3

In a way of defining a category called as single-sorted definition of a category which uses only one collection (representing the collection of morphisms) and thus is formulated as an untyped (or 1-sorted) first-order theory, the basic idea is that an object can be identified with its identity morphism.

Definition

A category (single-sorted version) is a collection C, whose elements are called morphisms, together with two functions s,t:C→C and a partial function ∘:C×C→C, such that:

  • s(s(x))=s(x)=t(s(x))
  • t(t(x))=t(x)=s(t(x))

The elements of their common image (the x such that s(x)=x, or equivalently t(x)=x) are called identities or objects.

https://ncatlab.org/nlab/show/single-sorted+definition+of+a+category

At the same time, the nLab article also mentions:

Specializations

A monoid is a single-sorted category in which s is a constant function (hence so is t, and they are equal). This works up to isomorphism of categories, not merely equivalence, so single-sorted categories may seem to be a more direct oidification of monoids than the usual categories.

I understood s and t are identity morphism s,t:C→C, but here it claims they are constant functions, and I feel it does not make sense.

Does this exactly mean, function composition such as

(C→5)∘(C→C)??

I am still confused.

In addition, for the mention:

This works up to isomorphism of categories, not merely equivalence, so single-sorted categories may seem to be a more direct oidification of monoids than the usual categories.

I don't quite understand the meaning.

Can someone who is familiar with single-sorted definition of a category explain? Thanks.

Perhaps, this topic is related to a Q&A: The $2$-category of monoids

smooth_writing
  • 229
  • You seem to be confused between identity morphisms and endomorphisms. The identity morphism is an endomorphism. Not every endomorphism is the identity morphism. – Zhen Lin Sep 01 '20 at 04:41
  • Thanks, but I understand that, and I think that is nothing to do with this topic. Perhaps, you didn't read the question well, this is about identity morphism. – smooth_writing Sep 01 '20 at 04:48
  • 2
    $s$ and $t$ are not the identity morphism $C \to C$ in general. I believe this is what @ZhenLin's comment is saying – if $s$ and $t$ are the identity, then every endomorphism would be the identity morphism. – diracdeltafunk Sep 01 '20 at 05:00
  • @diracdeltafunk Thanks, so "The elements of their common image (the x such that s(x)=x, or equivalently t(x)=x) are called identities or objects." in Definition in the article https://ncatlab.org/nlab/show/single-sorted+definition+of+a+category, are they saying x is called identities or objects? – smooth_writing Sep 01 '20 at 05:13
  • Correct. If $x$ is a morphism such that $s(x) = x$, then $x$ is called an identity (you might also call $x$ an "identity morphism" or "object", these all mean the same thing). – diracdeltafunk Sep 01 '20 at 06:28
  • @diracdeltafunk Thank you for making things clear! So, the 2 formulas of s and t is just to define the category needs to be closed under the composition operation? – smooth_writing Sep 01 '20 at 08:18
  • 1
    No, closure under composition is not related to the formulas $s(s(x)) = s(x) = t(s(x)), t(t(x)) = t(x) = s(t(x))$. These formulas just say (by definition) that $s(x)$ and $t(x)$ are always objects. – diracdeltafunk Sep 01 '20 at 23:06

1 Answers1

4

I understood s and t are identity morphism s,t:C→C

I don't know what you mean by this. The intended interpretation is that $C$ is the collection of all morphisms in a category and $s, t$ send a morphism $f : x \to y$ in the ordinary sense to the identity endomorphism of its source $s(f) = \text{id}_x$ and target $t(f) = \text{id}_y$ respectively.

If we have a monoid $M$, interpreted as a one-object category $BM$ (so there is one object $\bullet$ and it has endomorphisms $M$), then both $s$ and $t$ are necessarily constant, and their constant value is the identity endomorphism $\text{id}_{\bullet}$ of the unique object $\bullet$ of $BM$. Is it clear now?

Qiaochu Yuan
  • 468,795