Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Suppose we have a category $C$ such as $hom(C) \subset ob(C)$ (so every arrow $f: a \rightarrow b$ of $C$ is also an object of $C$). What is the name and the main properties of such category?

share|cite|improve this question
In my opinion, that's just not the sort of thing that should be going on in a category. There's nothing technically wrong about it, but for conceptual simplicity objects and morphisms should always be distinct, for the same reason it is usually required that $\mathrm{Hom}(a,b)\cap\mathrm{Hom}(c,d)=\varnothing$ unless $a=c$ and $b=d$. – Zev Chonoles Feb 15 '14 at 17:52
The more meaningful property is $\text{hom}_C(a,b)\in\text{Ob}(C)$ for every $a,b\in \text{Ob}(C)$(self-enriched categories, under certain assumptions). – Oskar Feb 15 '14 at 19:52
With the right version of set theory, it shouldn't even be possible to state this condition. (ZFC has the stupid property that the question "$a \subset b$?" makes sense for every pair of sets $a$ and $b$. Are you ever going to ask whether $\mathbb{R} \subset 3$? No.) – Qiaochu Yuan Feb 16 '14 at 6:04

I do not think there is a name for the property that you mention, for the simple reason that this property is not very meaningful.

Look at it this way:

Suppose you have a category $C$ which has the property that you mention, and then another category $D$ which has exactly the same objects, but the morphisms are written with red ink (just to enphasize some super-silly difference in naming) or have an overbar, if you prefer.

Now, we certainly agree that $C$ and $D$ should have the same mathematical properties, since they are effectively the same category, yet $C$ has your property, while $D$ does not, since its morphisms are written in red (or have an overbar) and the elements of $Ob(D)$ are all still in black (and have no overbar). So your property is very fragile and not respected even by isomorphic categories.

A mathematical property - to be considered meaningful - needs to respect equivalence of categories and even more isomorphism of categories. Yours does not. This is called the principle of equivalence in nlab

In his comment above @Oskar mentions the more interesting self-enriched categories, which is perhaps what you were looking for.

By the way, the collection of morphisms of $C$ is called $Mor(C)$

share|cite|improve this answer
In my reading, an injection $J:Mor(C)\to Ob(C)$ would suffice in the question. – Berci Feb 17 '14 at 0:30

This idea just came to my mind recently!

I haven't found anything about it, but wrote a short note, where I called such a thing a 'folded category', and I wanted to model infinite dimensional categories by them, starting out from an unbiased version of bicategories.

Then I posted it on MathOverflow, but nobody answered.

Finally, I found out that some elementary feature seems missing: namely I had to add one more composition for second dimension, but I kept on hoping that it will already generate all we need. Now, I think, there is no way to escape from introducing one more operation for each dimension, and with that, it seems that things get the same (or even more) complicated as with any original naive approach... :(

share|cite|improve this answer
Many thanks you for the link to your note. I think I should try to get all things in one piece before marking answers as accepted (as I'm not very good at such math). – Vladimir Feb 17 '14 at 19:24
I'm still not sure that this is what you were looking for. And, it didn't lead me nowhere.. – Berci Feb 19 '14 at 22:15

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.