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  "body": "\n## Statement\n\nLet $\\mathcal{C}$ be a [[category|A2DE66937D]] that has [[equalizers|CTJ5AD8ST5]] of all pairs of parallel morphisms.\n\n1. If $\\mathcal{C}$ has all small [[products|S5744QZV4N]], then $\\mathcal{C}$ is [[complete|FWGPKXCPB7]].\n2. If $\\mathcal{C}$ has all finite products, then $\\mathcal{C}$ is [[finitely complete|PYSBNWCHCF]].\n\n## Remarks\n\nLet $F \\colon J \\to \\mathcal{C}$ be a diagram with $J$ small, and write $u \\colon j \\to k$ for the morphisms of $J$.\nForm the products $P = \\prod_{j} F(j)$ over the objects and $Q = \\prod_{u \\colon j \\to k} F(k)$ over the morphisms.\nTwo 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 $\\pi_k$.\nThe equalizer of $s$ and $t$ is a [[limit|CHS5AFJGWE]] of $F$.\nIf $J$ is finite, only finite products are used.\n\nThe converse is clear, since products and equalizers are limits.\nFor example, in the category of groups the limit of a diagram is the subgroup of the product cut out by the equations $F(u)(x_j) = x_k$.\nDually, a category with all small coproducts and all coequalizers is cocomplete.\n\n## Explanations\n\n### L1 human\n\nEvery limit can be built from two simple ones.\nFirst take the product of all the objects of the diagram.\nThen cut it down with an equalizer to the families that agree with all the maps.\nSo to show that a category has all limits, it is enough to check products and equalizers.\n\n### L2 human\n\nThis is the usual way to prove that a category is complete.\nFor sets, groups, rings and topological spaces the products and equalizers are easy to write down, and the theorem does the rest.\nThe same construction shows that a functor which preserves products and equalizers preserves all limits.\nIt also gives the finite version: a terminal object, binary products and equalizers give all finite limits.\n",
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