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# construction of a Brandt groupoid

In the parent entry, we give an example of a Brandt groupoid. In the example, we started with a non-empty set $I$ and a group $G$, and showed that $I\times G\times I$ has the structure of a Brandt groupoid. In this entry, we show that every Brandt groupoid may be constructed this way.

###### Proposition 1.

If $B$ is a Brandt groupoid, then there is a non-empty set $I$, and a group $G$, such that $B$ is isomorphic to $I\times G\times I$. In other words, there is a bijection $\phi:B\to I\times G\times I$ such that $ab$ is defined in $B$ iff $\phi(a)\phi(b)$ is defined in $I\times G\times I$, and $\phi(ab)=\phi(a)\phi(b)$ whenever the multiplication is defined.

To prove this, let us observe the following series of facts: given a Brandt groupoid $B$, let $I$ be the set of idempotents in $B$.

###### Lemma 1.

Let $H(e,f)$ be the set consisting of all isomorphisms with source $e$ and target $f$. Then the set $K=\{H(e,f)\mid e,f\in I\}$ partitions $B$.

###### Proof.

This is clear from the previous discussion, as $B$ can be thought of as a category. Another way to see this is to define a binary relation $R$ on $B$ so that $aRb$ iff $a,b$ have the same source and target. Then $R$ is an equivalene relation, and its equivalence classes have the form $H(e,f)$. ∎

###### Lemma 2.

The cardinality of $H(e,f)$ is independent of $e$ and $f$.

###### Proof.

Define $\phi:H(e,f)\to H(e^{{\prime}},f^{{\prime}})$, by $\phi(a)=uav$, where $u\in H(f,f^{{\prime}})$ and $v\in H(e^{{\prime}},e)$. Notice that $u,v$ exist by condition 6 above. First, $\phi$ is well-defined, because both $ua$ and $av$ are defined by condition 3, and hence $uav=(ua)v=u(av)$ is defined. In addition, $\phi$ is a bijection, whose inverse is the map $b\mapsto u^{{-1}}bv^{{-1}}$. ∎

###### Lemma 3.

$H(e,e)$ is a group for every $e\in I$.

###### Proof.

The multiplication on $H(e,e)$ is just the multiplication on $B$ restricted to $H(e,e)$, which is total (defined for all of $H(e,e)$), and associative, with $e$ its multiplicative identity. For $a\in H(e,e)$, its inverse is guaranteed by condition 5 above. ∎

###### Lemma 4.

$H(e,e)$ is group isomorphic to $H(f,f)$ for every $e,f\in I$.

###### Proof.

The function $\phi:H(e,e)\to H(f,f)$ given by $\phi(a)=uau^{{-1}}$, where $u\in H(e,f)$, is a well-defined bijection according to the proof of the first observation. Furthermore, $\phi(ab)=u(ab)u^{{-1}}=u((ae)b)u^{{-1}}=u(a(u^{{-1}}u)b)u^{{-1}}=u(((au^{{-1}}% )u)b)u^{{-1}}=u((au^{{-1}})(ub))u^{{-1}}=(uau^{{-1}})(ubu^{{-1}})=\phi(a)\phi(b)$, hence $\phi$ is a group isomorphism. ∎

Set $G=H(e,e)$ for some $e\in I$. We are now ready to prove the proposition. Notice that the proof involves the axiom of choice.

###### Proof of Proposition 1..

By the axiom of choice, there is a function $\alpha:I\to B$ such that $\alpha(f)\in H(e,f)$ and $\alpha(e)=e$. For any $a\in B$, set

$\overline{a}:=\alpha(t(a))^{{-1}}a\alpha(s(a))\in G.$ |

If $ab$ is defined, then $s(a)=t(b)$, so that

$\displaystyle\overline{ab}$ | $\displaystyle=$ | $\displaystyle\alpha(t(ab))^{{-1}}ab\alpha(s(ab))$ | ||

$\displaystyle=$ | $\displaystyle\alpha(t(a))^{{-1}}ab\alpha(s(b))$ | |||

$\displaystyle=$ | $\displaystyle\alpha(t(a))^{{-1}}a\alpha(s(a))\alpha(s(a))^{{-1}}b\alpha(s(b)$ | |||

$\displaystyle=$ | $\displaystyle\alpha(t(a))^{{-1}}a\alpha(s(a))\alpha(t(b))^{{-1}}b\alpha(s(b)$ | |||

$\displaystyle=$ | $\displaystyle\overline{a}\overline{b}.$ |

Now, define $\phi:B\to I\times G\times I$ by

$\phi(a)=(t(a),\overline{a},s(a)).$ |

This is clearly a well-defined function. In addition, it is one-to-one: if $\phi(a)=\phi(b)$, then $s(a)=s(b):=f$, $t(a)=t(b):=g$ and $\alpha(g)^{{-1}}a\alpha(f)=\overline{a}=\overline{b}=\alpha(g)^{{-1}}b\alpha(f)$. As a result, $a=\alpha(g)\overline{a}\alpha(f)^{{-1}}=\alpha(g)\overline{b}\alpha(f)^{{-1}}=b$. It is also onto: given $(g,c,f)\in I\times G\times I$, then $\phi(d)=c$, where $d=\alpha(g)c\alpha(f)^{{-1}}$.

Finally, for $a,b\in B$, the multiplication $ab$ is defined in $B$ iff $s(a)=t(b)$ iff the multiplication

$\phi(a)\phi(b),\quad\mbox{or}\quad(t(a),\overline{a},s(a))(t(b),\overline{b},s% (b))$ |

is defined in $I\times G\times I$, which is equal to

$(t(a),\overline{a}\overline{b},s(b))=(t(a),\overline{ab},s(b))=(t(ab),% \overline{ab},s(ab))=\phi(ab),$ |

showing that $\phi$ preserves partial multiplications. ∎

## Mathematics Subject Classification

18B40*no label found*20L05

*no label found*

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