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# groupoid (category theoretic)

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.

## Mathematics Subject Classification

18B40*no label found*20L05

*no label found*

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## Comments

## Parenthetical classification in object name

I thought that this was bad practise. Doesn't the MSC make this superfluous? Or because this concept of groupoid is so close semantically (that is, algebra) to the one in group theory that you require the parenthetical remark?

- J"

## Re: Parenthetical classification in object name

It's for indexing purposes. Think of the search results screen: if you see

groupoid

blah dee blah

groupoid

blah de blah

Which one do you pick? It's even worse for the alphabetical listing.

If you put "hinting" in the actual title, but synonym the "real" title, this problem goes away.

apk

## Re: Parenthetical classification in object name

In that case,

http://aux.planetmath.org/doc/newuser.html

needs to be updated.

- J"

## Re: Parenthetical classification in object name

Oops, never mind. There is already a clause saying when this is appropriate. :P

- J"

## Re: Parenthetical classification in object name

Oops, never mind. There is already a clause saying when this is appropriate. :P

- J"