$. First we need to check that the relevant pullback diagram commutes, i.e. that we have

Indeed, given $\webleft (a,b\webright )\in A\times _{C}B$, we have

\begin{align*} \webleft [f\circ \text{pr}_{1}\webright ]\webleft (a,b\webright ) & = f\webleft (\text{pr}_{1}\webleft (a,b\webright )\webright )\\ & = f\webleft (a\webright )\\ & = g\webleft (b\webright )\\ & = g\webleft (\text{pr}_{2}\webleft (a,b\webright )\webright )\\ & = \webleft [g\circ \text{pr}_{2}\webright ]\webleft (a,b\webright ),\end{align*}

where $f\webleft (a\webright )=g\webleft (b\webright )$ since $\webleft (a,b\webright )\in A\times _{C}B$. Next, we prove that $A\times _{C}B$ satisfies the universal property of the pullback. Suppose we have a diagram of the form

in $\mathsf{Sets}$. Then there exists a unique map $\phi \colon P\to A\times _{C}B$ making the diagram

commute, being uniquely determined by the conditions

\begin{align*} \text{pr}_{1}\circ \phi & = p_{1},\\ \text{pr}_{2}\circ \phi & = p_{2}\end{align*}

via

\[ \phi \webleft (x\webright )=\webleft (p_{1}\webleft (x\webright ),p_{2}\webleft (x\webright )\webright ) \]

for each $x\in P$, where we note that $\webleft (p_{1}\webleft (x\webright ),p_{2}\webleft (x\webright )\webright )\in A\times B$ indeed lies in $A\times _{C}B$ by the condition

\[ f\circ p_{1}=g\circ p_{2}, \]

which gives

\[ f\webleft (p_{1}\webleft (x\webright )\webright )=g\webleft (p_{2}\webleft (x\webright )\webright ) \]

for each $x\in P$, so that $\webleft (p_{1}\webleft (x\webright ),p_{2}\webleft (x\webright )\webright )\in A\times _{C}B$.


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