Let be a diagram in some category and be an equalizer of and . We can form an associated diagram:
which commutes. Indeed, is a cone over the diagram , which follows only from the fact that is a cone over the parallel pair . However, is not necessarily a limit cone and thus a pullback. We can see this by working in . Let and . Let be the non-trivial (group) homomorphism between them and the trivial homomorphism. Intuitively, we should not expect the square to be a pullback, because the equalizer of and should be the subgroup , while the pullback of the aforementioned diagram should be . But, let’s suppose was the equalizer , and demonstrate a counterexample to the square being a pullback. Simply take the cone consisting of the singleton and maps picking out the elements and . It is a cone, because:
clearly commutes. But we have:
but there can be no map : there are only two possible maps. The map picking out makes the left triangle commute but not the top, and the map picking out makes the top triangle commute but not the left. Thus, cannot be a pullback because we’ve demonstrated a distinct cone without a viable map making the diagram commute.
In fact, the argument works in general. Take any cone over the diagram , such that and are distinct, and consider:
Since is a cone over , we have a unique map making the left triangle commute. But, simultaneously, since is a distinct cone over the same parallel pair, we have a distinct unique map making the top triangle commute. Even if both and made the entire diagram commute, they would be two non-equal such maps, meaning that fails to be a pullback.
However, the converse is true:
Theorem: If is a pullback of the diagram , then is an equalizer of .
Proof: Take an arbitrary cone over . It follows immediately that is a cone over . Then we have a unique with:
commuting. Then is also the unique map making the left and top (identical) triangles commute. But a unique along with the left or top commuting triangle are exactly the items needed for to be an equalizer.