lpetrich
Contributor
For inverse semigroups, the inverse of the inverse of an element is that element itself -- inversion is an involution.
The inverse of a product is the reverse of the inverses of each one: (a*b)-1 = b-1*a-1
(a*b)*(b-1*a-1)*(a*b) = a*(b*b-1)*(a-1*a)*b = a*(a-1*a)*(b*b-1)*b = a*b
(b-1*a-1)*(a*b)*(b-1*a-1) = b-1*(a-1*a)*(b*b-1)*a-1 = b-1*(b*b-1)*(a-1*a)*a-1 = b-1*a-1
If e is an idempotent, then a*e*a-1 is also an idempotent. a*e*a-1*a*e*a-1 = a*a-1*a*e*e*a-1 = a*e*a-1
For every idempotent e and element a, there is an idempotent f satisfying a*e = f*a, and one, f', satisfying e*a = a*f'. Proof of the first one, with the second one similar:
a*e = a*a-1*a*e = a*e*a-1*a = (a*e*a-1)*a = f*a
The inverse of a product is the reverse of the inverses of each one: (a*b)-1 = b-1*a-1
(a*b)*(b-1*a-1)*(a*b) = a*(b*b-1)*(a-1*a)*b = a*(a-1*a)*(b*b-1)*b = a*b
(b-1*a-1)*(a*b)*(b-1*a-1) = b-1*(a-1*a)*(b*b-1)*a-1 = b-1*(b*b-1)*(a-1*a)*a-1 = b-1*a-1
If e is an idempotent, then a*e*a-1 is also an idempotent. a*e*a-1*a*e*a-1 = a*a-1*a*e*e*a-1 = a*e*a-1
For every idempotent e and element a, there is an idempotent f satisfying a*e = f*a, and one, f', satisfying e*a = a*f'. Proof of the first one, with the second one similar:
a*e = a*a-1*a*e = a*e*a-1*a = (a*e*a-1)*a = f*a