Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 2

Name the upper objects $A,B,C$ and the lower objects $D,E,F$, so the morphisms are $b:A\to B$, $c:B\to C$, $a:A\to D$, $g:B\to E$, $d:C\to F$, $f:D\to E$, and $e:E\to F$. The commutative diagram gives $gb=fa$ and $dc=eg$.

Take $u:T\to D$ and $v:T\to C$ with $efu=dv$. The right pullback in a category gives a unique $w:T\to B$ with $gw=fu$ and $cw=v$. The left pullback in a category then gives a unique $z:T\to A$ with $az=u$ and $bz=w$. It follows that $cbz=v$, giving the required factorization through the outside rectangle. Any other factorization has the same middle morphism by uniqueness in the right pullback in a category, and therefore is the same $z$ by uniqueness in the left pullback in a category. The outside rectangle is a pullback.

Keep the notation of the preceding part. For $u:T\to D$ and $w:T\to B$ satisfying $fu=gw$, we have $efu=egw=dcw$. The outside pullback in a category gives a unique $z:T\to A$ with $az=u$ and $cbz=cw$. Moreover $gbz=faz=fu=gw$. The morphisms $bz,w:T\to B$ thus have the same composites with $g$ and $c$. Uniqueness in the right pullback in a category gives $bz=w$. Any factorization through the left square also factors through the outside rectangle, so it is unique there. The left square is a pullback. Together these two arguments prove the pullback pasting lemma.

Here is a precise base change of a pullback square statement. Suppose $A=B{\times }_{D}C$ is a pullback in a category and $t:D^{\prime }\to D$ is a morphism. Assume the pullbacks in a categoryexist, where $A\to D$ is the common composite. The induced square with vertices $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ is a pullback in a category; equivalently,Indeed, compatible morphisms $T\to B^{\prime }$ and $T\to C^{\prime }$ give morphisms $b:T\to B$, $c:T\to C$, and a common $v:T\to D^{\prime }$, with the composites of $b,c$ to $D$ both equal to $tv$. The original pullback in a category gives a unique morphism $h:T\to A$ with projections $b,c$. Its composite to $D$ is $tv$, so the defining pullback in a category for $A^{\prime }$ gives a unique morphism $T\to A^{\prime }$ above $h,v$. Its maps to $B^{\prime }$ and $C^{\prime }$ are the prescribed ones by uniqueness in those pullbacks in a category. Conversely, these projections recover $b,c,v$, so every possible factorization is the same. This proves the universal property without requiring arbitrary categorical limits in the ambient category.