Proved by reduction to absurdity, assuming that G is not injective, we assume that elements A and B in B are mapped from G to the same element C in C, then because F is surjective, elements D and F in A are mapped from F to A and B respectively, so D and F are mapped from F≥G to C, that is, F≥G is not injective. Contrary to the conditions, this assumption is not valid. So G must be injective.

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. . . . . . . . .