# Is there a name for a section and a retraction together?

I call $(s, r)$ such that $r\circ s =\operatorname{id}$ a split morphism. (Other names are “section-retraction pair”, “split pair.”) Notice that this is not a section, because any section is one morphism. I want to make sure that no official name exists thus I can invent it myself. Do you know the name?

The same question about isomorphisms. “Isomorphism pair”, “isomorphic pair”?

The distinction may be quite apparent. Every split morphism is a categorical diagram whereas a section is not.

• There is a notion of a split epi/monomorphism, but it isn't defined as a pair of morphisms. This seems more useful in a algebraic or constructive setting... – Zhen Lin Oct 17 '11 at 17:48
• @Zhen Lin: Split monomorphism is just a synonym for section. I read Wikipedia before. :) – beroal Oct 19 '11 at 11:33
• Have you settled on any terminology? I want to use it. For the second concept, I think "isomorphism pair" is a good name. – goblin Feb 12 '14 at 23:55
• @user18921: No. I leave it to you. – beroal Feb 14 '14 at 6:15

The points in a category $\mathcal A$ together with pairs of morphisms making the obvious diagrams commute form a category $\mathsf{Pt}(\mathcal A)$, the category of points of $\mathcal A$. Assuming $\mathcal A$ has pullbacks along split epis, the "codomain functor" $\mathsf{Pt}(\mathcal A)\to \mathcal A$ is called the fibration of points (it is a fibration in this case). The fibration of points is useful to classify various notions between (let's say) $\mathcal A$ being Mal'cev and $\mathcal A$ being essentially affine (Abelian minus [pointed & has cokernels]).
I suppose the name "point" comes from the fact that the fiber $\mathsf{Pt}_A$ of the codomain functor over $A$ is (essentially) the category of pointed objects of the slice category $\mathcal A \downarrow A$. The category of pointed objects of a category $\mathcal B$ with a terminal object $\mathbf{1}$ is the coslice category $\mathsf{1} \downarrow \mathcal B$, which is, you guessed it, pointed.