PlanetMath: construction of a Brandt groupoid

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.

Lemma 1 Let $H(e,f)$ be the set consisting of all isomorphisms with source $e$ and target $f$ . Then the set $K=\lbrace H(e,f)\mid e,f\in I\rbrace$ 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 $H(e,f)$ . $\qedsymbol$

Lemma 2 The cardinality of $H(e,f)$ is independent of $e$ and $f$ .

Proof. Define $\phi:H(e,f)\to H(e',f')$ , by $\phi(a)=uav$ , where $u\in H(f,f')$ and $v\in H(e',e)$ . 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 $b\mapsto u^{-1} b v^{-1}$ . $\qedsymbol$

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. $\qedsymbol$

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, $\phi(ab)= u(ab)u^{-1}= u((ae)b)u^{-1} = u(a(u^{-1}u)b) u^{-1} = u(((au^{-1})u)b)u^{-1}= u((au^{-1})(ub))u^{-1}=(u a u^{-1})(ub u^{-1})=\phi(a)\phi(b)$ , hence $\phi$ is a group isomorphism. $\qedsymbol$

Proof. [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 \begin{eqnarray*} \overline{ab} &=& \alpha(t(ab))^{-1} ab \alpha(s(ab)) \\ &=& \alpha(t(a))^{-1} ab \alpha(s(b)) \\ &=& \alpha(t(a))^{-1} a \alpha(s(a))\alpha(s(a))^{-1} b \alpha(s(b) \\ &=& \alpha(t(a))^{-1} a \alpha(s(a))\alpha(t(b))^{-1} b \alpha(s(b) \\ &=& \overline{a}\overline{b}. \end{eqnarray*}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. $\qedsymbol$

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