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September 14

Underlying structure for semigroupoid

Reflexive quivers are used as underlying structures in small category, and multiplicative graphs, but what about semigroupoid? Since each vertex doesn't need to have an identity and (but) the composition of pair of consecutive morphisms appears to always be defined, so I don't think it's based on a reflexive quiver. The statement "composition of pair of consecutive morphisms is always defined", I might be overlooking additional conditions. I'm worried about case of the composition of endomorphism (loop) and ordinary morphism with with same source or target.--SilverMatsu (talk) 04:21, 14 September 2026 (UTC)

I'm sorry, but I find the definition given for reflexive quivers hard to understand. Assuming that reflexive quivers are quivers, I think the definition should take the form of defining a reflexive quiver as being a quiver with an additional requirement, and use the notation of the Definition section to formulate that requirement. Also, the meanings of the notations $1_{G_{0}}$ and $\mathrm {loop} (x)$ are not given. I assume $1_{G_{0}}$ stands for the identity function ${\text{id}}_{G_{0}},$ but this may not be clear to all readers.  ​‑‑Lambiam 06:53, 14 September 2026 (UTC)

Thank you for pointing that out. I added a figure.--SilverMatsu (talk) 09:19, 14 September 2026 (UTC)

For example, I think for loop edge $\mathrm {id} _{B}$ to be an identity arrow, it requires that composition $g\circ \mathrm {id} _{B}$ and $\mathrm {id} _{B}\circ f$ are defined.(IMO) Also, quiver does not require composition.--SilverMatsu (talk) 04:35, 15 September 2026 (UTC)

What should we call the following axiom (in multiplicative graph#Definition)? Axiom 1: For each element $x$ of ${\underline {\Sigma }}$, the composites $x\circ \alpha (x)$ and $\beta (x)\circ x$ are defined, and we have $x\circ \alpha (x)=x=\beta (x)\circ x.$ Here, $\alpha (x)$ is the right identity of $x$ and is called the source of $x$, while $\beta (x)$ is the left identity of $x$ and is called the target of $x$. Should this axiom be called "existence of units" or "existence of composition for loops"? There might be a better term for "existence of composition for loops".--SilverMatsu (talk) 03:13, 18 September 2026 (UTC)

A semigroupoid is, in general, not a reflexive quiver. In particular, we can have a degenerate semigroupoid with just objects and only empty sets of morphisms – and therefore no loops.  ​‑‑Lambiam 08:53, 18 September 2026 (UTC)

Thank you for explaining with an example.--SilverMatsu (talk) 02:48, 20 September 2026 (UTC)