**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, itâ€™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.