Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 18 / 2

The diagonal satisfies $pd=qd=1_Y$, so it equalizes the two projections. If $h:Z\to Y\times Y$ satisfies $ph=qh$, put $u=ph=qh$. The universal property of the product in a category gives $h=(u,u)=du$. Any other factor $v$ with $dv=h$ must satisfy $v=pdv=ph=u$. Thus the diagonal is the equalizer of the two product projections.

Take the pullback in a category of $d_Y$ along $(f,g)$, with maps $e:E\to X$ and $k:E\to Y$ satisfying $(f,g)e=d_{Y}k$. Applying the two product projections gives $fe=k=ge$, so $e$ equalizes $f,g$.

Given $h:Z\to X$ with $fh=gh$, put $v=fh$. Then $(f,g)h=(v,v)=d_{Y}v$. The pullback property gives a unique $u:Z\to E$ with $eu=h$ and $ku=v$. Any map satisfying $eu=h$ automatically has $ku=feu=fh=v$, so uniqueness is exactly the equalizer uniqueness. Hence pulling back the diagonal constructs the equalizer of $f,g$.

A terminal object and pullbacks in a category give a binary product $X\times Y$ by pulling back $X\to 1\leftarrow Y$. Repeating this gives every finite product, with the empty product supplied by the terminal object. The preceding parts then construct equalizers of arbitrary parallel arrows.

For a diagram $D:J\to \mathcal{C}$ with finitely many objects and arrows, form $P={\prod }_{j\in \mathrm{Ob}J}D(j)$ and $Q={\prod }_{a:j\to k}D(k)$. Define $r,s:P\to Q$ by $r_a=D(a){\pi }_j$ and $s_a={\pi }_k$. Their equalizer $e:L\to P$ has projections ${\pi }_{j}e$ satisfying every cone equation. Conversely a categorical cone supplies a unique map to $P$, and its cone equations say that this map equalizes $r,s$, hence factors uniquely through $L$. Thus $L$ is the categorical limit. This proves a terminal object and all pullbacks give all finite limits, including the empty diagram.