Definition
$\left\langle S, \otimes \right\rangle$ is a regular semigroup means
- The operation $\otimes$ is closed on $S$: $\forall s_1, s_2 \in S: s_1 \otimes s_2 \in S$
- The operation $\otimes$ is associative: $\forall s_1, s_2, s_3 \in S: s_1 \otimes \left( s_2 \otimes s_3 \right) = \left( s_1 \otimes s_2 \right) \otimes s_3$
- For each $s \in S$ there is a $s' \in S$ (maybe not unique) such that $s \otimes s' \otimes s = s$ and $s' \otimes s \otimes s' = s'$
Being an inverse semigroup means uniqueness of the inverse.
Example
Let's take a look at the specific example. A set of square matrices of fixed size is closed under the matrix multiplication, so forms a regular semigroup (given the Moore-Penrose inverse as the inverse because not all matrices are invertible). The Moore-Penrose inverse (pseudoinverse) $A^+$ is defined for any matrix as $\forall n, m \in \mathbb{N}^+: \forall A \in \mathbb{R}^{n \times m}: A A^+ A = A, A^+ A A^+ = A^+$. It's unique so it forms an inverse semigroop as well as less restrictive regular semigroup.
Problem
In turn, the set of all real-valued matrices is not closed under the matrix multiplication, so it's not a regular semigroup. Though, it's associative and has a pseudo-inverse. I would call such structure regular semigroupoid (or inverse semigroupoid if being more restrictive and precise).
The problem: I can't find a lot about the regular semigroupoid (and inverse semigroupoid), and cannot find mentions of Moore-Penrose inverse and matrices as the example. The articles I've found are "E-free objects and e-locality for completely regular semigroups" by Peter R. Jones (referencing regular semigroupoid) and "Free inverse monoids and graph immersions" by S. W. Margolis and J. C. Meakin (referencing inverse semigroupoid).
Question
- Do matrices with multiplication and Moore-Penrose inverse form a regular semigroupoid?
- Is my understanding of regular and inverse semigroupoids correct?
- Why are regular and inverse semigroupoids not so popular?