Existence of limits from products and equalizers

Remarks

Let $F \colon J \to \mathcal{C}$ be a diagram with $J$ small, and write $u \colon j \to k$ for the morphisms of $J$. Form the products $P = \prod_{j} F(j)$ over the objects and $Q = \prod_{u \colon j \to k} F(k)$ over the morphisms. Two morphisms $s, t \colon P \to Q$ are given on the factor of $u \colon j \to k$ by $F(u) \circ \pi_j$ and by $\pi_k$. The equalizer of $s$ and $t$ is a limit of $F$. If $J$ is finite, only finite products are used.

The converse is clear, since products and equalizers are limits. For example, in the category of groups the limit of a diagram is the subgroup of the product cut out by the equations $F(u)(x_j) = x_k$. Dually, a category with all small coproducts and all coequalizers is cocomplete.

Going deeper

For a student

This is the usual way to prove that a category is complete. For sets, groups, rings and topological spaces the products and equalizers are easy to write down, and the theorem does the rest. The same construction shows that a functor which preserves products and equalizers preserves all limits. It also gives the finite version: a terminal object, binary products and equalizers give all finite limits.