4.1.4 Pullbacks
Let $A$, $B$, and $C$ be sets and let $f\colon A\to C$ and $g\colon B\to C$ be functions.
The pullback of $A$ and $B$ over $C$ along $f$ and $g$ is the pullback of $A$ and $B$ over $C$ along $f$ and $g$ in $\mathsf{Sets}$ as in
,
.
Concretely, the pullback of $A$ and $B$ over $C$ along $f$ and $g$ is the pair $(A\times _{C}B,\left\{ \operatorname {\mathrm{\mathrm{pr}}}_{1},\operatorname {\mathrm{\mathrm{pr}}}_{2}\right\} )$ consisting of:
-
1.
The Limit. The set $A\times _{C}B$ defined by
\[ A\times _{C}B\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ (a,b)\in A\times B\ \middle |\ f(a)=g(b)\right\} . \]
-
2.
The Cone. The maps
\begin{align*} \operatorname {\mathrm{\mathrm{pr}}}_{1} & \colon A\times _{C}B\to A,\\ \operatorname {\mathrm{\mathrm{pr}}}_{2} & \colon A\times _{C}B\to B \end{align*}
defined by
\begin{align*} \operatorname {\mathrm{\mathrm{pr}}}_{1}(a,b) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}a,\\ \operatorname {\mathrm{\mathrm{pr}}}_{2}(a,b) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}b \end{align*}
for each $(a,b)\in A\times _{C}B$.
We claim that $A\times _{C}B$ is the categorical pullback of $A$ and $B$ over $C$ with respect to $(f,g)$ in $\mathsf{Sets}$. First we need to check that the relevant pullback diagram commutes, i.e. that we have
Indeed, given $(a,b)\in A\times _{C}B$, we have
\begin{align*} [f\circ \operatorname {\mathrm{\mathrm{pr}}}_{1}](a,b) & = f(\operatorname {\mathrm{\mathrm{pr}}}_{1}(a,b))\\ & = f(a)\\ & = g(b)\\ & = g(\operatorname {\mathrm{\mathrm{pr}}}_{2}(a,b))\\ & = [g\circ \operatorname {\mathrm{\mathrm{pr}}}_{2}](a,b),\end{align*}
where $f(a)=g(b)$ since $(a,b)\in A\times _{C}B$. 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*} \operatorname {\mathrm{\mathrm{pr}}}_{1}\circ \phi & = p_{1},\\ \operatorname {\mathrm{\mathrm{pr}}}_{2}\circ \phi & = p_{2}\end{align*}
via
\[ \phi (x)=(p_{1}(x),p_{2}(x)) \]
for each $x\in P$, where we note that $(p_{1}(x),p_{2}(x))\in A\times B$ indeed lies in $A\times _{C}B$ by the condition
\[ f\circ p_{1}=g\circ p_{2}, \]
which gives
\[ f(p_{1}(x))=g(p_{2}(x)) \]
for each $x\in P$, so that $(p_{1}(x),p_{2}(x))\in A\times _{C}B$.
Here are some examples of pullbacks of sets.
-
1.
Unions via Intersections. Let $X$ be a set. We have
for each $A,B\in \mathcal{P}(X)$.
Item 1: Unions via Intersections
Indeed, we have
\begin{align*} A\times _{A\cup B}B & \cong \left\{ (x,y)\in A\times B\ \middle |\ x=y\right\} \\ & \cong A\cap B. \end{align*}
This finishes the proof.
This is a special case of functoriality of co/limits,
,
of
, with the explicit expression for $\xi $ following from the commutativity of the cube pullback diagram.
This is a repetition of
,
of
, and is proved there.
This follows from the universal property of the product (pullbacks are products in $\mathsf{Sets}_{/X}$).
We have
\begin{align*} (A\times _{X}B)\times _{Y}C & \cong \left\{ ((a,b),c)\in (A\times _{X}B)\times C\ \middle |\ h(b)=k(c)\right\} \\ & \cong \left\{ ((a,b),c)\in (A\times B)\times C\ \middle |\ \text{$f(a)=g(b)$ and $h(b)=k(c)$}\right\} \\ & \cong \left\{ (a,(b,c))\in A\times (B\times C)\ \middle |\ \text{$f(a)=g(b)$ and $h(b)=k(c)$}\right\} \\ & \cong \left\{ (a,(b,c))\in A\times (B\times _{Y}C)\ \middle |\ \text{$f(a)=g(b)$}\right\} \\ & \cong A\times _{X}(B\times _{Y}C) \end{align*}
and
\begin{align*} (A\times _{X}B)\times _{B}(B\times _{Y}C) & \cong \left\{ ((a,b),(b',c))\in (A\times _{X}B)\times (B\times _{Y}C)\ \middle |\ b=b'\right\} \\ & \cong \left\{ ((a,b),(b',c))\in (A\times B)\times (B\times C)\ \middle |\ \begin{aligned} & \text{$f(a)=g(b)$, $b=b'$,}\\ & \text{and $h(b')=k(c)$}\end{aligned}\right\} \\ & \cong \left\{ (a,(b,(b',c)))\in A\times (B\times (B\times C))\ \middle |\ \begin{aligned} & \text{$f(a)=g(b)$, $b=b'$,}\\ & \text{and $h(b')=k(c)$}\end{aligned}\right\} \\ & \cong \left\{ (a,((b,b'),c))\in A\times ((B\times B)\times C)\ \middle |\ \begin{aligned} & \text{$f(a)=g(b)$, $b=b'$,}\\ & \text{and $h(b')=k(c)$}\end{aligned}\right\} \\ & \cong \left\{ (a,((b,b'),c))\in A\times ((B\times _{B}B)\times C)\ \middle |\ \begin{aligned} & \text{$f(a)=g(b)$ and}\\ & \text{$h(b')=k(c)$}\end{aligned}\right\} \\ & \cong \left\{ (a,(b,c))\in A\times (B\times C)\ \middle |\ \text{$f(a)=g(b)$ and $h(b)=k(c)$}\right\} \\ & \cong A\times _{X}(B\times _{Y}C), \end{align*}
where we have used Item 6 for the isomorphism $B\times _{B}B\cong B$.
