Groupoids and skeletons of monoids and preorders

A monoid $(M,e,m)$, where $M$ is a set, $e$ an element of $M$, and $m$ an associative binary operation on $M$, can be viewed as a single object category. Take $M_0 = \\{M\\}$ or $M_0 = \emptyset$ and $M_1$ to be the elements of $M$, that is, morphisms in this category are elements of the monoid (where $M_0$ is the class of objects of the category and $M_1$ is the class of morphisms).

A groupoid is a category where every morphism is an isomorphism. Monoids are groupoids when every element of the monoid has an inverse. Since the morphisms of this category are elements of the monoid, if every element has an inverse, then every morphism is an isomorphism. So, monoids which are groupoids, are those monoids that are groups.

A preorder $(P,R)$, is a set $P$ together with a reflexive, transitive relation $R$. A preorder can be viewed as a category by letting $P_0 = P$ and $P_1=R$.

A preorder is a groupoid when the relation $R$ is symmetric. That is, an equivalence relation on a set is a groupoid. Since, for $x,y\in P$, if $(x,y)\in R$ then there is a morphism $x\to y$, and if $(y,x)\in R$ then there is a morphism $y\to x$, inverse to the morphism $x\to y$, so it is an isomorphism. If $(x,y)\in R \implies (y,x)\in R$ then every morphism $x\to y$ has an inverse morphism $y\to x$ and so every morphism is an isomorphism.

A skeleton is a category in which, for every object $A,B$, $A\approx B\implies A=B$. That is, if two objects are isomorphic then they are equal.

A preorder is a skeleton when the relation $R$ is antisymmetric. That is, an order is a skeleton. An antisymmetric relation is a relation in which $((x,y)\in R \land (y,x)\in R) \implies x=y$. Isomorphisms in preorders are those elements of the relation where $(x,y)\in R \implies (y,x)\in R$. So in an order, if $(x,y)\in R$ and $(y,x) \in R$, the morphism $x\to y$ is an isomorphism and the objects $x$ and $y$ are equal as a result of the antisymmetry of $R$.