Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 25 / 2 / a / Solution

Let $D:J\to \mathcal{C}$ be a finite diagram in a category. Form the finite products in a categoryDefine $s,t:P\rightrightarrows Q$ by ${\pi }_{a}s=D(a){\pi }_j$ and ${\pi }_{a}t={\pi }_k$. Take their equalizer $e:L\to P$. The maps ${\lambda }_j={\pi }_{j}e$ satisfy $D(a){\lambda }_j={\lambda }_k$, so they constitute a categorical cone.

Any other categorical cone $x_j:X\to D(j)$ gives a unique $x:X\to P$ by the product in a category property. Its cone equations say $sx=tx$, so the equalizer gives a unique $\bar{x}:X\to L$ with $e\bar{x}=x$. The projections then give ${\lambda }_j\bar{x}=x_j$, and uniqueness follows from both universal properties. Thus this equalizer of two product maps is the finite limit of the diagram in a category.

The empty indexing category is included: its empty product in a category is a terminal object, which is the empty categorical limit. Hence no nonemptiness restriction has entered the construction.