Exercise 5.2.25

A map is regular monic if is an equalizer of some parallel pair . is split monic if it admits a left inverse, i.e. a map with .

Theorem: Split monic regular monic monic

Proof: Assume is split monic, and let be its left inverse. Then, take the parallel pair . We claim is the equalizer of this parallel pair. Note that it is certainly a cone, since . Suppose that were another cone such that . Define as . Then certainly the appropriate triangle commutes: . Suppose were another such map, so that . But composing on the left by implies that and then . So, is unique and is the equalizer of the parallel pair, making regular monic.

Suppose is regular monic, and that it is the equalizer of . Take arbitrary maps with . Then, of course, since and are (the same) cone over , we have a unique map with . Taking and make this true, so we must have . So is monic.

An example of a monic which is not split occurs in . Consider the inclusion . Being inclusion, it’s clearly monic. However, its a simple exercise in group theory to show there are no group homomorphisms from in general, so there can certainly be no left inverse, so is not split monic.