A groupoid, also known as a virtual group, is a small category where every morphism is invertible. We can give a more explicit, algebraic definition: start with a set $G$ , and a partial binary operation $\circ$ on $G$ . Call a pair $(x,y)$ of elements of $G$ a composable pair if $(x,y)\in\operatorname{dom}(\circ)$ . A groupoid is the pair $(G,\circ)$ , together with two unary operations $e_L$ and $e_R$ on it, satisfying the following conditions:

1. $(x,y)$ is a composable pair iff $e_R(x)=e_L(y)$ .
2. $(x,y)$ and $(x\circ y, z)$ are composable pairs iff $(y,z)$ and $(x,y\circ z)$ are, and if one of these is true, then $(x\circ y)\circ z=x\circ (y\circ z)$ .
3. $(e_L(x),x)$ and $(x,e_R(x))$ are composable pairs and $x=e_L(x)\circ x=x\circ e_R(x)$ .
4. for each $x\in G$ , there exists $y\in G$ such that $(x,y)$ and $(y,x)$ are composable pairs, and $e_L(x)=x\circ y$ and $e_R(x)=y\circ x$ .

Below are some properties:

1. In condition 4 above, $e_L(x)=e_R(y)$ and $e_R(x)=e_L(y)$ . This is true by condition 1, since both $(x,y)$ and $(y,x)$ are composable pairs.
2. Again, in condition 4, $y$ is unique. To see this, suppose $z\in G$ satisfies condition 4 (in place of $y$ ). Then $y= y\circ e_R(y)= y\circ e_L(x) = y\circ (x\circ y)=y\circ (x\circ z)=(y\circ x)\circ z= (z\circ x)\circ z=e_R(x)\circ z = e_L(z)\circ z = z$ . Notice property 1 is used in the proof. We call $y$ the inverse of $x$ , and write $x^{-1}$ .
3. In view of condition 4, both $e_L$ and $e_R$ are unique. In other words, if $f_L,f_R:G\to G$ are unary operators on $G$ satisfying conditions 3 and 4 above (in place of $e_L$ and $e_R$ ), then $f_L=e_L$ and $f_R=e_R$ . In fact, $e_L(x)=x\circ x^{-1}$ and $e_R(x)=x^{-1}\circ x$ .
4. Since $x=e_L(x)\circ x=e_L(x)\circ (e_L(x)\circ x)=(e_L(x)\circ e_L(x))\circ x$ , we see that $e_L(x)$ is composable with itself, and that $e_L(x)\circ e_L(x)=e_L(x)$ by the previous property. Similarly, $e_R(x)\circ e_R(x)=e_R(x)$ . This shows that $e_R(x)$ and $e_L(x)$ are idempotent with respect to $\circ$ for every $x\in G$ .
5. Since $(e_L(x),x)$ is a composable pair, $e_R(e_L(x))=e_L(x)$ for any $x\in G$ . Similarly, $e_L(e_R(x))=e_R(x)$ . Hence $e_R(e_R(x))= e_R(e_L(e_R(x)))=e_L(e_R(x))=e_R(x)$ . Similarly, $e_L(e_L(x))=e_L(x)$ . This shows that $e_R$ and $e_L$ are idempotent with respect to functional compositions.
6. (Cancellation property): if $x\circ y= x\circ z$ , then $y=z$ ; if $y\circ x=z\circ x$ , then $y=z$ .
   Proof. Since $(x,y)$ is a composable pair, $e_R(x)=e_L(y)$ . But $e_R(e_R(x))=e_R(x)$ , we have $e_R(e_R(x))=e_L(y)$ so that $(e_R(x),y)=(x^{-1}\circ x,y)$ is a composable pair, hence $(x^{-1},x\circ y)$ is a composable pair and $x^{-1}\circ (x\circ y)=(x^{-1}\circ x)\circ y=e_R(x)\circ y$ . Since $(e_R(x),y)$ is a composable pair, $e_R(x)=e_R(e_R(x))=e_L(y)$ . As a result, $x^{-1}\circ (x\circ y)=e_L(y)\circ y =y$ . Similarly $x^{-1}\circ (x\circ z)=z$ . By assumption, we deduce that $y=z$ . The other statement is proved similarly.
7. The algebraic definition given can be easily turned into a categorical definition (using objects and morphisms). The details are left for the reader.

If $e_R$ and $e_L$ are constant functions, then $G$ is a group.

Remark. There is also a group-theoretic concept with the same name.