Item 5: Interaction With Composition
By Item 4, it suffices to construct only the isomorphism
\[ X\times ^{f\circ \phi ,g\circ \psi }_{K}Y\cong (X\times ^{\phi ,q_{1}}_{A}(A\times ^{f,g}_{K}B))\times ^{p_{2},p_{1}}_{A\times ^{f,g}_{K}B}((A\times ^{f,g}_{K}B)\times ^{q_{2},\psi }_{B}Y). \]
We have
so writing
\begin{align*} S & = (X\times ^{\phi ,q_{1}}_{A}(A\times ^{f,g}_{K}B))\\ S' & = ((A\times ^{f,g}_{K}B)\times ^{q_{2},\psi }_{B}Y), \end{align*}
we have
This finishes the proof.
We have
\begin{align*} X\times _{X}A & \cong \left\{ (x,a)\in X\times A\ \middle |\ f(a)=x\right\} ,\\ A\times _{X}X & \cong \left\{ (a,x)\in X\times A\ \middle |\ f(a)=x\right\} , \end{align*}
which are isomorphic to $A$ via the maps $(x,a)\mapsto a$ and $(a,x)\mapsto a$. The proof of the naturality of $\lambda ^{\mathsf{Sets}_{/X}}$ and $\rho ^{\mathsf{Sets}_{/X}}$ is omitted.
We have
\begin{align*} A\times _{C}B & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ (a,b)\in A\times B\ \middle |\ f(a)=g(b)\right\} \\ & = \left\{ (a,b)\in A\times B\ \middle |\ g(b)=f(a)\right\} \\ & \cong \left\{ (b,a)\in B\times A\ \middle |\ g(b)=f(a)\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}B\times _{C}A. \end{align*}
The proof of the naturality of $\sigma ^{\mathsf{Sets}_{/X}}$ is omitted.
Item 8: Distributivity Over Coproducts
We have
\begin{align*} A\times _{X}(B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ (a,z)\in A\times (B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C)\ \middle |\ \phi _{A}(a)=\phi _{B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C}(z)\right\} \\ & = \left\{ (a,z)\in A\times (B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C)\ \middle |\ \text{$z=(0,b)$ and $\phi _{A}(a)=\phi _{B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C}(z)$}\right\} \\ & \phantom{={}} \mkern 4mu\cup \left\{ (a,z)\in A\times (B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C)\ \middle |\ \text{$z=(1,c)$ and $\phi _{A}(a)=\phi _{B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C}(z)$}\right\} \\ & = \left\{ (a,z)\in A\times (B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C)\ \middle |\ \text{$z=(0,b)$ and $\phi _{A}(a)=\phi _{B}(b)$}\right\} \\ & \phantom{={}} \mkern 4mu\cup \left\{ (a,z)\in A\times (B\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}C)\ \middle |\ \text{$z=(1,c)$ and $\phi _{A}(a)=\phi _{C}(c)$}\right\} \\ & \cong \left\{ (a,b)\in A\times B\ \middle |\ \phi _{A}(a)=\phi _{B}(b)\right\} \\ & \phantom{={}} \mkern 4mu\cup \left\{ (a,c)\in A\times C\ \middle |\ \phi _{A}(a)=\phi _{C}(c)\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(A\times _{X}B)\cup (A\times _{X}C)\\ & \cong (A\times _{X}B)\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}(A\times _{X}C), \end{align*}
with the construction of the isomorphism
\[ \delta ^{\mathsf{Sets}_{/X}}_{r} \colon (A\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}B)\times _{X}C \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }(A\times _{X}C)\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}(B\times _{X}C) \]
being similar. The proof of the naturality of $\delta ^{\mathsf{Sets}_{/X}}_{\ell }$ and $\delta ^{\mathsf{Sets}_{/X}}_{r}$ is omitted.
Item 9: Annihilation With the Empty Set
We have
\begin{align*} A\times _{X}\text{Ø}& \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ (a,b)\in A\times \text{Ø}\ \middle |\ f(a)=g(b)\right\} \\ & = \left\{ k\in \text{Ø}\ \middle |\ f(a)=g(b)\right\} \\ & = \text{Ø}, \end{align*}
and similarly for $\text{Ø}\times _{X}A$, where we have used Item 8 of Proposition 4.1.3.1.3. The proof of the naturality of $\zeta ^{\mathsf{Sets}_{/X}}_{\ell }$ and $\zeta ^{\mathsf{Sets}_{/X}}_{r}$ is omitted.
We have
\begin{align*} A\times _{\mathrm{pt}}B & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ (a,b)\in A\times B\ \middle |\ \mathord {!}_{A}(a)\mathbin {=}\mathord {!}_{B}(b)\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ (a,b)\in A\times B\ \middle |\ \star =\star \right\} \\ & = \left\{ (a,b)\in A\times B\right\} \\ & = A\times B. \end{align*}
Omitted.