Pullback (category theory)

The pullback of $X\to Z\leftarrow Y$ is a universal commutative square with an object mapping to $X$ and $Y$. In the Category of sets, it is the set of pairs with equal images in $Z$.

Given a pullback in a category $A=B{\times }_{D}C$ and a morphism $D^{\prime }\to D$, assume the indicated pullbacks in a category exist. The four base-changed objects form a pullback in a category. A compatible pair of morphisms into $B{\times }_D{D}^{\prime }$ and $C{\times }_D{D}^{\prime }$ gives morphisms into $B,C$ and a common morphism into $D^{\prime }$. The first two morphisms factor uniquely through $A$, and their common image in $D$ permits unique factorization through $A{\times }_D{D}^{\prime }$. This proves the displayed canonical isomorphism without an assumption that the whole category is complete.

In a commutative diagram consisting of two horizontally adjacent squares, if both squares are pullbacks in a category, their outside rectangle is a pullback in a category. If the right square and the outside rectangle are pullbacks in a category, the left square is a pullback in a category. To prove the first assertion, factor a compatible pair of morphisms first through the right square and then through the left square. For the second assertion, factor through the rectangle and use uniqueness in the right square to recover the prescribed middle morphism. Each factorization is unique by the same universal properties.

The kernel pair of a morphism $f:A\to B$ is the pullback in a category $A{\times }_{B}A$ with its two projections to $A$. It is the universal parallel pair equalized by $f$.

Every epimorphism in an abelian category is the coequalizer of its kernel pair. Its kernel-pair square is a pullback in a category along an epimorphism and therefore also a pushout. A map equalizing the two projections supplies the same map on both pushout legs and so factors uniquely through the original epimorphism. In particular, all such epimorphisms are regular epimorphisms.