Determining Objects in a Semicategory

Question

Suppose $S$ is a small semicategory (or semigroupoid, if that's your preferred term) and $\cdot$ is the binary operation on $S$. Implicit in this definition is the set $\operatorname{Ob}(S)$ and two functions $\operatorname{dom}: S \to \operatorname{Ob}(S)$ and $\operatorname{codom}: S \to \operatorname{Ob}(S)$ such that for $a, b \in S$, $a \cdot b$ is defined if and only if $\operatorname{dom}(a) = \operatorname{codom}(b)$. (I view members of $S$ as morphisms of $S$, not objects of $S$.)

If we somehow lose $\operatorname{Ob}(S)$, $\operatorname{dom}$ and $\operatorname{codom}$, is there a canonical way to recover them from just $\cdot$? (There are definitely many possible choices as $\operatorname{Ob}(S)$ can be enlarged arbitrarily from any correct choice.)

My current idea is as follows. First, define $L \subseteq S$ and $R \subseteq S$ by $L = \{a \in S \mid b \cdot a \text{ is not defined for all } b \in S\}$ and $R = \{a \in S \mid a \cdot b \text{ is not defined for all } b \in S\}$. Intuitively, $L$ consists of left-only elements and $R$ consists of right-only elements.

Next, define $O \subseteq S^2$ by $O = \{(a, b) \in S^2 \mid a \cdot b \text{ is defined}\}$, and define an equivalence relation $\sim$ on $O$ such that $(a, b) \sim (a', b')$ if and only if $(a, b') \in O$ or $(a', b) \in O$ (in which case both would be true). (I believe that if $(S, \cdot)$ is not a semicategory, $\sim$ would not be an equivalence relation, but it's just a guess.)

Finally, define $\operatorname{Ob}(S) = (O / \sim) \amalg \{s, t\}$ where $s$ and $t$ are distinct constants. $\operatorname{dom}$ can be defined by $$ \operatorname{dom}(a) = \begin{cases} s & ; a \in R \\ [(a, b)]_{\sim} & ; \text{any $b \in S$}, a \notin R \end{cases} $$ $\operatorname{codom}$ can be defined similarly.

To make $\operatorname{Ob}(S)$ as small as possible, we can remove $s$ or $t$ or both if they are not needed. In other words, we can re-define $\operatorname{Ob}(S)$ to be the union of the images of $\operatorname{dom}$ and $\operatorname{codom}$.

I am not sure if this construction is canonical enough because the way it deals with left-only or right-only morphisms seems somewhat unnatural to me.

Here are my questions:

1. What do you think of this construction? Is it correct? If it is correct, is it natural? Can it be simplied?
2. Is there an existing work discussing this?
3. Are all choices that work equivalent? I have not formally defined equivalence here, but the approximate sense is that $\operatorname{dom}$ and $\operatorname{codom}$ are, in some sense, invariant.

Answer 1

1. Yes, it is correct.
2. I am not aware of it.
3. Well, we cannot totally recover the underlying graph structure in general from a semicategory: for an easy example, take the semicategory $\mathcal U$ with points $A,B,C,D$ and two arrows $f:A\to B$ and $g:C\to D$ (taking $S=\{f,g\}$ in this case). Now this semicategory can by no means be distingushed from its relatives like $\mathcal U'$ in which $A=C$, or $\mathcal U''$ in which $B=D$.
   And, the problem is mostly with your left-only and right-only arrows.

Let $\mathcal{sCat}$ denote the category of (small) semicategories without unnecessary objects (i.e. when ${\rm cod}$ and ${\rm dom}$ are jointly surjective).
And, for the moment, let us call a partial algebra $(S,\cdot)$ a graphless semicategory if whenever any two of $\ f\cdot g,\ \ g\cdot h,\ \ (f\cdot g)\cdot h,\ \ f\cdot(g\cdot h)\ $ are defined, then so are all four of them, and in this case $(f\cdot g)\cdot h\ =\ f\cdot(g\cdot h)$.
Let $\mathcal{gsCat}$ denote their category.

Then, the forgetful functor $\mathcal{sCat}\to\mathcal{gsCat}$ has a right adjoint, which is what you have described above.
But, it also has a left adjoint, which you can get if elements of $L$ and $R$ are 'not glued together' in one point [$s$ and $t$].
Alternatively, you can take the equivalence relation on $S\times\{0,1\}$ generated by
$$\quad (f,1)\sim (g,0)\quad \text{if}\quad \exists\, g\cdot f\,, $$ where $(f,0)$ wants to represent the domain of $f$ and $(g,1)$ the codomain of $g$.