Is a groupoid a universal algebra? I was trying to use the first isomorphism theorem on groupoids. From the wikipedia page I know that it holds for groups, rings and algebras.
So are groupoids algebras? Or, in other words, does the first isomorphism theorem hold for groupoids (with identity)?
 A: It appears the official answer is "no", since the "operation" on the groupoid is not necessarily defined for every pair of things in the groupoid. (Looking at things categorically, viewing a group or monoid as a category with one object, any two arrows are composable since they have the same domain and codomain. But with groupoids, there are in general more than one object, and so there would exist arrows that don't compose, i.e. "can't be multiplied.")
However, I did notice at the Wiki page that a pair of papers by Higgins was referenced as taking the ideas of universal algebra into sets with partial operations, and groupoids are called out specifically.
A: If you interpret a groupoid as a special kind of small category, you can use the construction summarized in the question Constructing a semigroup from a small category, to get a semigroup:

So for a small category $C$, let $S_C:=\{M\subset\operatorname{hom}(C):|M|<\infty \}$ and for $A,B\in S_C$ define $A*B:=\{f\circ g:f\in A, g\in B,\operatorname{source}(f)=\operatorname{target}(g)\}$. Then $S_C$ together with the operation $*$ is a semigroup. [...]
What I find even more interesting is that $S_c:=\{A\in S_C:|A|\leq1\}$ is a sub-semigroup of $S_C$. Because $S_c=\{\varnothing\}\cup\{\{f\}:f\in\operatorname{hom}(C)\}$, we have a "natural" correspondence between the semigroup $S_c$ and $\operatorname{hom}(C)\cup\{0\}$. Here $0\notin\operatorname{hom}(C)$ is an absorbing element corresponding to the empty set $\varnothing\in S_c$. The semigroup operation of $S_c$ is just the composition of the corresponding morphisms if this is defined, or the empty set otherwise.

However, the first isomorphism theorem $(G/M)/(N/M)\cong G/N$ for normal subgroups $M$ and $N$ of group $G$ turns into a statement about congruences $\varphi$ and $\theta$ of the semigroup $A$:

If $\varphi, \theta \in \operatorname{Con} A$ and $\theta \subseteq \varphi$, then the map $\alpha : (A/\theta)/(\varphi/\theta)\to A/\varphi$ defined by $\alpha((a/\theta)/(\varphi/\theta)) = a/\varphi$ is an isomorphism from $(A/\theta)/(\varphi/\theta)$ to $A/\varphi$.

If we set $A=S_c$, all that is left to do is check how congruences relate to the absorbing element $0$.
