MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading "Abstract and Concrete Categories" and, in the Chap. I (Definition 3.52 on Page 41), there's an ``Object-Free'' definition of a Category which, through the isomorphism $A \rightarrow \textit{id}_A$ turns out to be equivalent to the usual one. But all the other definitions in the book are only given in the usual "Objects+Morphisms" fashion. What is the definition of Initial&Terminal Objects in the "Object-Free" version of a Category?

share|cite|improve this question
If you choose to not include objects in the definition, you can (because of the equivalence you mention) introduce them later. Then the definition of, say, initial objects is simply the usual one! – Mariano Suárez-Alvarez Apr 6 '11 at 17:24
About this, I have a quotation in mind, but I'm not sure of the place (McLane?). It's some remark like "objects are just the domains of arrows, and categories should be named after arrows rather than objects. Yet nobody is so brave to call e.g. the category Top "category of continuous maps". – Pietro Majer Apr 7 '11 at 6:46
up vote 2 down vote accepted

In this view, objects are equated with morphisms that are identities, or "units" in their terminology. So a morphism $x$ is initial when it is a unit and for every unit $y$ there is a unique morphism $f$ for which $f \circ x$ and $y \circ f$ are defined. Similarly, a morphism $y$ is terminal when it is a unit and for every unit $x$ there is a unique morphism $f$ for which $f \circ x$ and $y \circ f$ are defined.

share|cite|improve this answer
I've found this paper which uses different axioms for the usual and object-free definitions. I'm not sure but maybe one could use the "undefined morphism" $\perp$ as initial obj (along the lines of $\emptyset$ in Set and the bottom type in type theory) and a unit as terminal?. so $\perp$ is initial if $f \circ \perp = \perp \circ f = \perp$ and $u$ is terminal iff it's a unit? – SHK Apr 7 '11 at 15:50
Those notes are only using the undefined morphism $\bot$ to make composition a binary operation on arrows, instead of just a partial binary operation. You can use this trick to make any partial operation into a fully defined one. That hasn't got much to do with initial objects at all. – Chris Heunen Apr 7 '11 at 17:27

As the others have said, the object-free definition can always define objects later and do everything normally. One might ask if there was a different-than-normal definition that was more arrow-like. For example:

A morphism $f$ is "terminating" if, for every morphism $g$, there exists a unique morphism $h$ such that $h \circ g$ and $f$ have the same target.

So, the target of $f$ is a terminal object, and $f$ itself is the unique projection from its source. But this seems like a superficial change to me. Maybe someone else knows something better?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.