Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 3

For a functor $D:\mathcal{J}\to \mathcal{C}$ with finite indexing category, form the finite products in a categoryDefine $s,t:P\rightrightarrows Q$ by requiring the coordinate indexed by $u:i\to j$ to be $D(u){\pi }_i$ for $s$ and ${\pi }_j$ for $t$. Let $e:L\to P$ be their equalizer, and put ${\lambda }_j={\pi }_{j}e$. The equality $se=te$ says exactly that $D(u){\lambda }_i={\lambda }_j$ for every morphism $u$, so these morphisms form a categorical cone.

For any other categorical cone $(x_j:X\to D(j))$, the product in a category gives a unique $x:X\to P$ with ${\pi }_{j}x=x_j$. The categorical cone equations give $sx=tx$, so the equalizer gives a unique $\bar{x}:X\to L$ with $e\bar{x}=x$. This is exactly the universal property of a categorical limit. Empty indexing products are terminal objects, so the construction includes the empty diagram in a category. Thus finite products and equalizers give all finite limits.

Let $1$ be a terminal object. The pullback in a category of $X\to 1\leftarrow Y$ has precisely the universal property of the binary product in a category $X\times Y$: every pair of morphisms to $X,Y$ is compatible over $1$. Repeated binary products in a category, together with $1$ as the empty product, give all finite products in a category.

For $f,g:X\rightrightarrows Y$, form the pullback in a category of the categorical diagonal ${\Delta }_Y:Y\to Y\times Y$ along $⟨f,g⟩:X\to Y\times Y$. Its projection $e:E\to X$ satisfies $fe=ge$. If $h:T\to X$ satisfies $fh=gh$, the pair $h,fh$ gives a unique morphism into this pullback in a category. Conversely any such morphism must project to that pair. Hence $e$ is the equalizer of $f,g$. The preceding construction now gives all finite limits.