certificate
(1) proves that the mapping f is injective.
For any b∈B, if a 1, a2∈A(a 1! =a2), so that g(a 1)=b and g(a2)=b, that is, g(a 1)=b= g(a2).
Because a1= ia (a1) = (g * f) (a1) = f (g (a1)) = f (g (a2)) = (g * f) (a2)
So f is injective.
(2) It is proved that the mapping G is surjective.
Because (g*f)(A)=IA(A)= A, g*f is injective.
For any c∈A, it is known that g*f is surjective and a∈A exists, so (g * f) (a) = c.
Then there is b∈B, so f(a) = b and g (b) = c.
So there is b∈B, so g(b) = c,
So g is surjective.
Just disprove by definition.
If F is not unique, then the element a 1≠ a2 of A exists so that f(a 1)=f(a2). ( 1)
G (f (a 1)) = g (f (a2)) is obtained from (1).
So g(f) is not injective, which contradicts that g(f) is injective. So f is simple.
On the other hand, if G is not satisfied, the element C of C exists so that g(b)≠c. (2) exists for any element B of B.
For any element A of A, f(a) is an element of B, so g(f(a))≠c is obtained from (2).
So g(f) is not surjective, which is contradictory to g(f) being bijective. So g is full.
Complete the certificate.
The others are similar. . . . . . . . .