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

There are several ways to reduce a category. The skeleton of a category is the category with isomorphic objects collapsed into one i.e. the only isomorphisms that remain are the identities.

What's the name of the category with parallel arrows collapsed into one?

Finally, consider a category of an appropriate "order type" that has undecomposable arrows, for example $(\mathbb{N},\leq)$. What's the name of the digraph (not a category then, of course) with undecomposable arrows only, e.g. $\mathbb{N}$ with the successor relation?

When I ask for names I mean the name of the construction, with respect to the original category (like in skeleton of).

share|cite|improve this question
It is not true that the only isomorphisms in the skeleton of a category are the identities. – Qiaochu Yuan Feb 7 '12 at 8:30
@Qiaochu: Why not? – Hans Stricker Feb 7 '12 at 8:36
That would defeat the point of the construction of the skeleton (it's supposed to be a category equivalent to the original category, and getting rid of all the isomorphisms doesn't accomplish this), and it's easy to find counterexamples (take e.g. the skeleton of the category of finite sets). – Qiaochu Yuan Feb 7 '12 at 8:39
Sorry, but Wikipedia on skeletons says that no two distinct objects are isomorphic. What do I miss? – Hans Stricker Feb 7 '12 at 8:44
You miss that an object may have nontrivial automorphisms. – Qiaochu Yuan Feb 7 '12 at 8:47

I guess I would call the first construction the "preorderization" since it is the universal map from a category to a preorder. I guess I would call the second construction the Hasse diagram.

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.